Chapter 9 Sequences And Series

(i) Given sequence definition: $a_n = 2n + 5$.

To find the first three terms, we substitute $n=1$, $n=2$, and $n=3$ into the formula for $a_n$.

For $n=1$: $a_1 = 2(1) + 5 = 2 + 5 = 7$.

For $n=2$: $a_2 = 2(2) + 5 = 4 + 5 = 9$.

For $n=3$: $a_3 = 2(3) + 5 = 6 + 5 = 11$.

The first three terms of the sequence are 7, 9, and 11.

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(ii) Given sequence definition: $a_n = \frac{n \;-\; 3}{4}$.

To find the first three terms, we substitute $n=1$, $n=2$, and $n=3$ into the formula for $a_n$.

For $n=1$: $a_1 = \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2}$.

For $n=2$: $a_2 = \frac{2 - 3}{4} = \frac{-1}{4}$.

For $n=3$: $a_3 = \frac{3 - 3}{4} = \frac{0}{4} = 0$.

The first three terms of the sequence are $-\frac{1}{2}$, $-\frac{1}{4}$, and 0.

Given sequence definition: $a_n = (n – 1) (2 – n) (3 + n)$.

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To find the 20^th term of the sequence, we need to find the value of $a_n$ when $n=20$.

Substitute $n=20$ into the formula for $a_n$:

$a_{20} = (20 – 1) (2 – 20) (3 + 20)$

$a_{20} = (19) (-18) (23)$

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Now, calculate the product:

$19 \times (-18) = -(19 \times 18)$

$19 \times 18 = 19 \times (20 - 2) = 19 \times 20 - 19 \times 2 = 380 - 38 = 342$.

So, $19 \times (-18) = -342$.

Now multiply this by 23:

$a_{20} = -342 \times 23$

Calculate $342 \times 23$:

$\begin{array}{cccc}& & 3 & 4 & 2 \\ \times & & & 2 & 3 \\ \hline & 1 & 0 & 2 & 6 \\ 6 & 8 & 4 & \times \\ \hline 7 & 8 & 6 & 6 \\ \hline \end{array}$

$342 \times 23 = 7866$.

So, $a_{20} = -7866$.

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The 20^th term of the sequence is $\mathbf{-7866}$.

Given sequence definition:

$a_1 = 1$

$a_n = a_{n – 1} + 2$ for $n \geq 2$.

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This is a recursive definition of a sequence. The first term is given, and each subsequent term is defined based on the previous term by adding 2.

We can find the terms of the sequence by using the given formula for $n=2, 3, 4, \dots$. Let's find the first few terms (assuming the task is to list some initial terms, similar to Example 1).

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The first term is already given:

$a_1 = 1$

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For $n=2$, using the recursive formula $a_n = a_{n – 1} + 2$:

$a_2 = a_{2 – 1} + 2 = a_1 + 2$

Substitute the value of $a_1$:

$a_2 = 1 + 2 = 3$

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For $n=3$, using the recursive formula:

$a_3 = a_{3 – 1} + 2 = a_2 + 2$

Substitute the value of $a_2$:

$a_3 = 3 + 2 = 5$

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For $n=4$, using the recursive formula:

$a_4 = a_{4 – 1} + 2 = a_3 + 2$

Substitute the value of $a_3$:

$a_4 = 5 + 2 = 7$

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This sequence starts with 1, and each subsequent term is obtained by adding 2. This is an arithmetic progression with the first term 1 and common difference 2. The terms are 1, 3, 5, 7, 9, ...

The first few terms of the sequence are 1, 3, 5, ... (If specifically the first three terms are needed, they are 1, 3, 5).

Given the n^th term of the sequence: $a_n = n (n + 2)$.

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To find the first five terms, we substitute $n=1, 2, 3, 4, 5$ into the formula for $a_n$.

For $n=1$: $a_1 = 1 (1 + 2) = 1 \times 3 = 3$.

For $n=2$: $a_2 = 2 (2 + 2) = 2 \times 4 = 8$.

For $n=3$: $a_3 = 3 (3 + 2) = 3 \times 5 = 15$.

For $n=4$: $a_4 = 4 (4 + 2) = 4 \times 6 = 24$.

For $n=5$: $a_5 = 5 (5 + 2) = 5 \times 7 = 35$.

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The first five terms of the sequence are 3, 8, 15, 24, and 35.

Given the n^th term of the sequence: $a_n = \frac{n}{n \;+\; 1}$.

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To find the first five terms, we substitute $n=1, 2, 3, 4, 5$ into the formula for $a_n$.

For $n=1$: $a_1 = \frac{1}{1 + 1} = \frac{1}{2}$.

For $n=2$: $a_2 = \frac{2}{2 + 1} = \frac{2}{3}$.

For $n=3$: $a_3 = \frac{3}{3 + 1} = \frac{3}{4}$.

For $n=4$: $a_4 = \frac{4}{4 + 1} = \frac{4}{5}$.

For $n=5$: $a_5 = \frac{5}{5 + 1} = \frac{5}{6}$.

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The first five terms of the sequence are $\mathbf{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \text{ and } \frac{5}{6}}$.

Given the n^th term of the sequence: $a_n = 2^n$.

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To find the first five terms, we substitute $n=1, 2, 3, 4, 5$ into the formula for $a_n$.

For $n=1$: $a_1 = 2^1 = 2$.

For $n=2$: $a_2 = 2^2 = 4$.

For $n=3$: $a_3 = 2^3 = 8$.

For $n=4$: $a_4 = 2^4 = 16$.

For $n=5$: $a_5 = 2^5 = 32$.

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The first five terms of the sequence are 2, 4, 8, 16, and 32.

Given the n^th term of the sequence: $a_n = \frac{2n \;-\; 3}{6}$.

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To find the first five terms, we substitute $n=1, 2, 3, 4, 5$ into the formula for $a_n$.

For $n=1$: $a_1 = \frac{2(1) - 3}{6} = \frac{2 - 3}{6} = \frac{-1}{6}$.

For $n=2$: $a_2 = \frac{2(2) - 3}{6} = \frac{4 - 3}{6} = \frac{1}{6}$.

For $n=3$: $a_3 = \frac{2(3) - 3}{6} = \frac{6 - 3}{6} = \frac{3}{6} = \frac{1}{2}$.

For $n=4$: $a_4 = \frac{2(4) - 3}{6} = \frac{8 - 3}{6} = \frac{5}{6}$.

For $n=5$: $a_5 = \frac{2(5) - 3}{6} = \frac{10 - 3}{6} = \frac{7}{6}$.

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The first five terms of the sequence are $\mathbf{-\frac{1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \text{ and } \frac{7}{6}}$.

Given the n^th term of the sequence: $a_n = (–1)^{n–1} 5^{n+1}$.

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To find the first five terms, we substitute $n=1, 2, 3, 4, 5$ into the formula for $a_n$.

For $n=1$: $a_1 = (-1)^{1-1} 5^{1+1} = (-1)^0 5^2 = 1 \times 25 = 25$.

For $n=2$: $a_2 = (-1)^{2-1} 5^{2+1} = (-1)^1 5^3 = -1 \times 125 = -125$.

For $n=3$: $a_3 = (-1)^{3-1} 5^{3+1} = (-1)^2 5^4 = 1 \times 625 = 625$.

For $n=4$: $a_4 = (-1)^{4-1} 5^{4+1} = (-1)^3 5^5 = -1 \times 3125 = -3125$.

For $n=5$: $a_5 = (-1)^{5-1} 5^{5+1} = (-1)^4 5^6 = 1 \times 15625 = 15625$.

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The first five terms of the sequence are 25, -125, 625, -3125, and 15625.

Given the n^th term of the sequence: $a_n = n\frac{n^{2}\;+\;5}{4}$.

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To find the first five terms, we substitute $n=1, 2, 3, 4, 5$ into the formula for $a_n$.

For $n=1$: $a_1 = 1 \times \frac{1^2 + 5}{4} = 1 \times \frac{1 + 5}{4} = 1 \times \frac{6}{4} = \frac{6}{4} = \frac{3}{2}$.

For $n=2$: $a_2 = 2 \times \frac{2^2 + 5}{4} = 2 \times \frac{4 + 5}{4} = 2 \times \frac{9}{4} = \frac{18}{4} = \frac{9}{2}$.

For $n=3$: $a_3 = 3 \times \frac{3^2 + 5}{4} = 3 \times \frac{9 + 5}{4} = 3 \times \frac{14}{4} = \frac{42}{4} = \frac{21}{2}$.

For $n=4$: $a_4 = 4 \times \frac{4^2 + 5}{4} = 4 \times \frac{16 + 5}{4} = 4 \times \frac{21}{4} = 21$.

For $n=5$: $a_5 = 5 \times \frac{5^2 + 5}{4} = 5 \times \frac{25 + 5}{4} = 5 \times \frac{30}{4} = \frac{150}{4} = \frac{75}{2}$.

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The first five terms of the sequence are $\mathbf{\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21, \text{ and } \frac{75}{2}}$.

Given the n^th term of the sequence: $a_n = 4n - 3$.

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To find the 17^th term ($a_{17}$), we substitute $n=17$ into the formula:

$a_{17} = 4(17) - 3$

$a_{17} = 68 - 3$

$a_{17} = 65$

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To find the 24^th term ($a_{24}$), we substitute $n=24$ into the formula:

$a_{24} = 4(24) - 3$

$a_{24} = 96 - 3$

$a_{24} = 93$

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The 17^th term is $\mathbf{65}$ and the 24^th term is $\mathbf{93}$.

Given the n^th term of the sequence: $a_n = \frac{n^{2}}{2^{n}}$.

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To find the 7^th term ($a_{7}$), we substitute $n=7$ into the formula:

$a_{7} = \frac{7^2}{2^7}$

$a_{7} = \frac{49}{128}$

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The 7^th term is $\mathbf{\frac{49}{128}}$.

Given the n^th term of the sequence: $a_n = (–1)^{n – 1} n^3$.

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To find the 9^th term ($a_{9}$), we substitute $n=9$ into the formula:

$a_{9} = (-1)^{9-1} 9^3$

$a_{9} = (-1)^8 9^3$

$a_{9} = 1 \times (9 \times 9 \times 9)$

$a_{9} = 1 \times (81 \times 9)$

$a_{9} = 729$

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The 9^th term is $\mathbf{729}$.

Given the n^th term of the sequence: $a_n = \frac{n (n \;-\; 2)}{n \;+\; 3}$.

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To find the 20^th term ($a_{20}$), we substitute $n=20$ into the formula:

$a_{20} = \frac{20 (20 - 2)}{20 + 3}$

$a_{20} = \frac{20 (18)}{23}$

$a_{20} = \frac{360}{23}$

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The 20^th term is $\mathbf{\frac{360}{23}}$.

Given sequence definition:

$a_1 = 3$

$a_n = 3a_{n – 1} + 2$ for all $n > 1$ (which means $n \geq 2$).

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To find the first five terms, we use the given recursive formula starting from $n=2$.

The first term is already given:

$a_1 = 3$

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For $n=2$: $a_2 = 3a_{2-1} + 2 = 3a_1 + 2$. Substitute $a_1=3$: $a_2 = 3(3) + 2 = 9 + 2 = 11$.

For $n=3$: $a_3 = 3a_{3-1} + 2 = 3a_2 + 2$. Substitute $a_2=11$: $a_3 = 3(11) + 2 = 33 + 2 = 35$.

For $n=4$: $a_4 = 3a_{4-1} + 2 = 3a_3 + 2$. Substitute $a_3=35$: $a_4 = 3(35) + 2 = 105 + 2 = 107$.

For $n=5$: $a_5 = 3a_{5-1} + 2 = 3a_4 + 2$. Substitute $a_4=107$: $a_5 = 3(107) + 2 = 321 + 2 = 323$.

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The first five terms of the sequence are 3, 11, 35, 107, and 323.

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The corresponding series is the sum of the terms of the sequence.

The series is the sum of the first five terms: $a_1 + a_2 + a_3 + a_4 + a_5$.

Series = $3 + 11 + 35 + 107 + 323$.

Sum = $14 + 35 + 107 + 323 = 49 + 107 + 323 = 156 + 323 = 479$.

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The first five terms of the sequence are 3, 11, 35, 107, 323.

The corresponding series is $\mathbf{3 + 11 + 35 + 107 + 323}$. The sum of the first five terms of the series is 479.

Given sequence definition:

$a_1 = -1$

$a_n = \frac{a_{n-1}}{n}$ for $n \geq 2$.

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To find the first five terms, we use the given recursive formula starting from $n=2$.

The first term is already given:

$a_1 = -1$

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For $n=2$: $a_2 = \frac{a_{2-1}}{2} = \frac{a_1}{2}$. Substitute $a_1=-1$: $a_2 = \frac{-1}{2}$.

For $n=3$: $a_3 = \frac{a_{3-1}}{3} = \frac{a_2}{3}$. Substitute $a_2 = -\frac{1}{2}$: $a_3 = \frac{-\frac{1}{2}}{3} = -\frac{1}{2 \times 3} = -\frac{1}{6}$.

For $n=4$: $a_4 = \frac{a_{4-1}}{4} = \frac{a_3}{4}$. Substitute $a_3 = -\frac{1}{6}$: $a_4 = \frac{-\frac{1}{6}}{4} = -\frac{1}{6 \times 4} = -\frac{1}{24}$.

For $n=5$: $a_5 = \frac{a_{5-1}}{5} = \frac{a_4}{5}$. Substitute $a_4 = -\frac{1}{24}$: $a_5 = \frac{-\frac{1}{24}}{5} = -\frac{1}{24 \times 5} = -\frac{1}{120}$.

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The first five terms of the sequence are $-1$, $-\frac{1}{2}$, $-\frac{1}{6}$, $-\frac{1}{24}$, and $-\frac{1}{120}$.

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The corresponding series is the sum of the terms of the sequence.

The series is the sum of the first five terms: $a_1 + a_2 + a_3 + a_4 + a_5$.

Series = $-1 + \left(-\frac{1}{2}\right) + \left(-\frac{1}{6}\right) + \left(-\frac{1}{24}\right) + \left(-\frac{1}{120}\right)$

Series = $-1 - \frac{1}{2} - \frac{1}{6} - \frac{1}{24} - \frac{1}{120}$

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The first five terms of the sequence are $\mathbf{-1, -\frac{1}{2}, -\frac{1}{6}, -\frac{1}{24}, -\frac{1}{120}}$.

The corresponding series is $\mathbf{-1 - \frac{1}{2} - \frac{1}{6} - \frac{1}{24} - \frac{1}{120}}$.

Given sequence definition:

$a_1 = 2$

$a_2 = 2$

$a_n = a_{n – 1} – 1$ for $n > 2$ (which means $n \geq 3$).

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To find the first five terms, we use the given values for the first two terms and the recursive formula starting from $n=3$.

The first two terms are already given:

$a_1 = 2$

$a_2 = 2$

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For $n=3$, using the recursive formula $a_n = a_{n – 1} – 1$:

$a_3 = a_{3 – 1} – 1 = a_2 – 1$. Substitute $a_2=2$: $a_3 = 2 – 1 = 1$.

For $n=4$, using the recursive formula:

$a_4 = a_{4 – 1} – 1 = a_3 – 1$. Substitute $a_3=1$: $a_4 = 1 – 1 = 0$.

For $n=5$, using the recursive formula:

$a_5 = a_{5 – 1} – 1 = a_4 – 1$. Substitute $a_4=0$: $a_5 = 0 – 1 = -1$.

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The first five terms of the sequence are 2, 2, 1, 0, and -1.

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The corresponding series is the sum of the terms of the sequence.

The series is the sum of the first five terms: $a_1 + a_2 + a_3 + a_4 + a_5$.

Series = $2 + 2 + 1 + 0 + (-1)$

Series = $4 + 1 + 0 - 1 = 5 - 1 = 4$.

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The first five terms of the sequence are 2, 2, 1, 0, -1.

The corresponding series is $\mathbf{2 + 2 + 1 + 0 - 1}$. The sum of the first five terms of the series is 4.

Given the Fibonacci sequence definition:

$a_1 = 1$

$a_2 = 1$

$a_n = a_{n – 1} + a_{n – 2}$ for $n > 2$ (i.e., $n \geq 3$).

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First, we need to find the terms of the sequence up to $a_6$ to compute the ratios $\frac{a_{n+1}}{a_n}$ for $n=1, 2, 3, 4, 5$.

$a_1 = 1$

$a_2 = 1$

For $n=3$: $a_3 = a_{3-1} + a_{3-2} = a_2 + a_1 = 1 + 1 = 2$.

For $n=4$: $a_4 = a_{4-1} + a_{4-2} = a_3 + a_2 = 2 + 1 = 3$.

For $n=5$: $a_5 = a_{5-1} + a_{5-2} = a_4 + a_3 = 3 + 2 = 5$.

For $n=6$: $a_6 = a_{6-1} + a_{6-2} = a_5 + a_4 = 5 + 3 = 8$.

The first six terms of the sequence are 1, 1, 2, 3, 5, 8.

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Now, we find the ratio $\frac{a_{n+1}}{a_n}$ for $n=1, 2, 3, 4, 5$:

For $n=1$: $\frac{a_{1+1}}{a_1} = \frac{a_2}{a_1} = \frac{1}{1} = 1$.

For $n=2$: $\frac{a_{2+1}}{a_2} = \frac{a_3}{a_2} = \frac{2}{1} = 2$.

For $n=3$: $\frac{a_{3+1}}{a_3} = \frac{a_4}{a_3} = \frac{3}{2}$.

For $n=4$: $\frac{a_{4+1}}{a_4} = \frac{a_5}{a_4} = \frac{5}{3}$.

For $n=5$: $\frac{a_{5+1}}{a_5} = \frac{a_6}{a_5} = \frac{8}{5}$.

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The values of $\frac{a_{n+1}}{a_n}$ for $n=1, 2, 3, 4, 5$ are $1, 2, \frac{3}{2}, \frac{5}{3}, \text{ and } \frac{8}{5}$, respectively.

Given:

The sequence is an Arithmetic Progression (A.P.).

The m^th term ($a_m$) is n.

The n^th term ($a_n$) is m.

$m \neq n$.

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To Find:

The p^th term ($a_p$).

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Solution:

Let the first term of the A.P. be $a$ and the common difference be $d$.

The formula for the k^th term of an A.P. is $a_k = a + (k-1)d$.

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Using the given information:

The m^th term is n:

The n^th term is m:

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We have a system of two linear equations with two variables $a$ and $d$.

Subtract equation (ii) from equation (i):

$(a + (m-1)d) - (a + (n-1)d) = n - m$

$a + (m-1)d - a - (n-1)d = n - m$

$(m-1)d - (n-1)d = n - m$

$d[(m-1) - (n-1)] = n - m$

$d[m - 1 - n + 1] = n - m$

$d(m - n) = n - m$

$d(m - n) = -(m - n)$

Since $m \neq n$, $m-n \neq 0$. We can divide both sides by $(m-n)$.

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Now substitute the value of $d = -1$ into equation (i) to find $a$:

$a + (m-1)(-1) = n$

$a - (m-1) = n$

$a - m + 1 = n$

$a = n + m - 1$

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Now we need to find the p^th term, $a_p$.

Using the formula $a_p = a + (p-1)d$, substitute the values of $a$ and $d$ from equations (iv) and (iii).

$a_p = (m + n - 1) + (p-1)(-1)$

$a_p = m + n - 1 - (p-1)$

$a_p = m + n - 1 - p + 1$

$a_p = m + n - p$

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The p^th term of the A.P. is $\mathbf{m + n - p}$.

Given:

The sum of the first $n$ terms of an A.P., denoted by $S_n$, is given by the formula:

$S_n = nP + \frac{1}{2} n (n - 1)Q$

where $P$ and $Q$ are constants.

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To Find:

The common difference of the A.P.

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Solution:

Let the A.P. be $a_1, a_2, a_3, \dots$, with first term $a_1$ and common difference $d$.

The sum of the first $n$ terms $S_n$ is related to the individual terms $a_n$ by the property:

$a_n = S_n - S_{n-1}$ for $n \geq 2$.

Also, the first term $a_1 = S_1$.

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Let's find the first term $a_1$ using the given formula for $S_n$ with $n=1$:

$S_1 = 1 \times P + \frac{1}{2} \times 1 (1 - 1)Q$

$S_1 = P + \frac{1}{2} \times 1 \times 0 \times Q$

$S_1 = P + 0 = P$

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Now let's find the second term $a_2$. We can use $a_2 = S_2 - S_1$.

First, find $S_2$ using the given formula with $n=2$:

$S_2 = 2P + \frac{1}{2} \times 2 (2 - 1)Q$

$S_2 = 2P + 1 \times 1 \times Q$

$S_2 = 2P + Q$

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Now calculate $a_2$:

$a_2 = S_2 - S_1 = (2P + Q) - P = 2P + Q - P = P + Q$

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In an A.P., the common difference $d$ is the difference between consecutive terms, i.e., $d = a_2 - a_1$.

Using the values from equations (i) and (ii):

$d = (P + Q) - P$

$d = P + Q - P = Q$

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Alternatively, we can derive the general term $a_n = S_n - S_{n-1}$ for $n \geq 2$.

$S_{n-1} = (n-1)P + \frac{1}{2} (n-1) ((n-1) - 1)Q$

$S_{n-1} = (n-1)P + \frac{1}{2} (n-1) (n-2)Q$

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$a_n = S_n - S_{n-1} = \left(nP + \frac{1}{2} n (n - 1)Q\right) - \left((n-1)P + \frac{1}{2} (n-1) (n-2)Q\right)$

$a_n = nP + \frac{1}{2} (n^2 - n)Q - (n-1)P - \frac{1}{2} (n^2 - 3n + 2)Q$

$a_n = P(n - (n-1)) + \frac{Q}{2} ((n^2 - n) - (n^2 - 3n + 2))$

$a_n = P(n - n + 1) + \frac{Q}{2} (n^2 - n - n^2 + 3n - 2)$

$a_n = P(1) + \frac{Q}{2} (2n - 2)$

$a_n = P + \frac{Q}{2} \times 2(n - 1)$

$a_n = P + Q(n - 1)$

This is the formula for the n^th term of an A.P. with first term $P$ and common difference $Q$.

$a_n = P + (n-1)Q$.

The common difference is the coefficient of $(n-1)$, which is $Q$.

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The common difference of the A.P. is $\mathbf{Q}$.

Given:

Let the two A.P.s be denoted by A.P. 1 and A.P. 2.

Let the first term of A.P. 1 be $a_1$ and its common difference be $d_1$.

Let the first term of A.P. 2 be $a_2$ and its common difference be $d_2$.

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The sum of the first $n$ terms of an A.P. is given by $S_n = \frac{n}{2} [2a + (n-1)d]$.

The sum of the first $n$ terms of A.P. 1 is $S_{n,1} = \frac{n}{2} [2a_1 + (n-1)d_1]$.

The sum of the first $n$ terms of A.P. 2 is $S_{n,2} = \frac{n}{2} [2a_2 + (n-1)d_2]$.

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We are given the ratio of their sums of n terms:

$\frac{S_{n,1}}{S_{n,2}} = \frac{\frac{n}{2} [2a_1 + (n-1)d_1]}{\frac{n}{2} [2a_2 + (n-1)d_2]} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2}$

We are also given that this ratio is $\frac{3n + 8}{7n + 15}$.

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To Find:

The ratio of their 12^th terms ($a_{12,1} : a_{12,2}$).

The k^th term of an A.P. is $a_k = a + (k-1)d$.

The 12^th term of A.P. 1 is $a_{12,1} = a_1 + (12-1)d_1 = a_1 + 11d_1$.

The 12^th term of A.P. 2 is $a_{12,2} = a_2 + (12-1)d_2 = a_2 + 11d_2$.

The ratio of the 12^th terms is $\frac{a_1 + 11d_1}{a_2 + 11d_2}$.

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Compare the structure of the ratio of the k^th terms $\frac{a_1 + (k-1)d_1}{a_2 + (k-1)d_2}$ with the ratio of the sums $\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2}$.

The ratio of the k^th terms can be written as $\frac{2a_1 + 2(k-1)d_1}{2a_2 + 2(k-1)d_2}$.

We want the term $11d_1$ and $11d_2$ in the ratio of the 12^th terms. This means we want $2(k-1) = 2 \times 11 = 22$ in the numerator and denominator of the ratio of terms.

In the ratio of sums, the coefficient of $d_1$ and $d_2$ is $(n-1)$. We need $(n-1) = 22$ to match the structure.

So, we need $n-1 = 22$, which means $n = 22 + 1 = 23$.

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Substitute $n=23$ into the given ratio of sums (equation i):

$\frac{2a_1 + (23-1)d_1}{2a_2 + (23-1)d_2} = \frac{3(23) + 8}{7(23) + 15}$

$\frac{2a_1 + 22d_1}{2a_2 + 22d_2} = \frac{69 + 8}{161 + 15}$

$\frac{2a_1 + 22d_1}{2a_2 + 22d_2} = \frac{77}{176}$

---

Factor out 2 from the numerator and denominator of the left side:

$\frac{2(a_1 + 11d_1)}{2(a_2 + 11d_2)} = \frac{77}{176}$

$\frac{a_1 + 11d_1}{a_2 + 11d_2} = \frac{77}{176}$

---

Simplify the fraction $\frac{77}{176}$. Both are divisible by 11.

$\frac{77 \div 11}{176 \div 11} = \frac{7}{16}$

So, $\frac{a_1 + 11d_1}{a_2 + 11d_2} = \frac{7}{16}$.

---

The ratio of the 12^th terms is $\frac{a_{12,1}}{a_{12,2}} = \frac{7}{16}$.

The ratio of their 12^th terms is 7 : 16.

---

Alternate approach:

The k^th term of an A.P. can be obtained from the sum of terms using $a_k = S_k - S_{k-1}$ for $k \geq 2$ and $a_1 = S_1$. Or, we can express the k^th term using the average of two sums.

Consider the relation $a_k = \frac{S_{2k-1}}{2k-1} + \frac{S_{2k-3}}{2k-3}$? No, this is not standard.

Let's use the idea that $a_k = S_k - S_{k-1}$.

The ratio of the k^th terms is $\frac{a_{k,1}}{a_{k,2}} = \frac{S_{k,1} - S_{k-1,1}}{S_{k,2} - S_{k-1,2}}$. This seems complicated to evaluate directly.

Consider the formula $S_n = \frac{n}{2} [2a + (n-1)d]$. The k^th term is $a_k = a + (k-1)d$.

The ratio of sums is $\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{3n+8}{7n+15}$.

We want the ratio of the 12^th terms: $\frac{a_1 + 11d_1}{a_2 + 11d_2}$.

We can observe that $a_1 + 11d_1$ is part of the expression $2a_1 + (n-1)d_1$ if we choose $n-1$ correctly.

We want $a_1 + 11d_1$. If we take the expression $2a_1 + (n-1)d_1$ and divide by 2, we get $a_1 + \frac{(n-1)}{2}d_1$.

So, if we set $\frac{n-1}{2} = 11$, we can match the terms. This means $n-1 = 22$, so $n=23$.

If we substitute $n=23$ into the ratio of sums, we get:

$\frac{2a_1 + (23-1)d_1}{2a_2 + (23-1)d_2} = \frac{3(23)+8}{7(23)+15}$

$\frac{2a_1 + 22d_1}{2a_2 + 22d_2} = \frac{69+8}{161+15}$

$\frac{2(a_1 + 11d_1)}{2(a_2 + 11d_2)} = \frac{77}{176}$

$\frac{a_1 + 11d_1}{a_2 + 11d_2} = \frac{77}{176} = \frac{7}{16}$

The ratio of the 12^th terms is $\frac{a_{12,1}}{a_{12,2}} = \frac{a_1 + 11d_1}{a_2 + 11d_2} = \frac{7}{16}$.

This confirms the previous method and is arguably the intended approach for such problems.

Given:

Income in the first year = $\textsf{₹} \; 3,00,000$.

Annual increase in income = $\textsf{₹} \; 10,000$ for the next 19 years.

Total number of years = 20.

---

To Find:

The total amount received in 20 years.

---

Solution:

Let the income in the n^th year be denoted by $a_n$.

The income in the first year is $a_1 = \textsf{₹} \; 3,00,000$.

The income increases by a fixed amount ($\textsf{₹} \; 10,000$) each year. This indicates that the sequence of annual incomes forms an Arithmetic Progression (A.P.).

The common difference of the A.P. is $d = \textsf{₹} \; 10,000$.

The number of years is $n = 20$.

---

We need to find the total amount received in 20 years, which is the sum of the incomes for the first 20 years. This is the sum of the first 20 terms of the A.P., $S_{20}$.

The formula for the sum of the first $n$ terms of an A.P. is $S_n = \frac{n}{2} [2a + (n-1)d]$.

Substitute $n=20$, $a=a_1=3,00,000$, and $d=10,000$ into the formula for $S_{20}$:

$S_{20} = \frac{20}{2} [2(3,00,000) + (20-1)(10,000)]$

$S_{20} = 10 [6,00,000 + 19(10,000)]$

$S_{20} = 10 [6,00,000 + 1,90,000]$

$S_{20} = 10 [7,90,000]$

$S_{20} = 79,00,000$

---

The total amount received in 20 years is $\textsf{₹} \; 79,00,000$.

Given:

We need to insert 6 numbers between 3 and 24 to form an Arithmetic Progression (A.P.).

---

To Find:

The 6 numbers to be inserted.

---

Solution:

Let the resulting A.P. be $3, x_1, x_2, x_3, x_4, x_5, x_6, 24$.

In this A.P., the first term is $a_1 = 3$ and the last term is the $(6+2)^{\text{th}} = 8^{\text{th}}$ term, $a_8 = 24$.

Let the common difference of the A.P. be $d$.

---

The formula for the n^th term of an A.P. is $a_n = a_1 + (n-1)d$.

Using the 8^th term:

$a_8 = a_1 + (8-1)d$

$24 = 3 + 7d$

---

Solve for the common difference $d$:

$24 - 3 = 7d$

$21 = 7d$

$d = \frac{21}{7} = 3$

---

Now that we have the common difference $d=3$, we can find the 6 inserted numbers.

The inserted numbers are $x_1, x_2, x_3, x_4, x_5, x_6$, which correspond to the 2^nd, 3^rd, 4^th, 5^th, 6^th, and 7^th terms of the A.P.

$x_1 = a_2 = a_1 + 1d = 3 + 1(3) = 3 + 3 = 6$.

$x_2 = a_3 = a_1 + 2d = 3 + 2(3) = 3 + 6 = 9$.

$x_3 = a_4 = a_1 + 3d = 3 + 3(3) = 3 + 9 = 12$.

$x_4 = a_5 = a_1 + 4d = 3 + 4(3) = 3 + 12 = 15$.

$x_5 = a_6 = a_1 + 5d = 3 + 5(3) = 3 + 15 = 18$.

$x_6 = a_7 = a_1 + 6d = 3 + 6(3) = 3 + 18 = 21$.

---

The 6 numbers inserted between 3 and 24 are 6, 9, 12, 15, 18, and 21.

The resulting A.P. is 3, 6, 9, 12, 15, 18, 21, 24.

---

The 6 numbers to be inserted are 6, 9, 12, 15, 18, and 21.

Given:

We need to find the sum of odd integers from 1 to 2001.

The sequence of odd integers starting from 1 is 1, 3, 5, 7, ..., 2001.

---

To Find:

The sum of this sequence.

---

Solution:

The given sequence of odd integers forms an Arithmetic Progression (A.P.) because the difference between consecutive terms is constant (i.e., $3-1=2$, $5-3=2$, etc.).

The first term is $a_1 = 1$.

The last term is $a_n = 2001$.

The common difference is $d = 2$.

---

To find the sum of an A.P., we first need to determine the number of terms ($n$) in the sequence.

We use the formula for the n^th term of an A.P.: $a_n = a_1 + (n-1)d$.

Substitute the known values:

$2001 = 1 + (n-1)2$

Subtract 1 from both sides:

$2001 - 1 = (n-1)2$

$2000 = (n-1)2$

Divide by 2:

$\frac{2000}{2} = n-1$

$1000 = n-1$

Add 1 to both sides:

$n = 1000 + 1 = 1001$

There are 1001 terms in the sequence.

---

Now, we find the sum of the first $n$ terms of the A.P. using the formula $S_n = \frac{n}{2} (a_1 + a_n)$.

Substitute $n=1001$, $a_1=1$, and $a_n=2001$:

$S_{1001} = \frac{1001}{2} (1 + 2001)$

$S_{1001} = \frac{1001}{2} (2002)$

$S_{1001} = 1001 \times \frac{2002}{2}$

$S_{1001} = 1001 \times 1001$

$S_{1001} = (1001)^2$

$(1001)^2 = (1000 + 1)^2 = 1000^2 + 2 \times 1000 \times 1 + 1^2 = 1000000 + 2000 + 1 = 1002001$.

---

The sum of the odd integers from 1 to 2001 is $\mathbf{1002001}$.

Given:

We need to find the sum of all natural numbers lying between 100 and 1000 which are multiples of 5.

"Between 100 and 1000" means numbers greater than 100 and less than 1000.

---

The multiples of 5 greater than 100 are 105, 110, 115, ...

The multiples of 5 less than 1000 are ..., 985, 990, 995.

The sequence of multiples of 5 between 100 and 1000 is 105, 110, 115, ..., 995.

---

To Find:

The sum of this sequence.

---

Solution:

The given sequence forms an Arithmetic Progression (A.P.) because the difference between consecutive terms is constant (i.e., 5).

The first term is $a_1 = 105$.

The last term is $a_n = 995$.

The common difference is $d = 5$.

---

To find the sum of this A.P., we first need to determine the number of terms ($n$) in the sequence.

We use the formula for the n^th term of an A.P.: $a_n = a_1 + (n-1)d$.

Substitute the known values:

$995 = 105 + (n-1)5$

Subtract 105 from both sides:

$995 - 105 = (n-1)5$

$890 = (n-1)5$

Divide by 5:

$\frac{890}{5} = n-1$

$178 = n-1$

Add 1 to both sides:

$n = 178 + 1 = 179$

There are 179 terms in the sequence.

---

Now, we find the sum of the first $n$ terms of the A.P. using the formula $S_n = \frac{n}{2} (a_1 + a_n)$.

Substitute $n=179$, $a_1=105$, and $a_n=995$:

$S_{179} = \frac{179}{2} (105 + 995)$

$S_{179} = \frac{179}{2} (1100)$

$S_{179} = 179 \times \frac{1100}{2}$

$S_{179} = 179 \times 550$

Calculate the product $179 \times 550$:

$179 \times 550 = 179 \times 55 \times 10$

$\begin{array}{cccc}& & 1 & 7 & 9 \\ \times & & & 5 & 5 \\ \hline & & 8 & 9 & 5 \\ 8 & 9 & 5 & \times \\ \hline 9 & 8 & 4 & 5 \\ \hline \end{array}$

$179 \times 55 = 9845$.

So, $179 \times 550 = 98450$.

---

The sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is $\mathbf{98450}$.

Given:

The sequence is an Arithmetic Progression (A.P.).

The first term is $a_1 = 2$.

The sum of the first five terms ($S_5$) is one-fourth of the sum of the next five terms.

---

To Show:

The 20^th term ($a_{20}$) is -112.

---

Solution:

Let the common difference of the A.P. be $d$.

The sum of the first $n$ terms of an A.P. is $S_n = \frac{n}{2} [2a_1 + (n-1)d]$.

---

The sum of the first five terms is $S_5 = \frac{5}{2} [2(2) + (5-1)d] = \frac{5}{2} [4 + 4d] = \frac{5}{2} \times 4(1 + d) = 10(1 + d) = 10 + 10d$.

---

The next five terms are the 6^th, 7^th, 8^th, 9^th, and 10^th terms ($a_6, a_7, a_8, a_9, a_{10}$).

The sum of the next five terms is $a_6 + a_7 + a_8 + a_9 + a_{10}$.

This sum can also be calculated as the sum of the first 10 terms minus the sum of the first 5 terms: $S_{10} - S_5$.

The sum of the first 10 terms is $S_{10} = \frac{10}{2} [2(2) + (10-1)d] = 5 [4 + 9d] = 20 + 45d$.

---

The sum of the next five terms = $S_{10} - S_5 = (20 + 45d) - (10 + 10d) = 20 + 45d - 10 - 10d = 10 + 35d$.

---

We are given that the sum of the first five terms is one-fourth of the sum of the next five terms.

$S_5 = \frac{1}{4} (S_{10} - S_5)$

Substitute the expressions for $S_5$ and $S_{10} - S_5$:

$10 + 10d = \frac{1}{4} (10 + 35d)$

Multiply both sides by 4:

$4(10 + 10d) = 10 + 35d$

$40 + 40d = 10 + 35d$

Rearrange to solve for $d$:

$40d - 35d = 10 - 40$

$5d = -30$

$d = \frac{-30}{5} = -6$

---

The common difference of the A.P. is $d = -6$. The first term is $a_1 = 2$.

---

We need to find the 20^th term ($a_{20}$). The formula for the k^th term is $a_k = a_1 + (k-1)d$.

Substitute $k=20$, $a_1=2$, and $d=-6$:

$a_{20} = 2 + (20-1)(-6)$

$a_{20} = 2 + (19)(-6)$

$19 \times 6 = 114$. So $19 \times (-6) = -114$.

$a_{20} = 2 + (-114)$

$a_{20} = 2 - 114 = -112$

---

The 20^th term is -112.

We have shown that the 20^th term is -112.

Hence Showed.

Given A.P.:

-6, $-\frac{11}{2}$, -5, ...

The sum of the first $n$ terms ($S_n$) is -25.

---

To Find:

The number of terms ($n$) required to get a sum of -25.

---

Solution:

The first term of the A.P. is $a_1 = -6$.

The common difference $d$ can be found by subtracting any term from its subsequent term.

$d = a_2 - a_1 = -\frac{11}{2} - (-6) = -\frac{11}{2} + 6 = -\frac{11}{2} + \frac{12}{2} = \frac{-11+12}{2} = \frac{1}{2}$.

Alternatively, $d = a_3 - a_2 = -5 - \left(-\frac{11}{2}\right) = -5 + \frac{11}{2} = -\frac{10}{2} + \frac{11}{2} = \frac{-10+11}{2} = \frac{1}{2}$.

The common difference is $d = \frac{1}{2}$.

---

The formula for the sum of the first $n$ terms of an A.P. is $S_n = \frac{n}{2} [2a_1 + (n-1)d]$.

We are given $S_n = -25$, $a_1 = -6$, and $d = \frac{1}{2}$. Substitute these values into the formula:

$-25 = \frac{n}{2} [2(-6) + (n-1)\frac{1}{2}]$

$-25 = \frac{n}{2} [-12 + \frac{n-1}{2}]$

Multiply both sides by 2:

$2(-25) = n [-12 + \frac{n-1}{2}]$

$-50 = n [-12 + \frac{n-1}{2}]$

Multiply the term inside the bracket by 2 to remove the fraction inside, and multiply the outside $n$ by $\frac{1}{2}$. No, multiply the entire right side by 2 and the left side by 2 to clear the fraction inside.

$-50 = n \left[\frac{-24 + n - 1}{2}\right]$

$-50 = n \left[\frac{n - 25}{2}\right]$

Multiply both sides by 2:

$2(-50) = n (n - 25)$

$-100 = n^2 - 25n$

Rearrange the equation to form a quadratic equation:

$n^2 - 25n + 100 = 0$

---

Solve this quadratic equation for $n$. We can try factoring or use the quadratic formula.

We need two numbers that multiply to 100 and add up to -25. The pairs of factors of 100 are (1, 100), (2, 50), (4, 25), (5, 20), (10, 10). To get a sum of -25, both numbers should be negative. The pair (-5, -20) multiplies to 100 and adds up to -25.

So, we can factor the quadratic equation:

$(n - 5)(n - 20) = 0$

---

Setting each factor to zero gives the possible values for $n$:

$n - 5 = 0 \implies n = 5$

$n - 20 = 0 \implies n = 20$

---

Since $n$ represents the number of terms, it must be a positive integer. Both 5 and 20 are positive integers.

Let's check if both values of $n$ give a sum of -25.

If $n=5$: $S_5 = \frac{5}{2} [2(-6) + (5-1)\frac{1}{2}] = \frac{5}{2} [-12 + 4 \times \frac{1}{2}] = \frac{5}{2} [-12 + 2] = \frac{5}{2} [-10] = 5 \times (-5) = -25$. Correct.

If $n=20$: $S_{20} = \frac{20}{2} [2(-6) + (20-1)\frac{1}{2}] = 10 [-12 + 19 \times \frac{1}{2}] = 10 [-12 + \frac{19}{2}] = 10 [\frac{-24 + 19}{2}] = 10 [\frac{-5}{2}] = 5 \times (-5) = -25$. Correct.

---

Both 5 terms and 20 terms result in a sum of -25. This happens when the terms after the 5th term are such that their sum is zero. The terms are -6, -5.5, -5, -4.5, -4, -3.5, -3, -2.5, -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5. The 6th to 20th terms are -3.5, -3, ..., 3.5. The sum of an arithmetic progression with an odd number of terms where the middle term is 0, or with terms symmetric around 0, can be 0.

The question asks "How many terms... are needed". This implies finding all possible values of $n$.

---

The possible numbers of terms needed are 5 or 20.

Given:

The sequence is an Arithmetic Progression (A.P.).

The p^th term is $a_p = \frac{1}{q}$.

The q^th term is $a_q = \frac{1}{p}$.

$p$ and $q$ are integers and $p \neq q$.

---

To Prove:

The sum of the first pq terms ($S_{pq}$) is $\frac{1}{2} (pq + 1)$.

---

Proof:

Let the first term of the A.P. be $a_1$ and the common difference be $d$.

The formula for the n^th term of an A.P. is $a_n = a_1 + (n-1)d$.

---

Using the given information about the p^th and q^th terms:

For the p^th term:

For the q^th term:

---

Subtract equation (ii) from equation (i) to eliminate $a_1$:

$(a_1 + (p-1)d) - (a_1 + (q-1)d) = \frac{1}{q} - \frac{1}{p}$

$(p-1)d - (q-1)d = \frac{p}{pq} - \frac{q}{pq}$

$d(p-1 - (q-1)) = \frac{p-q}{pq}$

$d(p - 1 - q + 1) = \frac{p-q}{pq}$

$d(p - q) = \frac{p-q}{pq}$

Since $p \neq q$, $p-q \neq 0$. Divide both sides by $(p-q)$:

---

Substitute the value of $d = \frac{1}{pq}$ into equation (i) to find $a_1$:

$a_1 + (p-1)\left(\frac{1}{pq}\right) = \frac{1}{q}$

$a_1 + \frac{p-1}{pq} = \frac{1}{q}$

$a_1 = \frac{1}{q} - \frac{p-1}{pq}$

Find a common denominator on the right side ($pq$):

$a_1 = \frac{p}{pq} - \frac{p-1}{pq}$

$a_1 = \frac{p - (p-1)}{pq} = \frac{p - p + 1}{pq} = \frac{1}{pq}$

---

Now we need to find the sum of the first pq terms, $S_{pq}$.

The formula for the sum of the first $n$ terms of an A.P. is $S_n = \frac{n}{2} [2a_1 + (n-1)d]$.

Substitute $n=pq$, $a_1 = \frac{1}{pq}$, and $d = \frac{1}{pq}$ into the formula for $S_{pq}$:

$S_{pq} = \frac{pq}{2} \left[2\left(\frac{1}{pq}\right) + (pq-1)\left(\frac{1}{pq}\right)\right]$

$S_{pq} = \frac{pq}{2} \left[\frac{2}{pq} + \frac{pq-1}{pq}\right]$

Combine the terms inside the square bracket (they have a common denominator $pq$):

$S_{pq} = \frac{pq}{2} \left[\frac{2 + (pq-1)}{pq}\right]$

$S_{pq} = \frac{pq}{2} \left[\frac{2 + pq - 1}{pq}\right]$

$S_{pq} = \frac{pq}{2} \left[\frac{pq + 1}{pq}\right]$

Assuming $p \neq 0$ and $q \neq 0$ (which must be true since $\frac{1}{q}$ and $\frac{1}{p}$ exist), we can cancel the $pq$ term in the numerator and denominator:

$S_{pq} = \frac{1}{2} (pq + 1)$

---

We have shown that the sum of the first pq terms is $\frac{1}{2}(pq+1)$.

Hence Proved.

Given A.P.:

25, 22, 19, ...

The sum of a certain number of terms ($S_n$) is 116.

---

To Find:

The last term of the sum (i.e., the n^th term $a_n$, where $n$ is the number of terms whose sum is 116).

---

Solution:

The first term of the A.P. is $a_1 = 25$.

The common difference $d$ can be found by subtracting any term from its subsequent term.

$d = a_2 - a_1 = 22 - 25 = -3$.

The common difference is $d = -3$.

---

The formula for the sum of the first $n$ terms of an A.P. is $S_n = \frac{n}{2} [2a_1 + (n-1)d]$.

We are given $S_n = 116$, $a_1 = 25$, and $d = -3$. Substitute these values into the formula:

$116 = \frac{n}{2} [2(25) + (n-1)(-3)]$

$116 = \frac{n}{2} [50 - 3n + 3]$

$116 = \frac{n}{2} [53 - 3n]$

Multiply both sides by 2:

$2(116) = n (53 - 3n)$

$232 = 53n - 3n^2$

Rearrange the equation to form a quadratic equation:

$3n^2 - 53n + 232 = 0$

---

Solve this quadratic equation for $n$. We can use the quadratic formula: $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Here, $a=3$, $b=-53$, $c=232$.

$n = \frac{-(-53) \pm \sqrt{(-53)^2 - 4(3)(232)}}{2(3)}$

$n = \frac{53 \pm \sqrt{2809 - 12 \times 232}}{6}$

$12 \times 232 = 2784$.

$n = \frac{53 \pm \sqrt{2809 - 2784}}{6}$

$n = \frac{53 \pm \sqrt{25}}{6}$

$n = \frac{53 \pm 5}{6}$

---

We have two possible values for $n$:

$n_1 = \frac{53 + 5}{6} = \frac{58}{6} = \frac{29}{3}$

$n_2 = \frac{53 - 5}{6} = \frac{48}{6} = 8$

---

Since $n$ represents the number of terms, it must be a positive integer. $n=\frac{29}{3}$ is not an integer, while $n=8$ is a positive integer.

So, the number of terms is $n=8$.

---

We need to find the last term, which is the 8^th term ($a_8$).

The formula for the n^th term of an A.P. is $a_n = a_1 + (n-1)d$.

Substitute $n=8$, $a_1=25$, and $d=-3$:

$a_8 = 25 + (8-1)(-3)$

$a_8 = 25 + (7)(-3)$

$a_8 = 25 - 21$

$a_8 = 4$

---

The last term of the A.P. that gives a sum of 116 is the 8^th term, which is 4.

The last term is $\mathbf{4}$.

Given:

The k^th term of the A.P. is $a_k = 5k + 1$.

---

To Find:

The sum to n terms of the A.P., $S_n$.

---

Solution:

The k^th term of the A.P. is given by $a_k = 5k + 1$.

To find the first term, we substitute $k=1$:

$a_1 = 5(1) + 1 = 5 + 1 = 6$.

To find the second term, we substitute $k=2$:

$a_2 = 5(2) + 1 = 10 + 1 = 11$.

The common difference $d$ of the A.P. is the difference between consecutive terms:

$d = a_2 - a_1$

$d = 11 - 6 = 5$.

Alternatively, since the k^th term is a linear expression in $k$, $a_k = 5k + 1$, the coefficient of $k$ is the common difference. Thus, $d = 5$.

The sum of the first n terms of an A.P. is given by the formula:

$S_n = \frac{n}{2}[2a_1 + (n-1)d]$

Substitute the values of $a_1 = 6$ and $d = 5$ into the formula:

$S_n = \frac{n}{2}[2(6) + (n-1)5]$

$S_n = \frac{n}{2}[12 + 5n - 5]$

$S_n = \frac{n}{2}[7 + 5n]$

$S_n = \frac{n(5n + 7)}{2}$

The sum to n terms of the given A.P. is $\frac{n(5n + 7)}{2}$.

Given:

The sum of the first n terms of an A.P. is $S_n = pn + qn^2$, where p and q are constants.

---

To Find:

The common difference of the A.P.

---

Solution:

The sum of the first n terms of an A.P. is given by $S_n = pn + qn^2$.

The first term of the A.P., $a_1$, is equal to the sum of the first term, i.e., $S_1$.

Substitute $n=1$ in the expression for $S_n$:

$S_1 = p(1) + q(1)^2 = p + q$

So, the first term is $a_1 = p + q$.

---

The sum of the first two terms of the A.P. is $S_2$.

Substitute $n=2$ in the expression for $S_n$:

$S_2 = p(2) + q(2)^2 = 2p + 4q$

---

The second term of the A.P., $a_2$, can be found by subtracting the sum of the first term ($S_1$) from the sum of the first two terms ($S_2$).

$a_2 = S_2 - S_1$

$a_2 = (2p + 4q) - (p + q)$

$a_2 = 2p + 4q - p - q$

$a_2 = p + 3q$

---

The common difference, $d$, of an A.P. is the difference between any term and its preceding term.

$d = a_2 - a_1$

$d = (p + 3q) - (p + q)$

$d = p + 3q - p - q$

$d = 2q$

Thus, the common difference of the A.P. is $2q$.

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Alternate Method:

The k^th term of an A.P., $a_k$, can be found using the relation $a_k = S_k - S_{k-1}$ for $k > 1$.

$S_k = pk + qk^2$

$S_{k-1} = p(k-1) + q(k-1)^2 = p(k-1) + q(k^2 - 2k + 1) = pk - p + qk^2 - 2qk + q$

$a_k = S_k - S_{k-1} = (pk + qk^2) - (pk - p + qk^2 - 2qk + q)$

$a_k = pk + qk^2 - pk + p - qk^2 + 2qk - q$

$a_k = 2qk + p - q$

$a_k = (2q)k + (p-q)$

This is the expression for the k^th term of the A.P. Since the k^th term is a linear function of $k$, the coefficient of $k$ is the common difference.

The common difference is $d = 2q$.

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Answer: The common difference of the A.P. is $2q$.

Given:

Let the two arithmetic progressions be denoted by AP$_1$ and AP$_2$.

Let the first term and common difference of AP$_1$ be $a_1$ and $d_1$ respectively.

Let the first term and common difference of AP$_2$ be $a_2$ and $d_2$ respectively.

The sum of the first n terms of AP$_1$ is $S_{n,1}$.

The sum of the first n terms of AP$_2$ is $S_{n,2}$.

The ratio of their sums of n terms is given by:

---

To Find:

The ratio of their 18^th terms, i.e., $\frac{a_{18,1}}{a_{18,2}}$.

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Solution:

The formula for the sum of the first n terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$.

Using this formula, we can write the ratio of the sums as:

$\frac{S_{n,1}}{S_{n,2}} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]}$

$\frac{S_{n,1}}{S_{n,2}} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2}$

---

The formula for the k^th term of an AP is $a_k = a + (k-1)d$.

The 18^th term of AP$_1$ is $a_{18,1} = a_1 + (18-1)d_1 = a_1 + 17d_1$.

The 18^th term of AP$_2$ is $a_{18,2} = a_2 + (18-1)d_2 = a_2 + 17d_2$.

We want to find the ratio $\frac{a_{18,1}}{a_{18,2}} = \frac{a_1 + 17d_1}{a_2 + 17d_2}$.

---

Comparing the expression for the ratio of sums with the expression for the ratio of the 18^th terms:

Ratio of sums: $\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2}$

Ratio of 18^th terms: $\frac{a_1 + 17d_1}{a_2 + 17d_2}$

---

We can rewrite the ratio of sums by dividing the numerator and denominator by 2:

$\frac{S_{n,1}}{S_{n,2}} = \frac{a_1 + \frac{(n-1)}{2}d_1}{a_2 + \frac{(n-1)}{2}d_2}$

To make this expression equal to the ratio of the 18^th terms $\frac{a_1 + 17d_1}{a_2 + 17d_2}$, we need to equate the coefficients of $d_1$ and $d_2$ in the numerators and denominators respectively.

We set $\frac{(n-1)}{2} = 17$.

$n - 1 = 17 \times 2$

$n - 1 = 34$

$n = 34 + 1$

$n = 35$.

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So, the ratio of the 18^th terms is obtained by substituting $n=35$ into the given ratio of sums (Equation 1).

$\frac{a_{18,1}}{a_{18,2}} = \left. \frac{5n + 4}{9n + 6} \right|_{n=35}$

Substitute $n=35$:

$\frac{a_{18,1}}{a_{18,2}} = \frac{5(35) + 4}{9(35) + 6}$

$\frac{a_{18,1}}{a_{18,2}} = \frac{175 + 4}{315 + 6}$

$\frac{a_{18,1}}{a_{18,2}} = \frac{179}{321}$

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Answer:

The ratio of their 18^th terms is 179 : 321.

Given:

The sum of the first p terms of an A.P. is equal to the sum of the first q terms.

Let the first term of the A.P. be $a$ and the common difference be $d$.

The sum of the first n terms of an A.P. is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.

Given $S_p = S_q$, where $p \neq q$.

---

To Find:

The sum of the first (p + q) terms, i.e., $S_{p+q}$.

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Solution:

Using the formula for the sum of n terms, we have:

$S_p = \frac{p}{2}[2a + (p-1)d]$

$S_q = \frac{q}{2}[2a + (q-1)d]$

Since $S_p = S_q$, we can write:

$\frac{p}{2}[2a + (p-1)d] = \frac{q}{2}[2a + (q-1)d]$

Multiply both sides by 2:

$p[2a + (p-1)d] = q[2a + (q-1)d]$

Expand both sides:

$2ap + p(p-1)d = 2aq + q(q-1)d$

Rearrange the terms to group terms with $2a$ and terms with $d$:

$2ap - 2aq = q(q-1)d - p(p-1)d$

$2a(p - q) = [q(q-1) - p(p-1)]d$

$2a(p - q) = [q^2 - q - (p^2 - p)]d$

$2a(p - q) = [q^2 - q - p^2 + p]d$

$2a(p - q) = [q^2 - p^2 - (q - p)]d$

$2a(p - q) = [(q - p)(q + p) - (q - p)]d$

Factor out $(q - p)$ from the right side:

$2a(p - q) = (q - p)[(q + p) - 1]d$

$2a(p - q) = -(p - q)(p + q - 1)d$

Move all terms to the left side:

$2a(p - q) + (p - q)(p + q - 1)d = 0$

Factor out $(p - q)$:

$(p - q)[2a + (p + q - 1)d] = 0$

Since it is given that $p \neq q$, it means $p - q \neq 0$.

Therefore, the term in the square bracket must be zero:

---

Now, we need to find the sum of the first (p + q) terms, $S_{p+q}$.

Using the formula for the sum of n terms with $n = p+q$:

$S_{p+q} = \frac{p+q}{2}[2a + ((p+q)-1)d]$

$S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d]$

Substitute the value from Equation (i) into this expression:

$S_{p+q} = \frac{p+q}{2}[0]$

$S_{p+q} = 0$

---

Answer:

The sum of the first (p + q) terms of the A.P. is 0.

Given:

Sum of the first p terms of an A.P. is $a$, i.e., $S_p = a$.

Sum of the first q terms of the same A.P. is $b$, i.e., $S_q = b$.

Sum of the first r terms of the same A.P. is $c$, i.e., $S_r = c$.

Let the first term of the A.P. be $A$ and the common difference be $D$.

---

To Prove:

$\frac{a}{p} (q - r) + \frac{b}{q} (r - p) + \frac{c}{r} (p - q) = 0$

---

Proof:

The formula for the sum of the first n terms of an A.P. is $S_n = \frac{n}{2}[2A + (n-1)D]$.

Using this formula, we can write the given information as:

$S_p = a = \frac{p}{2}[2A + (p-1)D]$

$S_q = b = \frac{q}{2}[2A + (q-1)D]$

$S_r = c = \frac{r}{2}[2A + (r-1)D]$

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From these equations, we can express $\frac{a}{p}$, $\frac{b}{q}$, and $\frac{c}{r}$:

$\frac{a}{p} = \frac{1}{2}[2A + (p-1)D]$

$\frac{b}{q} = \frac{1}{2}[2A + (q-1)D]$

$\frac{c}{r} = \frac{1}{2}[2A + (r-1)D]$

---

Consider the left-hand side (LHS) of the equation we need to prove:

LHS $= \frac{a}{p} (q - r) + \frac{b}{q} (r - p) + \frac{c}{r} (p - q)$

Substitute the expressions for $\frac{a}{p}$, $\frac{b}{q}$, and $\frac{c}{r}$ into the LHS:

LHS $= \frac{1}{2}[2A + (p-1)D](q - r) + \frac{1}{2}[2A + (q-1)D](r - p) + \frac{1}{2}[2A + (r-1)D](p - q)$

Factor out $\frac{1}{2}$:

LHS $= \frac{1}{2} \left\{ [2A + (p-1)D](q - r) + [2A + (q-1)D](r - p) + [2A + (r-1)D](p - q) \right\}$

Expand the terms inside the curly braces:

$= \frac{1}{2} \left\{ [2A(q - r) + (p-1)D(q - r)] + [2A(r - p) + (q-1)D(r - p)] + [2A(p - q) + (r-1)D(p - q)] \right\}$

Group the terms containing $2A$ and the terms containing $D$:

$= \frac{1}{2} \left\{ [2A(q - r) + 2A(r - p) + 2A(p - q)] + [(p-1)D(q - r) + (q-1)D(r - p) + (r-1)D(p - q)] \right\}$

Consider the sum of the terms containing $2A$:

$2A(q - r) + 2A(r - p) + 2A(p - q) = 2A(q - r + r - p + p - q) = 2A(0) = 0$

Consider the sum of the terms containing $D$:

$D [ (p-1)(q - r) + (q-1)(r - p) + (r-1)(p - q) ]$

Expand the products inside the square brackets:

$(p-1)(q - r) = pq - pr - q + r$

$(q-1)(r - p) = qr - qp - r + p$

$(r-1)(p - q) = rp - rq - p + q$

Summing these expanded terms:

$(pq - pr - q + r) + (qr - pq - r + p) + (pr - qr - p + q)$

$= pq - pr - q + r + qr - pq - r + p + pr - qr - p + q$

$= (pq - pq) + (-pr + pr) + (-q + q) + (r - r) + (qr - qr) + (p - p)$

$= 0 + 0 + 0 + 0 + 0 + 0 = 0$

So, the sum of the terms containing $D$ is $D \times 0 = 0$.

---

Substitute these sums back into the expression for LHS:

LHS $= \frac{1}{2} \left\{ 0 + 0 \right\}$

LHS $= \frac{1}{2} \times 0$

LHS $= 0$

Since LHS = RHS (which is 0), the equation is proved.

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Answer:

Hence, $\frac{a}{p}$ (q - r) + $\frac{b}{q}$ (r - p) + $\frac{c}{r}$ (p - q) = 0 is proved.

Given:

Let the first term of the A.P. be $a$ and the common difference be $d$.

The sum of the first m terms is $S_m$.

The sum of the first n terms is $S_n$.

The ratio of the sums of m and n terms is given by:

---

To Show:

The ratio of the m^th term ($a_m$) and the n^th term ($a_n$) is $\frac{a_m}{a_n} = \frac{2m - 1}{2n - 1}$.

---

Proof:

The formula for the sum of the first k terms of an A.P. is $S_k = \frac{k}{2}[2a + (k-1)d]$.

Using this formula, we can write $S_m$ and $S_n$ as:

$S_m = \frac{m}{2}[2a + (m-1)d]$

$S_n = \frac{n}{2}[2a + (n-1)d]$

---

Substitute these expressions into the given ratio (Equation 1):

$\frac{\frac{m}{2}[2a + (m-1)d]}{\frac{n}{2}[2a + (n-1)d]} = \frac{m^2}{n^2}$

Cancel $\frac{1}{2}$ from the numerator and denominator of the left side:

$\frac{m[2a + (m-1)d]}{n[2a + (n-1)d]} = \frac{m^2}{n^2}$

Cancel one $m$ from the numerator of the left side and the $m^2$ from the right side. Cancel one $n$ from the denominator of the left side and the $n^2$ from the right side:

$\frac{2a + (m-1)d}{2a + (n-1)d} = \frac{m}{n}$

Cross-multiply:

$n[2a + (m-1)d] = m[2a + (n-1)d]$

Expand both sides:

$2an + n(m-1)d = 2am + m(n-1)d$

$2an + (mn - n)d = 2am + (mn - m)d$

Rearrange the terms to group terms with $2a$ and terms with $d$:

$2an - 2am = (mn - m)d - (mn - n)d$

$2a(n - m) = (mn - m - mn + n)d$

$2a(n - m) = (n - m)d$

Since the ratio of sums of m and n terms is given, we assume $m \neq n$. Thus, $n - m \neq 0$. We can divide both sides by $(n - m)$:

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Now, let's consider the ratio of the m^th and n^th terms.

The formula for the k^th term of an A.P. is $a_k = a + (k-1)d$.

The m^th term is $a_m = a + (m-1)d$.

The n^th term is $a_n = a + (n-1)d$.

The ratio of the m^th and n^th terms is:

$\frac{a_m}{a_n} = \frac{a + (m-1)d}{a + (n-1)d}$

Substitute the relation $d = 2a$ (from Equation 2) into this ratio:

$\frac{a_m}{a_n} = \frac{a + (m-1)(2a)}{a + (n-1)(2a)}$

Expand the terms:

$\frac{a_m}{a_n} = \frac{a + (2m - 2)a}{a + (2n - 2)a}$

Factor out $a$ from the numerator and the denominator:

$\frac{a_m}{a_n} = \frac{a[1 + (2m - 2)]}{a[1 + (2n - 2)]}$

Assuming $a \neq 0$ (otherwise all terms are 0, and the ratio is undefined in this context), cancel $a$ from the numerator and denominator:

$\frac{a_m}{a_n} = \frac{1 + 2m - 2}{1 + 2n - 2}$

Simplify the expressions in the numerator and denominator:

$\frac{a_m}{a_n} = \frac{2m - 1}{2n - 1}$

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This shows that the ratio of the m^th and n^th terms is $(2m - 1) : (2n - 1)$.

---

Answer:

Hence, the ratio of m^th and n^th terms is $(2m - 1) : (2n - 1)$.

Given:

The sum of the first n terms of an A.P. is $S_n = 3n^2 + 5n$.

The m^th term of the A.P. is $a_m = 164$.

---

To Find:

The value of $m$.

---

Solution:

The sum of the first n terms of the A.P. is $S_n = 3n^2 + 5n$.

The n^th term of an A.P. can be found using the relation $a_n = S_n - S_{n-1}$ for $n > 1$.

First, find $S_{n-1}$ by substituting $(n-1)$ for $n$ in the expression for $S_n$:

$S_{n-1} = 3(n-1)^2 + 5(n-1)$

$S_{n-1} = 3(n^2 - 2n + 1) + 5n - 5$

$S_{n-1} = 3n^2 - 6n + 3 + 5n - 5$

$S_{n-1} = 3n^2 - n - 2$

---

Now, find the n^th term, $a_n$:

$a_n = S_n - S_{n-1}$

$a_n = (3n^2 + 5n) - (3n^2 - n - 2)$

$a_n = 3n^2 + 5n - 3n^2 + n + 2$

$a_n = 6n + 2$

---

To check the formula for $n=1$, we find the first term $a_1$ using the formula and $S_1$ from the sum expression.

$S_1 = 3(1)^2 + 5(1) = 3 + 5 = 8$.

$a_1$ from $a_n = 6n + 2$ is $a_1 = 6(1) + 2 = 8$.

Since $a_1 = S_1$, the formula $a_n = 6n + 2$ is valid for all $n \geq 1$.

---

We are given that the m^th term of the A.P. is 164.

Using the formula for the n^th term, the m^th term is $a_m = 6m + 2$.

We are given $a_m = 164$.

Equating the two expressions for $a_m$:

$6m + 2 = 164$

Subtract 2 from both sides:

$6m = 164 - 2$

$6m = 162$

Divide by 6:

$m = \frac{162}{6}$

$\frac{162}{6} = \frac{\cancel{162}^{27}}{\cancel{6}_{1}} = 27$

$m = 27$

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Answer:

The value of m is 27.

Given:

We need to insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

The sequence will be of the form 8, $x_1$, $x_2$, $x_3$, $x_4$, $x_5$, 26.

---

To Find:

The five numbers $x_1, x_2, x_3, x_4, x_5$ to be inserted.

---

Solution:

The resulting sequence 8, $x_1$, $x_2$, $x_3$, $x_4$, $x_5$, 26 is an A.P.

The total number of terms in this A.P. is the initial term (8) + the five inserted terms + the final term (26).

Total number of terms, $n = 1 + 5 + 1 = 7$.

The first term of the A.P. is $a_1 = 8$.

The last term of the A.P. is the 7^th term, $a_7 = 26$.

---

Let the common difference of the A.P. be $d$.

The formula for the n^th term of an A.P. is $a_n = a_1 + (n-1)d$.

Using this for the 7^th term ($n=7$), we have:

$a_7 = a_1 + (7-1)d$

$26 = 8 + 6d$

Subtract 8 from both sides:

$26 - 8 = 6d$

$18 = 6d$

Divide by 6:

$d = \frac{18}{6}$

$d = 3$

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The five numbers to be inserted are the terms $a_2, a_3, a_4, a_5, a_6$.

$x_1 = a_2 = a_1 + d = 8 + 3 = 11$.

$x_2 = a_3 = a_1 + 2d = 8 + 2(3) = 8 + 6 = 14$.

$x_3 = a_4 = a_1 + 3d = 8 + 3(3) = 8 + 9 = 17$.

$x_4 = a_5 = a_1 + 4d = 8 + 4(3) = 8 + 12 = 20$.

$x_5 = a_6 = a_1 + 5d = 8 + 5(3) = 8 + 15 = 23$.

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The resulting A.P. is 8, 11, 14, 17, 20, 23, 26.

The five numbers inserted between 8 and 26 are 11, 14, 17, 20, and 23.

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Answer:

The five numbers inserted are 11, 14, 17, 20, and 23.

Given:

The expression $\frac{a^{n} + b^{n}}{a^{n-1} + b^{n-1}}$ is the arithmetic mean (A.M.) between $a$ and $b$.

The formula for the A.M. between two numbers $a$ and $b$ is $\frac{a+b}{2}$.

---

To Find:

The value of $n$.

---

Solution:

According to the problem statement, we equate the given expression to the A.M. formula:

$\frac{a^{n} + b^{n}}{a^{n-1} + b^{n-1}} = \frac{a+b}{2}$

Cross-multiply the equation:

$2(a^n + b^n) = (a+b)(a^{n-1} + b^{n-1})$

Expand the right side of the equation:

$2a^n + 2b^n = a \cdot a^{n-1} + a \cdot b^{n-1} + b \cdot a^{n-1} + b \cdot b^{n-1}$

$2a^n + 2b^n = a^n + ab^{n-1} + ba^{n-1} + b^n$

Rearrange the terms by moving all terms to the left side:

$2a^n - a^n + 2b^n - b^n - ab^{n-1} - ba^{n-1} = 0$

$a^n + b^n - ab^{n-1} - a^{n-1}b = 0$

Group the terms with common factors:

$(a^n - a^{n-1}b) + (b^n - ab^{n-1}) = 0$

Factor out $a^{n-1}$ from the first group and $b^{n-1}$ from the second group:

$a^{n-1}(a - b) + b^{n-1}(b - a) = 0$

Rewrite $(b - a)$ as $-(a - b)$:

$a^{n-1}(a - b) - b^{n-1}(a - b) = 0$

Factor out the common term $(a - b)$:

$(a - b)(a^{n-1} - b^{n-1}) = 0$

---

For this equation to hold, either $(a - b) = 0$ or $(a^{n-1} - b^{n-1}) = 0$.

If $a - b = 0$, then $a = b$. In this case, the expression becomes $\frac{a^n + a^n}{a^{n-1} + a^{n-1}} = \frac{2a^n}{2a^{n-1}} = a$ (assuming $a \neq 0$), and the A.M. is $\frac{a+a}{2} = a$. So, the equality holds for any value of $n$ if $a=b$ (and $a, b \neq 0$). However, the question usually implies that the expression represents the A.M. for arbitrary $a$ and $b$ (where the expression is defined and $a \neq b$).

Assuming $a \neq b$, we must have the second factor equal to zero:

$a^{n-1} - b^{n-1} = 0$

$a^{n-1} = b^{n-1}$

Assuming $b \neq 0$, we can divide both sides by $b^{n-1}$:

$\frac{a^{n-1}}{b^{n-1}} = 1$

$\left(\frac{a}{b}\right)^{n-1} = 1$

For this equality to hold for arbitrary values of $a$ and $b$ (such that $a \neq b$ and $b \neq 0$, meaning $\frac{a}{b}$ is a positive number not equal to 1), the exponent must be zero.

$n - 1 = 0$

$n = 1$

---

We can verify this result by substituting $n=1$ back into the original expression:

$\frac{a^{1} + b^{1}}{a^{1-1} + b^{1-1}} = \frac{a+b}{a^{0} + b^{0}}$

Assuming $a \neq 0$ and $b \neq 0$, we have $a^0 = 1$ and $b^0 = 1$.

$\frac{a+b}{1 + 1} = \frac{a+b}{2}$

This is indeed the A.M. between $a$ and $b$.

---

Answer:

The value of n is 1.

Given:

Two numbers are 1 and 31.

m numbers are inserted between 1 and 31 to form an A.P.

The resulting sequence is 1, $x_1, x_2, \dots, x_m$, 31.

The ratio of the 7^th inserted number to the (m – 1)^th inserted number is 5 : 9.

---

To Find:

The value of m.

---

Solution:

The resulting sequence 1, $x_1, x_2, \dots, x_m$, 31 is an A.P.

Let the first term of this A.P. be $A_1$ and the common difference be $d$.

The first term is $A_1 = 1$.

The total number of terms in the A.P. is the sum of the initial term, the m inserted terms, and the final term.

Total number of terms, $n = 1 + m + 1 = m+2$.

The last term (31) is the (m+2)^th term of the A.P.

So, $A_{m+2} = 31$.

---

The formula for the n^th term of an A.P. is $A_n = A_1 + (n-1)d$.

Using this for the (m+2)^th term:

$A_{m+2} = A_1 + ((m+2)-1)d$

$31 = 1 + (m+1)d$

Subtract 1 from both sides:

$30 = (m+1)d$

Thus, the common difference is $d = \frac{30}{m+1}$.

---

The inserted numbers are $x_1, x_2, \dots, x_m$.

These correspond to the terms $A_2, A_3, \dots, A_{m+1}$ in the A.P.

The 7^th inserted number is $x_7$, which is the $A_{7+1} = A_8$ term of the A.P.

$A_8 = A_1 + (8-1)d = A_1 + 7d = 1 + 7d$.

The (m – 1)^th inserted number is $x_{m-1}$, which is the $A_{(m-1)+1} = A_m$ term of the A.P.

$A_m = A_1 + (m-1)d = 1 + (m-1)d$.

---

The ratio of the 7^th and (m – 1)^th inserted numbers is given as 5 : 9.

$\frac{\text{7}^{th} \text{ inserted number}}{\text{(m-1)}^{th} \text{ inserted number}} = \frac{A_8}{A_m} = \frac{5}{9}$

Substitute the expressions for $A_8$ and $A_m$:

$\frac{1 + 7d}{1 + (m-1)d} = \frac{5}{9}$

Substitute the expression for $d = \frac{30}{m+1}$:

$\frac{1 + 7\left(\frac{30}{m+1}\right)}{1 + (m-1)\left(\frac{30}{m+1}\right)} = \frac{5}{9}$

Simplify the numerator and the denominator:

Numerator $= 1 + \frac{210}{m+1} = \frac{m+1}{m+1} + \frac{210}{m+1} = \frac{m+1+210}{m+1} = \frac{m+211}{m+1}$

Denominator $= 1 + \frac{30(m-1)}{m+1} = \frac{m+1}{m+1} + \frac{30m-30}{m+1} = \frac{m+1+30m-30}{m+1} = \frac{31m-29}{m+1}$

The ratio becomes:

$\frac{\frac{m+211}{m+1}}{\frac{31m-29}{m+1}} = \frac{5}{9}$

$\frac{m+211}{31m-29} = \frac{5}{9}$

Cross-multiply:

$9(m+211) = 5(31m-29)$

Expand both sides:

$9m + 9 \times 211 = 5 \times 31m - 5 \times 29$

$9m + 1899 = 155m - 145$

Collect terms with m on one side and constant terms on the other side:

$1899 + 145 = 155m - 9m$

$2044 = 146m$

Solve for m:

$m = \frac{2044}{146}$

Performing the division:

$\begin{array}{r} 14\phantom{)} \\ 146{\overline{\smash{\big)}\,2044\phantom{)}}} \\ \underline{-~\phantom{(}146\downarrow\phantom{)}} \\ 584\phantom{)} \\ \underline{-~\phantom{(}584\phantom{)}} \\ 0\phantom{)} \end{array}$

$m = 14$

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Answer:

The value of m is 14.

Given:

First instalment amount = $\textsf{₹}100$.

Monthly increase in instalment amount = $\textsf{₹}5$.

This scenario forms an arithmetic progression (A.P.) where:

The first term, $a_1 = \textsf{₹}100$.

The common difference, $d = \textsf{₹}5$.

---

To Find:

The amount paid in the 30^th instalment.

---

Solution:

The amount paid in the 30^th instalment is the 30^th term of the A.P.

The formula for the n^th term of an A.P. is $a_n = a_1 + (n-1)d$.

To find the 30^th instalment ($n=30$), we substitute the values of $a_1$ and $d$ into the formula:

$a_{30} = a_1 + (30-1)d$

$a_{30} = 100 + (29) \times 5$

Calculate the product of 29 and 5:

$29 \times 5 = 145$

Substitute this value back into the expression for $a_{30}$:

$a_{30} = 100 + 145$

$a_{30} = 245$

The amount paid in the 30^th instalment is $\textsf{₹}245$.

---

Answer:

The amount the man will pay in the 30^th instalment is $\textsf{₹}245$.

Given:

The interior angles of the polygon form an arithmetic progression (A.P.).

The smallest interior angle is $120^\circ$. This is the first term of the A.P.

Let $a_1 = 120^\circ$.

The difference between any two consecutive interior angles is $5^\circ$. This is the common difference of the A.P.

Since the smallest angle is $120^\circ$ and the angles form an A.P. with a difference of $5^\circ$, the angles are increasing. Thus, the common difference $d = 5^\circ$.

Let the number of sides of the polygon be $n$. The polygon has $n$ interior angles, which are the terms of the A.P.: $a_1, a_2, \dots, a_n$.

---

To Find:

The number of sides of the polygon, $n$.

---

Solution:

The sum of the interior angles of a polygon with $n$ sides is given by the formula:

Sum $= (n-2) \times 180^\circ$

---

The interior angles form an A.P. with first term $a_1 = 120^\circ$ and common difference $d = 5^\circ$. The sum of the first $n$ terms of this A.P. is given by the formula:

$S_n = \frac{n}{2}[2a_1 + (n-1)d]$

Substitute the values of $a_1$ and $d$ into the formula for $S_n$:

$S_n = \frac{n}{2}[2(120) + (n-1)5]$

$S_n = \frac{n}{2}[240 + 5n - 5]$

$S_n = \frac{n}{2}[235 + 5n]$

---

Equating the two expressions for the sum of the interior angles:

$\frac{n}{2}[235 + 5n] = (n-2)180$

Multiply both sides by 2:

$n(235 + 5n) = 2(180n - 360)$

$235n + 5n^2 = 360n - 720$

Rearrange the terms to form a quadratic equation:

$5n^2 + 235n - 360n + 720 = 0$

$5n^2 - 125n + 720 = 0$

Divide the entire equation by 5:

$n^2 - 25n + 144 = 0$

---

Solve the quadratic equation by factoring:

We need two numbers that multiply to 144 and add up to -25. These numbers are -9 and -16.

$(n - 9)(n - 16) = 0$

This gives two possible values for $n$:

$n - 9 = 0 \implies n = 9$

$n - 16 = 0 \implies n = 16$

---

For a polygon, the number of sides $n$ must be an integer greater than or equal to 3. Both $n=9$ and $n=16$ satisfy this condition.

However, for a convex polygon, all interior angles must be less than $180^\circ$. The angles in this A.P. are $a_k = 120 + (k-1)5$. Since $d=5 > 0$, the angles are increasing. The largest angle is the last term, $a_n$.

Check the largest angle for each possible value of $n$:

If $n=9$, the largest angle is $a_9 = a_1 + (9-1)d = 120 + 8 \times 5 = 120 + 40 = 160^\circ$. Since $160^\circ < 180^\circ$, $n=9$ is a valid number of sides for a convex polygon.

If $n=16$, the largest angle is $a_{16} = a_1 + (16-1)d = 120 + 15 \times 5 = 120 + 75 = 195^\circ$. Since $195^\circ > 180^\circ$, $n=16$ is not a valid number of sides for a convex polygon (it corresponds to a non-convex polygon with a re-entrant angle).

In typical problems involving polygons at this level, it is usually implied that the polygon is convex unless stated otherwise. Therefore, we consider only the solution that results in all interior angles being less than $180^\circ$.

Thus, the only valid number of sides for a convex polygon under these conditions is $n=9$.

---

Answer:

The number of sides of the polygon is 9.

Given:

The geometric progression (G.P.) is 5, 25, 125, ...

---

To Find:

The 10^th term and the n^th term of the given G.P.

---

Solution:

The given sequence is 5, 25, 125, ...

This is a G.P. with the first term $a = 5$.

To find the common ratio, $r$, we divide a term by its preceding term:

$r = \frac{25}{5} = 5$

Checking with the next pair of terms:

$r = \frac{125}{25} = 5$

The common ratio is $r = 5$.

---

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

---

To find the 10^th term ($a_{10}$), substitute $n=10$, $a=5$, and $r=5$ into the formula:

$a_{10} = 5 \times 5^{10-1}$

$a_{10} = 5 \times 5^9$

$a_{10} = 5^{1} \times 5^9$

$a_{10} = 5^{1+9}$

$a_{10} = 5^{10}$

---

To find the n^th term ($a_n$), substitute $a=5$ and $r=5$ into the formula:

$a_n = 5 \times 5^{n-1}$

$a_n = 5^{1} \times 5^{n-1}$

$a_n = 5^{1 + (n-1)}$

$a_n = 5^n$

---

Answer:

The 10^th term of the G.P. is $5^{10}$.

The n^th term of the G.P. is $5^n$.

Given:

The geometric progression (G.P.) is 2, 8, 32, ...

A term in the G.P. is 131072.

---

To Find:

The term number of the term 131072.

---

Solution:

The given sequence is 2, 8, 32, ...

This is a G.P. with the first term $a = 2$.

The common ratio, $r$, is found by dividing any term by its preceding term:

$r = \frac{8}{2} = 4$

We can verify this with the next pair of terms: $r = \frac{32}{8} = 4$.

So, the common ratio is $r = 4$.

---

Let the n^th term of the G.P. be $a_n$. We are given that $a_n = 131072$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Substitute the known values into the formula:

$131072 = 2 \times 4^{n-1}$

Divide both sides by 2:

$\frac{131072}{2} = 4^{n-1}$

$65536 = 4^{n-1}$

We need to express 65536 as a power of 4.

Let's find the powers of 4:

$4^1 = 4$

$4^2 = 16$

$4^3 = 64$

$4^4 = 256$

$4^5 = 1024$

$4^6 = 4096$

$4^7 = 16384$

$4^8 = 65536$

So, we have $65536 = 4^8$.

---

Substitute this back into the equation:

$4^8 = 4^{n-1}$

Since the bases are equal, the exponents must be equal:

$8 = n - 1$

Add 1 to both sides to solve for $n$:

$n = 8 + 1$

$n = 9$

---

The 9^th term of the given G.P. is 131072.

---

Answer:

The 9^th term of the G.P. is 131072.

Given:

The 3^rd term of a G.P. is $a_3 = 24$.

The 6^th term of the same G.P. is $a_6 = 192$.

---

To Find:

The 10^th term of the G.P., $a_{10}$.

---

Solution:

Let the first term of the G.P. be $a$ and the common ratio be $r$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Using this formula, we can write the given information as equations:

---

To find the common ratio $r$, divide Equation (2) by Equation (1):

$\frac{ar^5}{ar^2} = \frac{192}{24}$

$r^{5-2} = 8$

$r^3 = 8$

Taking the cube root of both sides:

$r = \sqrt[3]{8}$

$r = 2$

---

Now, substitute the value of $r=2$ back into Equation (1) to find the first term $a$:

$ar^2 = 24$

$a(2)^2 = 24$

$a \times 4 = 24$

Divide both sides by 4:

$a = \frac{24}{4}$

$a = 6$

---

The first term is $a=6$ and the common ratio is $r=2$.

Now, find the 10^th term ($a_{10}$) using the formula $a_n = ar^{n-1}$ with $n=10$:

$a_{10} = ar^{10-1}$

$a_{10} = a r^9$

Substitute the values of $a=6$ and $r=2$:

$a_{10} = 6 \times 2^9$

Calculate $2^9$:

$2^9 = 512$

Substitute the value of $2^9$:

$a_{10} = 6 \times 512$

Multiply 6 by 512:

$a_{10} = 3072$

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Answer:

The 10^th term of the G.P. is 3072.

Given:

The geometric series is $1 + \frac{2}{3} + \frac{4}{9} + \dots$.

---

To Find:

The sum of the first n terms ($S_n$) and the sum of the first 5 terms ($S_5$) of the given geometric series.

---

Solution:

The given series is a geometric series.

The first term is $a = 1$.

The common ratio, $r$, is found by dividing a term by its preceding term:

$r = \frac{2/3}{1} = \frac{2}{3}$

We can verify with the next pair of terms: $r = \frac{4/9}{2/3} = \frac{4}{9} \times \frac{3}{2} = \frac{12}{18} = \frac{2}{3}$.

The common ratio is $r = \frac{2}{3}$. Since $|r| = \left|\frac{2}{3}\right| < 1$, the sum to infinity exists, but we are asked for the sum of the first n terms.

---

The formula for the sum of the first n terms of a G.P. with first term $a$ and common ratio $r$ is $S_n = \frac{a(1 - r^n)}{1 - r}$.

---

Substitute the values $a=1$ and $r=\frac{2}{3}$ into the formula for $S_n$:

$S_n = \frac{1 \left(1 - \left(\frac{2}{3}\right)^n\right)}{1 - \frac{2}{3}}$

$S_n = \frac{1 - \left(\frac{2}{3}\right)^n}{\frac{3}{3} - \frac{2}{3}}$

$S_n = \frac{1 - \left(\frac{2}{3}\right)^n}{\frac{1}{3}}$

$S_n = 3 \left(1 - \left(\frac{2}{3}\right)^n\right)$

$S_n = 3 \left(1 - \frac{2^n}{3^n}\right)$

$S_n = 3 - \frac{3 \times 2^n}{3^n}$

$S_n = 3 - \frac{2^n}{3^{n-1}}$

---

To find the sum of the first 5 terms ($S_5$), substitute $n=5$ into the formula for $S_n$:

$S_5 = 3 \left(1 - \left(\frac{2}{3}\right)^5\right)$

$S_5 = 3 \left(1 - \frac{2^5}{3^5}\right)$

Calculate $2^5 = 32$ and $3^5 = 243$:

$S_5 = 3 \left(1 - \frac{32}{243}\right)$

Find a common denominator for the term inside the parenthesis:

$S_5 = 3 \left(\frac{243}{243} - \frac{32}{243}\right)$

$S_5 = 3 \left(\frac{243 - 32}{243}\right)$

$S_5 = 3 \left(\frac{211}{243}\right)$

$S_5 = \frac{\cancel{3}^{1} \times 211}{\cancel{243}_{81}}$

$S_5 = \frac{211}{81}$

---

Answer:

The sum of the first n terms is $S_n = 3 - \frac{2^n}{3^{n-1}}$.

The sum of the first 5 terms is $S_5 = \frac{211}{81}$.

Given:

The geometric progression (G.P.) is 3, $\frac{3}{2}$, $\frac{3}{4}$, ...

The sum of the first n terms is $S_n = \frac{3069}{512}$.

---

To Find:

The number of terms, $n$.

---

Solution:

The given sequence is 3, $\frac{3}{2}$, $\frac{3}{4}$, ...

This is a G.P. with the first term $a = 3$.

The common ratio, $r$, is found by dividing any term by its preceding term:

$r = \frac{3/2}{3} = \frac{3}{2} \times \frac{1}{3} = \frac{1}{2}$.

The common ratio is $r = \frac{1}{2}$. Since $|r| = \left|\frac{1}{2}\right| < 1$, we use the formula for the sum of the first n terms:

$S_n = \frac{a(1 - r^n)}{1 - r}$

---

We are given $S_n = \frac{3069}{512}$. Substitute the values of $a$, $r$, and $S_n$ into the formula:

$\frac{3069}{512} = \frac{3 \left(1 - \left(\frac{1}{2}\right)^n\right)}{1 - \frac{1}{2}}$

Simplify the denominator on the right side:

$1 - \frac{1}{2} = \frac{2}{2} - \frac{1}{2} = \frac{1}{2}$

So the equation becomes:

$\frac{3069}{512} = \frac{3 \left(1 - \left(\frac{1}{2}\right)^n\right)}{\frac{1}{2}}$

Simplify the right side by dividing by $\frac{1}{2}$ (which is the same as multiplying by 2):

$\frac{3069}{512} = 3 \times 2 \times \left(1 - \left(\frac{1}{2}\right)^n\right)$

$\frac{3069}{512} = 6 \left(1 - \left(\frac{1}{2}\right)^n\right)$

Divide both sides by 6:

$\frac{3069}{512 \times 6} = 1 - \left(\frac{1}{2}\right)^n$

Simplify the fraction on the left side. Divide the numerator and denominator by 3:

$\frac{\cancel{3069}^{1023}}{512 \times \cancel{6}_{2}} = 1 - \left(\frac{1}{2}\right)^n$

$\frac{1023}{512 \times 2} = 1 - \left(\frac{1}{2}\right)^n$

$\frac{1023}{1024} = 1 - \left(\frac{1}{2}\right)^n$

Rearrange the equation to isolate the term with $n$:

$\left(\frac{1}{2}\right)^n = 1 - \frac{1023}{1024}$

Calculate the difference on the right side:

$1 - \frac{1023}{1024} = \frac{1024}{1024} - \frac{1023}{1024} = \frac{1}{1024}$

So, we have:

$\left(\frac{1}{2}\right)^n = \frac{1}{1024}$

We need to express 1024 as a power of 2:

$1024 = 2^{10}$

So, $\frac{1}{1024} = \frac{1}{2^{10}} = \left(\frac{1}{2}\right)^{10}$.

The equation becomes:

$\left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^{10}$

Since the bases are equal and positive, the exponents must be equal:

$n = 10$

---

Answer:

10 terms of the G.P. are needed to give the sum $\frac{3069}{512}$.

Given:

The sum of the first three terms of a G.P. is $\frac{13}{12}$.

The product of the first three terms of the G.P. is $-1$.

---

To Find:

The common ratio and the first three terms of the G.P.

---

Solution:

Let the first three terms of the G.P. be $\frac{a}{r}$, $a$, and $ar$, where $a$ is the first term and $r$ is the common ratio. (Note: Using this representation simplifies the product calculation. The terms in standard notation would be $a_1 = \frac{a}{r}$, $a_2 = a$, $a_3 = ar$).

---

According to the given information, the product of the first three terms is $-1$:

$\left(\frac{a}{r}\right) \times (a) \times (ar) = -1$

$a^{1} \times a^{1} \times a^{1} \times r^{-1} \times r^{1} = -1$

$a^{1+1+1} \times r^{-1+1} = -1$

$a^3 \times r^0 = -1$

Since $r^0 = 1$ (assuming $r \neq 0$), we have:

Taking the cube root of both sides, we get $a = -1$.

So, the second term of the G.P. is $-1$. The three terms are $\frac{-1}{r}$, $-1$, and $-r$.

---

According to the given information, the sum of the first three terms is $\frac{13}{12}$:

$\frac{-1}{r} + (-1) + (-r) = \frac{13}{12}$

$\frac{-1}{r} - 1 - r = \frac{13}{12}$

Multiply the entire equation by $12r$ to clear the denominators (assuming $r \neq 0$):

$12r \left(\frac{-1}{r}\right) - 12r(1) - 12r(r) = 12r \left(\frac{13}{12}\right)$

$-12 - 12r - 12r^2 = 13r$

Move all terms to one side to form a quadratic equation in $r$:

$12r^2 + 12r + 13r + 12 = 0$

---

We can solve this quadratic equation by factoring. We look for two numbers that multiply to $12 \times 12 = 144$ and add up to 25. These numbers are 9 and 16.

Rewrite the middle term:

$12r^2 + 9r + 16r + 12 = 0$

Group the terms and factor by grouping:

$(12r^2 + 9r) + (16r + 12) = 0$

$3r(4r + 3) + 4(4r + 3) = 0$

Factor out the common binomial $(4r + 3)$:

$(4r + 3)(3r + 4) = 0$

This gives two possible values for the common ratio $r$:

Case 1: $4r + 3 = 0 \implies 4r = -3 \implies r = -\frac{3}{4}$

Case 2: $3r + 4 = 0 \implies 3r = -4 \implies r = -\frac{4}{3}$

---

Now, we find the three terms for each case, using $a = -1$ and the respective common ratio $r$. The terms are $\frac{a}{r}, a, ar$, which simplifies to $\frac{-1}{r}, -1, -r$ since $a=-1$.

---

Case 1: Common ratio $r = -\frac{3}{4}$

The terms are:

First term: $\frac{-1}{r} = \frac{-1}{-3/4} = -1 \times (-\frac{4}{3}) = \frac{4}{3}$

Second term: $-1$

Third term: $-r = -(-\frac{3}{4}) = \frac{3}{4}$

The terms are $\frac{4}{3}, -1, \frac{3}{4}$.

---

Case 2: Common ratio $r = -\frac{4}{3}$

The terms are:

First term: $\frac{-1}{r} = \frac{-1}{-4/3} = -1 \times (-\frac{3}{4}) = \frac{3}{4}$

Second term: $-1$

Third term: $-r = -(-\frac{4}{3}) = \frac{4}{3}$

The terms are $\frac{3}{4}, -1, \frac{4}{3}$.

---

Both cases satisfy the given conditions. The common ratio can be $-\frac{3}{4}$ or $-\frac{4}{3}$, leading to two possible sets of terms.

---

Answer:

The common ratio is $-\frac{3}{4}$ or $-\frac{4}{3}$.

The terms of the G.P. are $\frac{4}{3}, -1, \frac{3}{4}$ (when $r = -\frac{3}{4}$) or $\frac{3}{4}, -1, \frac{4}{3}$ (when $r = -\frac{4}{3}$).

Given:

The sequence is 7, 77, 777, 7777, ... to n terms.

---

To Find:

The sum of the first n terms of the given sequence.

---

Solution:

Let the sum of the first n terms be $S_n$.

The terms of the sequence are $a_1 = 7$, $a_2 = 77$, $a_3 = 777$, and so on.

The n^th term, $a_n$, is a number consisting of n digits, all equal to 7.

We can write each term by factoring out 7:

$a_1 = 7 = 7 \times 1$

$a_2 = 77 = 7 \times 11$

$a_3 = 777 = 7 \times 111$

$a_n = 7 \times \underbrace{11\dots1}_{\text{n times}}$

The number $\underbrace{11\dots1}_{\text{n times}}$ can be expressed as $\frac{10^n - 1}{9}$.

So, the n^th term is $a_n = 7 \times \frac{10^n - 1}{9} = \frac{7}{9}(10^n - 1)$.

---

The sum of the first n terms is $S_n = a_1 + a_2 + \dots + a_n = \sum_{k=1}^{n} a_k$.

$S_n = \sum_{k=1}^{n} \frac{7}{9}(10^k - 1)$

$S_n = \frac{7}{9} \sum_{k=1}^{n} (10^k - 1)$

Separate the summation:

$S_n = \frac{7}{9} \left( \sum_{k=1}^{n} 10^k - \sum_{k=1}^{n} 1 \right)$

---

The second summation is simply the sum of n ones:

$\sum_{k=1}^{n} 1 = 1 + 1 + \dots + 1$ (n times) $= n$.

---

The first summation is the sum of a geometric series:

$\sum_{k=1}^{n} 10^k = 10^1 + 10^2 + 10^3 + \dots + 10^n$.

This is a geometric progression with the first term $A = 10$, the common ratio $R = 10$, and the number of terms is $n$.

The sum of the first n terms of a G.P. is $S_n = \frac{A(R^n - 1)}{R - 1}$.

Sum of this G.P. $= \frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9}$.

---

Substitute these results back into the expression for $S_n$:

$S_n = \frac{7}{9} \left( \frac{10(10^n - 1)}{9} - n \right)$

This can be further simplified:

$S_n = \frac{7}{9} \times \frac{10(10^n - 1)}{9} - \frac{7}{9}n$

$S_n = \frac{70(10^n - 1)}{81} - \frac{7n}{9}$

---

Answer:

The sum of the first n terms of the sequence is $S_n = \frac{70(10^n - 1)}{81} - \frac{7n}{9}$.

Given:

The number of ancestors in successive generations preceding a person's own are 2 (parents), 4 (grandparents), 8 (great grandparents), and so on.

We need to find the total number of ancestors in the first ten generations preceding the person's own.

---

To Find:

The sum of the number of ancestors in the first ten generations.

---

Solution:

The number of ancestors in the generations preceding the person forms a sequence: 2, 4, 8, ...

This sequence is a geometric progression (G.P.) because each term is obtained by multiplying the previous term by a constant value.

The first term of this G.P. is $a = 2$ (number of parents in the 1^st generation).

The common ratio, $r$, is found by dividing the second term by the first term:

$r = \frac{4}{2} = 2$.

The sequence represents the number of ancestors in the 1^st, 2^nd, 3^rd, ..., 10^th generations preceding the person.

We need to find the total number of ancestors in the first 10 generations, which means finding the sum of the first 10 terms of this G.P.

---

The formula for the sum of the first n terms of a G.P. with first term $a$ and common ratio $r$ is $S_n = \frac{a(r^n - 1)}{r - 1}$ (for $r \neq 1$).

In this case, $n = 10$ (number of generations), $a = 2$, and $r = 2$.

Substitute these values into the formula for the sum $S_{10}$:

$S_{10} = \frac{2(2^{10} - 1)}{2 - 1}$

$S_{10} = \frac{2(2^{10} - 1)}{1}$

$S_{10} = 2(2^{10} - 1)$

Calculate $2^{10}$:

$2^{10} = 1024$

Substitute the value of $2^{10}$:

$S_{10} = 2(1024 - 1)$

$S_{10} = 2(1023)$

Multiply 2 by 1023:

$S_{10} = 2046$

The total number of ancestors during the ten generations preceding the person's own is 2046.

---

Answer:

The number of his ancestors during the ten generations preceding his own is 2046.

Given:

We need to insert three numbers between 1 and 256 such that the resulting sequence is a Geometric Progression (G.P.).

The sequence will be of the form 1, $g_1, g_2, g_3$, 256.

---

To Find:

The three numbers $g_1, g_2, g_3$ to be inserted.

---

Solution:

The resulting sequence 1, $g_1, g_2, g_3$, 256 is a G.P.

The total number of terms in this G.P. is the initial term (1) + the three inserted terms + the final term (256).

Total number of terms, $n = 1 + 3 + 1 = 5$.

The first term of the G.P. is $a = 1$.

The last term of the G.P. is the 5^th term, $a_5 = 256$.

---

Let the common ratio of the G.P. be $r$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Using this for the 5^th term ($n=5$), we have:

$a_5 = ar^{5-1}$

$256 = 1 \times r^4$

$r^4 = 256$

We need to find the value(s) of $r$ such that $r^4 = 256$.

$r^4 - 256 = 0$

$(r^2)^2 - 16^2 = 0$

$(r^2 - 16)(r^2 + 16) = 0$

From $r^2 - 16 = 0$, we get $r^2 = 16$, so $r = \pm \sqrt{16}$, which means $r = 4$ or $r = -4$.

From $r^2 + 16 = 0$, we get $r^2 = -16$. This equation has no real solutions for $r$. Since we are inserting real numbers between 1 and 256, we consider only the real values of $r$.

So, the possible values for the common ratio are $r = 4$ and $r = -4$.

---

The three numbers to be inserted are the 2^nd, 3^rd, and 4^th terms of the G.P. ($a_2, a_3, a_4$).

$a_2 = ar^1 = ar$

$a_3 = ar^2$

$a_4 = ar^3$

Since $a = 1$, the inserted terms are $r, r^2, r^3$.

---

Case 1: Common ratio $r = 4$

The inserted numbers are:

$g_1 = a_2 = 1 \times 4 = 4$

$g_2 = a_3 = 1 \times 4^2 = 16$

$g_3 = a_4 = 1 \times 4^3 = 64$

The resulting G.P. is 1, 4, 16, 64, 256.

---

Case 2: Common ratio $r = -4$

The inserted numbers are:

$g_1 = a_2 = 1 \times (-4) = -4$

$g_2 = a_3 = 1 \times (-4)^2 = 16$

$g_3 = a_4 = 1 \times (-4)^3 = -64$

The resulting G.P. is 1, -4, 16, -64, 256.

---

Both sets of numbers form a G.P. between 1 and 256.

---

Answer:

The three numbers inserted between 1 and 256 are either 4, 16, 64 (common ratio 4) or -4, 16, -64 (common ratio -4).

Given:

The two positive numbers are $a$ and $b$.

The Arithmetic Mean (A.M.) of $a$ and $b$ is 10.

The Geometric Mean (G.M.) of $a$ and $b$ is 8.

---

To Find:

The values of the two positive numbers $a$ and $b$.

---

Solution:

The formula for the Arithmetic Mean (A.M.) of two numbers $a$ and $b$ is $\frac{a+b}{2}$.

According to the given information:

Multiply both sides of Equation (1) by 2:

---

The formula for the Geometric Mean (G.M.) of two positive numbers $a$ and $b$ is $\sqrt{ab}$.

According to the given information:

Square both sides of Equation (3):

---

We now have a system of two equations with two variables:

$a + b = 20$

$ab = 64$

We can solve this system. From Equation (2), express $b$ in terms of $a$:

$b = 20 - a$

Substitute this expression for $b$ into Equation (4):

$a(20 - a) = 64$

Expand the left side:

$20a - a^2 = 64$

Rearrange the terms to form a quadratic equation:

$a^2 - 20a + 64 = 0$

---

Solve the quadratic equation for $a$. We can factor the quadratic expression. We look for two numbers that multiply to 64 and add up to -20. These numbers are -4 and -16.

$(a - 4)(a - 16) = 0$

This gives two possible values for $a$:

$a - 4 = 0 \implies a = 4$

$a - 16 = 0 \implies a = 16$

---

If $a = 4$, substitute this value back into Equation (2) to find $b$:

$4 + b = 20$

$b = 20 - 4 = 16$

If $a = 16$, substitute this value back into Equation (2) to find $b$:

$16 + b = 20$

$b = 20 - 16 = 4$

---

In both cases, the pair of numbers is $\{4, 16\}$. The problem asks for "the numbers", which are 4 and 16.

We verify that these numbers are positive, which they are.

---

Answer:

The two positive numbers are 4 and 16.

Given:

The geometric progression (G.P.) is $\frac{5}{2}$, $\frac{5}{4}$, $\frac{5}{8}$, ...

---

To Find:

The 20^th term and the n^th term of the given G.P.

---

Solution:

The given sequence is $\frac{5}{2}$, $\frac{5}{4}$, $\frac{5}{8}$, ...

This is a G.P. with the first term $a = \frac{5}{2}$.

The common ratio, $r$, is found by dividing any term by its preceding term:

$r = \frac{\frac{5}{4}}{\frac{5}{2}} = \frac{5}{4} \times \frac{2}{5} = \frac{10}{20} = \frac{1}{2}$.

The common ratio is $r = \frac{1}{2}$.

---

The formula for the n^th term of a G.P. with first term $a$ and common ratio $r$ is $a_n = ar^{n-1}$.

---

To find the n^th term ($a_n$), substitute $a = \frac{5}{2}$ and $r = \frac{1}{2}$ into the formula:

$a_n = \frac{5}{2} \left(\frac{1}{2}\right)^{n-1}$

$a_n = \frac{5}{2^1} \times \frac{1^{n-1}}{2^{n-1}}$

$a_n = \frac{5}{2^1} \times \frac{1}{2^{n-1}}$

$a_n = \frac{5}{2^{1 + (n-1)}}$

$a_n = \frac{5}{2^n}$

---

To find the 20^th term ($a_{20}$), substitute $n=20$ into the formula for $a_n$:

$a_{20} = \frac{5}{2^{20}}$

---

Answer:

The n^th term of the G.P. is $a_n = \frac{5}{2^n}$.

The 20^th term of the G.P. is $a_{20} = \frac{5}{2^{20}}$.

Given:

The 8^th term of a G.P. is $a_8 = 192$.

The common ratio is $r = 2$.

---

To Find:

The 12^th term of the G.P., $a_{12}$.

---

Solution:

Let the first term of the G.P. be $a$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Using the given 8^th term ($n=8$) and common ratio ($r=2$):

$a_8 = ar^{8-1} = ar^7$

Substitute the value of $a_8$:

$192 = a(2)^7$

Calculate $2^7$:

$2^7 = 128$

So, $192 = a \times 128$

Solve for $a$:

$a = \frac{192}{128}$

Simplify the fraction:

$a = \frac{\cancel{192}^{3}}{\cancel{128}_{2}}$

$a = \frac{3}{2}$

The first term is $a = \frac{3}{2}$ and the common ratio is $r = 2$.

Now, find the 12^th term ($a_{12}$) using the formula $a_n = ar^{n-1}$ with $n=12$:

$a_{12} = ar^{12-1}$

$a_{12} = a r^{11}$

Substitute the values of $a$ and $r$:

$a_{12} = \frac{3}{2} \times 2^{11}$

$a_{12} = \frac{3}{2^1} \times 2^{11}$

$a_{12} = 3 \times 2^{11-1}$

$a_{12} = 3 \times 2^{10}$

Calculate $2^{10}$:

$2^{10} = 1024$

Substitute the value of $2^{10}$:

$a_{12} = 3 \times 1024$

$a_{12} = 3072$

---

Answer:

The 12^th term of the G.P. is 3072.

Given:

The 5^th term of a G.P. is $a_5 = p$.

The 8^th term of the same G.P. is $a_8 = q$.

The 11^th term of the same G.P. is $a_{11} = s$.

---

To Show:

$q^2 = ps$

---

Proof:

Let the first term of the G.P. be $a$ and the common ratio be $r$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Using this formula, we can express the given terms:

---

Consider the product $ps$ from Equations (1) and (3):

$ps = (ar^4)(ar^{10})$

$ps = a \cdot a \cdot r^4 \cdot r^{10}$

Using the exponent rule $x^m \cdot x^n = x^{m+n}$:

---

Now consider $q^2$ from Equation (2):

$q = ar^7$

$q^2 = (ar^7)^2$

Using the exponent rule $(xy)^m = x^m y^m$ and $(x^m)^n = x^{mn}$:

---

Comparing Equation (4) and Equation (5), we see that $ps = a^2 r^{14}$ and $q^2 = a^2 r^{14}$.

Therefore, $q^2 = ps$.

---

Answer:

Hence, it is shown that $q^2 = ps$.

Given:

The first term of the G.P. is $a_1 = -3$.

The 4^th term is the square of the second term, i.e., $a_4 = (a_2)^2$.

---

To Find:

The 7^th term of the G.P., $a_7$.

---

Solution:

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio.

We are given the first term $a_1 = a = -3$.

The second term is $a_2 = ar^{2-1} = ar = -3r$.

The fourth term is $a_4 = ar^{4-1} = ar^3 = -3r^3$.

---

According to the given condition, the 4^th term is the square of the second term:

Substitute the expressions for $a_2$ and $a_4$ into Equation (1):

$-3r^3 = (-3r)^2$

$-3r^3 = (-3)^2 r^2$

$-3r^3 = 9r^2$

Move all terms to one side to form an equation equal to zero:

$9r^2 + 3r^3 = 0$

Factor out the common term $3r^2$:

$3r^2(3 + r) = 0$

This equation holds true if either $3r^2 = 0$ or $3 + r = 0$.

Case 1: $3r^2 = 0 \implies r^2 = 0 \implies r = 0$.

Case 2: $3 + r = 0 \implies r = -3$.

---

For a standard geometric progression, the common ratio $r$ must be non-zero. If $r=0$ and $a \neq 0$, the sequence is $a, 0, 0, 0, \dots$. The ratio of terms after the first term becomes undefined ($0/0$). Therefore, we consider the valid common ratio $r = -3$.

---

The common ratio is $r = -3$. The first term is $a = -3$.

We need to find the 7^th term ($a_7$). Using the formula $a_n = ar^{n-1}$ with $n=7$:

$a_7 = ar^{7-1}$

$a_7 = ar^6$

Substitute the values of $a = -3$ and $r = -3$:

$a_7 = (-3)(-3)^6$

Calculate $(-3)^6$: $(-3)^6 = (-3 \times -3) \times (-3 \times -3) \times (-3 \times -3) = 9 \times 9 \times 9 = 729$.

$a_7 = (-3)(729)$

Perform the multiplication:

$a_7 = -2187$

---

Answer:

The 7^th term of the G.P. is -2187.

(a) For the G.P. 2, $2\sqrt{2}$, 4, … is 128?

---

Given:

The geometric progression (G.P.) is 2, $2\sqrt{2}$, 4, ...

A term in the G.P. is 128.

---

To Find:

The term number of the term 128.

---

Solution:

The given sequence is 2, $2\sqrt{2}$, 4, ...

This is a G.P. with the first term $a = 2$.

The common ratio, $r$, is found by dividing any term by its preceding term:

$r = \frac{2\sqrt{2}}{2} = \sqrt{2}$.

Let the n^th term of the G.P. be $a_n$. We are given that $a_n = 128$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Substitute the known values into the formula:

$128 = 2 (\sqrt{2})^{n-1}$

Divide both sides by 2:

$\frac{128}{2} = (\sqrt{2})^{n-1}$

$64 = (\sqrt{2})^{n-1}$

Express both sides with the same base, 2. We know that $64 = 2^6$ and $\sqrt{2} = 2^{1/2}$.

$2^6 = (2^{1/2})^{n-1}$

Using the exponent rule $(x^m)^n = x^{mn}$:

$2^6 = 2^{\frac{n-1}{2}}$

Since the bases are equal, the exponents must be equal:

$6 = \frac{n-1}{2}$

Multiply both sides by 2:

$12 = n - 1$

Add 1 to both sides to solve for $n$:

$n = 12 + 1$

$n = 13$

---

Answer:

The 13^th term of the G.P. is 128.

---

(b) For the G.P. $\sqrt{3}$, 3, 3$\sqrt{3}$ , … is 729?

---

Given:

The geometric progression (G.P.) is $\sqrt{3}$, 3, $3\sqrt{3}$, ...

A term in the G.P. is 729.

---

To Find:

The term number of the term 729.

---

Solution:

The given sequence is $\sqrt{3}$, 3, $3\sqrt{3}$, ...

This is a G.P. with the first term $a = \sqrt{3}$.

The common ratio, $r$, is found by dividing any term by its preceding term:

$r = \frac{3}{\sqrt{3}} = \sqrt{3}$.

Let the n^th term of the G.P. be $a_n$. We are given that $a_n = 729$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Substitute the known values into the formula:

$729 = \sqrt{3} (\sqrt{3})^{n-1}$

Using the exponent rule $x^m \cdot x^n = x^{m+n}$ ($ \sqrt{3} = (\sqrt{3})^1$):

$729 = (\sqrt{3})^{1 + (n-1)}$

$729 = (\sqrt{3})^n$

Express both sides with the same base, 3. We know that $729 = 3^6$ and $\sqrt{3} = 3^{1/2}$.

$3^6 = (3^{1/2})^n$

Using the exponent rule $(x^m)^n = x^{mn}$:

$3^6 = 3^{\frac{n}{2}}$

Since the bases are equal, the exponents must be equal:

$6 = \frac{n}{2}$

Multiply both sides by 2:

$n = 12$

---

Answer:

The 12^th term of the G.P. is 729.

---

(c) For the G.P. $\frac{1}{3}$ , $\frac{1}{9}$ , $\frac{1}{27}$, … is $\frac{1}{19683}$?

---

Given:

The geometric progression (G.P.) is $\frac{1}{3}$, $\frac{1}{9}$, $\frac{1}{27}$, ...

A term in the G.P. is $\frac{1}{19683}$.

---

To Find:

The term number of the term $\frac{1}{19683}$.

---

Solution:

The given sequence is $\frac{1}{3}$, $\frac{1}{9}$, $\frac{1}{27}$, ...

This is a G.P. with the first term $a = \frac{1}{3}$.

The common ratio, $r$, is found by dividing any term by its preceding term:

$r = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{9} \times 3 = \frac{3}{9} = \frac{1}{3}$.

Let the n^th term of the G.P. be $a_n$. We are given that $a_n = \frac{1}{19683}$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Substitute the known values into the formula:

$\frac{1}{19683} = \frac{1}{3} \left(\frac{1}{3}\right)^{n-1}$

Using the exponent rule $x^m \cdot x^n = x^{m+n}$:

$\frac{1}{19683} = \left(\frac{1}{3}\right)^{1 + (n-1)}$

$\frac{1}{19683} = \left(\frac{1}{3}\right)^n$

To find $n$, we need to determine what power of 3 gives 19683 in the denominator. This is equivalent to finding $n$ such that $3^n = 19683$.

Let's calculate powers of 3:

$3^1 = 3$

$3^2 = 9$

$3^3 = 27$

$3^4 = 81$

$3^5 = 243$

$3^6 = 729$

$3^7 = 2187$

$3^8 = 6561$

$3^9 = 19683$

So, $19683 = 3^9$.

The equation is $\frac{1}{3^9} = \left(\frac{1}{3}\right)^n$, which means $\left(\frac{1}{3}\right)^9 = \left(\frac{1}{3}\right)^n$.

Since the bases are equal, the exponents must be equal:

$n = 9$

---

Answer:

The 9^th term of the G.P. is $\frac{1}{19683}$.

Given:

The three numbers $-\frac{2}{7}$, $x$, $-\frac{7}{2}$ are in Geometric Progression (G.P.).

---

To Find:

The values of $x$ for which the given numbers form a G.P.

---

Solution:

For three numbers, $a$, $b$, and $c$ to be in G.P., the ratio of consecutive terms must be the same. That is, $\frac{b}{a} = \frac{c}{b}$.

This property can be written as $b^2 = ac$. The middle term $b$ is the geometric mean of the other two terms $a$ and $c$.

---

In this case, the three terms are $a = -\frac{2}{7}$, $b = x$, and $c = -\frac{7}{2}$.

Using the property $b^2 = ac$, we have:

$x^2 = \left(-\frac{2}{7}\right) \times \left(-\frac{7}{2}\right)$

Multiply the fractions on the right side:

$x^2 = \left(\frac{-2 \times -7}{7 \times 2}\right)$

$x^2 = \frac{14}{14}$

$x^2 = 1$

To find the values of $x$, take the square root of both sides:

$x = \pm \sqrt{1}$

$x = \pm 1$

So, there are two possible values for $x$: $1$ and $-1$.

---

We can verify these values:

If $x = 1$, the sequence is $-\frac{2}{7}$, 1, $-\frac{7}{2}$.

The ratio of the second term to the first term is $\frac{1}{-2/7} = 1 \times (-\frac{7}{2}) = -\frac{7}{2}$.

The ratio of the third term to the second term is $\frac{-7/2}{1} = -\frac{7}{2}$.

Since the ratios are equal ($-\frac{7}{2}$), the sequence is a G.P. with $r = -\frac{7}{2}$.

---

If $x = -1$, the sequence is $-\frac{2}{7}$, -1, $-\frac{7}{2}$.

The ratio of the second term to the first term is $\frac{-1}{-2/7} = -1 \times (-\frac{7}{2}) = \frac{7}{2}$.

The ratio of the third term to the second term is $\frac{-7/2}{-1} = \frac{7}{2}$.

Since the ratios are equal ($\frac{7}{2}$), the sequence is a G.P. with $r = \frac{7}{2}$.

---

Answer:

The values of x for which the numbers $-\frac{2}{7}$, x, $-\frac{7}{2}$ are in G.P. are $x = 1$ and $x = -1$.

Given:

The geometric progression (G.P.) is 0.15, 0.015, 0.0015, ...

The number of terms is $n = 20$.

---

To Find:

The sum of the first 20 terms of the given G.P.

---

Solution:

The given sequence is 0.15, 0.015, 0.0015, ...

The first term is $a = 0.15$.

The common ratio, $r$, is found by dividing a term by its preceding term:

$r = \frac{0.015}{0.15} = \frac{15/1000}{15/100} = \frac{15}{1000} \times \frac{100}{15} = \frac{1}{10} = 0.1$.

Since $|r| = |0.1| < 1$, the formula for the sum of the first $n$ terms of a G.P. is $S_n = \frac{a(1 - r^n)}{1 - r}$.

We need to find the sum of the first 20 terms, so $n=20$. Substitute the values of $a$, $r$, and $n$ into the formula:

$S_{20} = \frac{0.15(1 - (0.1)^{20})}{1 - 0.1}$

Simplify the denominator:

$1 - 0.1 = 0.9$

So the expression becomes:

$S_{20} = \frac{0.15(1 - (0.1)^{20})}{0.9}$

Simplify the fraction $\frac{0.15}{0.9}$:

$\frac{0.15}{0.9} = \frac{15/100}{9/10} = \frac{15}{100} \times \frac{10}{9} = \frac{15 \times 10}{100 \times 9} = \frac{150}{900} = \frac{15}{90} = \frac{1}{6}$.

The expression for $S_{20}$ is:

$S_{20} = \frac{1}{6} (1 - (0.1)^{20})$

This can also be written as:

$S_{20} = \frac{1}{6} (1 - 10^{-20})$

Or, distributing the $\frac{1}{6}$:

$S_{20} = \frac{1}{6} - \frac{1}{6} \times 10^{-20}$

---

Answer:

The sum of the first 20 terms of the G.P. is $\frac{1}{6} (1 - (0.1)^{20})$ or $\frac{1}{6}(1 - 10^{-20})$.

Given:

The geometric progression (G.P.) is $\sqrt{7}$, $\sqrt{21}$, $3\sqrt{7}$, ...

The number of terms is $n$.

---

To Find:

The sum of the first n terms of the given G.P.

---

Solution:

The given sequence is $\sqrt{7}$, $\sqrt{21}$, $3\sqrt{7}$, ...

The first term is $a = \sqrt{7}$.

The common ratio, $r$, is found by dividing a term by its preceding term:

$r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{\frac{21}{7}} = \sqrt{3}$.

The common ratio is $r = \sqrt{3}$. Since $|r| = |\sqrt{3}| > 1$, the formula for the sum of the first $n$ terms of a G.P. is $S_n = \frac{a(r^n - 1)}{r - 1}$.

Substitute the values of $a$ and $r$ into the formula for $S_n$:

$S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)}{\sqrt{3} - 1}$

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} + 1$:

$S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$

$S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}$

$S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)(\sqrt{3} + 1)}{3 - 1}$

$S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)(\sqrt{3} + 1)}{2}$

We can also write $(\sqrt{3})^n$ as $(3^{1/2})^n = 3^{n/2}$.

$S_n = \frac{\sqrt{7}(3^{n/2} - 1)(\sqrt{3} + 1)}{2}$

---

Answer:

The sum of the first n terms of the G.P. is $S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)(\sqrt{3} + 1)}{2}$.

Given:

The geometric progression (G.P.) is 1, – a, a², – a³, ...

The number of terms is $n$.

Condition: $a \neq -1$.

---

To Find:

The sum of the first n terms of the given G.P.

---

Solution:

The given sequence is 1, – a, a², – a³, ...

The first term is $A = 1$.

The common ratio, $r$, is found by dividing a term by its preceding term:

$r = \frac{-\text{a}}{1} = -\text{a}$.

The common ratio is $r = -a$.

The condition $a \neq -1$ implies $r = -a \neq -(-1) = 1$. Thus, $r \neq 1$.

The formula for the sum of the first n terms of a G.P. with first term $A$ and common ratio $r \neq 1$ is $S_n = \frac{A(1 - r^n)}{1 - r}$.

Substitute the values $A=1$ and $r=-a$ into the formula for $S_n$:

$S_n = \frac{1(1 - (-a)^n)}{1 - (-a)}$

$S_n = \frac{1 - (-a)^n}{1 + a}$

We can write $(-a)^n$ as $(-1 \times a)^n = (-1)^n a^n$.

So the sum is:

$S_n = \frac{1 - (-1)^n a^n}{1 + a}$

---

Answer:

The sum of the first n terms of the G.P. is $S_n = \frac{1 - (-1)^n a^n}{1 + a}$.

Given:

The geometric progression (G.P.) is x³, x⁵, x⁷, ...

The number of terms is $n$.

Condition: $x \neq \pm 1$.

---

To Find:

The sum of the first n terms of the given G.P.

---

Solution:

The given sequence is x³, x⁵, x⁷, ...

The first term is $a = x^3$.

The common ratio, $r$, is found by dividing a term by its preceding term:

$r = \frac{x^5}{x^3} = x^{5-3} = x^2$.

We can verify this with the next pair of terms: $r = \frac{x^7}{x^5} = x^{7-5} = x^2$.

The common ratio is $r = x^2$.

The condition $x \neq \pm 1$ implies $x^2 \neq (\pm 1)^2$, so $x^2 \neq 1$. Thus, the common ratio $r \neq 1$.

The formula for the sum of the first n terms of a G.P. with first term $a$ and common ratio $r \neq 1$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.

Substitute the values $a=x^3$ and $r=x^2$ into the formula for $S_n$:

$S_n = \frac{x^3((x^2)^n - 1)}{x^2 - 1}$

Using the exponent rule $(p^m)^n = p^{mn}$:

$(x^2)^n = x^{2 \times n} = x^{2n}$.

So the sum is:

$S_n = \frac{x^3(x^{2n} - 1)}{x^2 - 1}$

---

Answer:

The sum of the first n terms of the G.P. is $S_n = \frac{x^3(x^{2n} - 1)}{x^2 - 1}$.

Given:

The summation expression $\sum\limits_{k=1}^{11} (2 + 3^k)$.

---

To Evaluate:

The value of the given summation.

---

Solution:

The summation can be split into two parts using the property of summation $\sum (a_k + b_k) = \sum a_k + \sum b_k$:

$\sum\limits_{k=1}^{11} (2 + 3^k) = \sum\limits_{k=1}^{11} 2 + \sum\limits_{k=1}^{11} 3^k$

---

Consider the first part of the summation: $\sum\limits_{k=1}^{11} 2$.

This is the sum of a constant term (2) repeated 11 times:

$\sum\limits_{k=1}^{11} 2 = 2 + 2 + \dots + 2$ (11 times)

$\sum\limits_{k=1}^{11} 2 = 11 \times 2 = 22$

---

Consider the second part of the summation: $\sum\limits_{k=1}^{11} 3^k$.

This expands to $3^1 + 3^2 + 3^3 + \dots + 3^{11}$.

This is a finite geometric series with:

First term, $a = 3^1 = 3$.

Common ratio, $r = \frac{3^2}{3^1} = 3$.

Number of terms, $n = 11$.

Since the common ratio $r = 3 \neq 1$, the formula for the sum of the first $n$ terms of a G.P. is $S_n = \frac{a(r^n - 1)}{r - 1}$.

Substitute the values $a=3$, $r=3$, and $n=11$ into the formula for the sum of this G.P.:

$\sum\limits_{k=1}^{11} 3^k = \frac{3(3^{11} - 1)}{3 - 1}$

$= \frac{3(3^{11} - 1)}{2}$

Calculate $3^{11}$:

$3^{11} = 177147$

Substitute the value of $3^{11}$:

$= \frac{3(177147 - 1)}{2}$

$= \frac{3(177146)}{2}$

$= 3 \times \frac{177146}{2}$

$= 3 \times 88573$

$= 265719$

---

Now, add the sums of the two parts:

$\sum\limits_{k=1}^{11} (2 + 3^k) = \sum\limits_{k=1}^{11} 2 + \sum\limits_{k=1}^{11} 3^k$

$= 22 + 265719$

$= 265741$

---

Answer:

The value of the summation $\sum\limits_{k=1}^{11} (2 + 3^k)$ is 265741.

Given:

The sum of the first three terms of a G.P. is $\frac{39}{10}$.

The product of the first three terms of the G.P. is 1.

---

To Find:

The common ratio and the first three terms of the G.P.

---

Solution:

Let the first three terms of the G.P. be $\frac{a}{r}$, $a$, and $ar$, where $a$ is the first term and $r$ is the common ratio. (Note: Using this representation simplifies the product calculation. The terms in standard notation would be $a_1 = \frac{a}{r}$, $a_2 = a$, $a_3 = ar$).

---

According to the given information, the product of the first three terms is 1:

$\left(\frac{a}{r}\right) \times (a) \times (ar) = 1$

$a^{1} \times a^{1} \times a^{1} \times r^{-1} \times r^{1} = 1$

$a^{1+1+1} \times r^{-1+1} = 1$

$a^3 \times r^0 = 1$

Since $r^0 = 1$ (assuming $r \neq 0$), we have:

Taking the cube root of both sides, we get $a = 1$.

So, the second term of the G.P. is 1. The three terms are $\frac{1}{r}$, $1$, and $r$.

---

According to the given information, the sum of the first three terms is $\frac{39}{10}$:

$\frac{1}{r} + 1 + r = \frac{39}{10}$

Combine the terms on the left side by finding a common denominator $r$:

$\frac{1}{r} + \frac{r}{r} + \frac{r^2}{r} = \frac{39}{10}$

$\frac{1 + r + r^2}{r} = \frac{39}{10}$

Cross-multiply:

$10(1 + r + r^2) = 39r$

Expand the left side:

$10 + 10r + 10r^2 = 39r$

Move all terms to one side to form a quadratic equation in $r$:

$10r^2 + 10r - 39r + 10 = 0$

---

We can solve this quadratic equation by factoring. We look for two numbers that multiply to $10 \times 10 = 100$ and add up to -29. These numbers are -4 and -25.

Rewrite the middle term:

$10r^2 - 4r - 25r + 10 = 0$

Group the terms and factor by grouping:

$(10r^2 - 4r) + (-25r + 10) = 0$

$2r(5r - 2) - 5(5r - 2) = 0$

Factor out the common binomial $(5r - 2)$:

$(5r - 2)(2r - 5) = 0$

This gives two possible values for the common ratio $r$:

Case 1: $5r - 2 = 0 \implies 5r = 2 \implies r = \frac{2}{5}$

Case 2: $2r - 5 = 0 \implies 2r = 5 \implies r = \frac{5}{2}$

---

Now, we find the three terms for each case, using $a = 1$ and the respective common ratio $r$. The terms are $\frac{a}{r}, a, ar$, which simplifies to $\frac{1}{r}, 1, r$ since $a=1$.

---

Case 1: Common ratio $r = \frac{2}{5}$

The terms are:

First term: $\frac{1}{r} = \frac{1}{2/5} = 1 \times \frac{5}{2} = \frac{5}{2}$

Second term: $1$

Third term: $r = \frac{2}{5}$

The terms are $\frac{5}{2}, 1, \frac{2}{5}$.

Check sum: $\frac{5}{2} + 1 + \frac{2}{5} = \frac{25}{10} + \frac{10}{10} + \frac{4}{10} = \frac{25+10+4}{10} = \frac{39}{10}$. (Correct)

Check product: $\frac{5}{2} \times 1 \times \frac{2}{5} = 1$. (Correct)

---

Case 2: Common ratio $r = \frac{5}{2}$

The terms are:

First term: $\frac{1}{r} = \frac{1}{5/2} = 1 \times \frac{2}{5} = \frac{2}{5}$

Second term: $1$

Third term: $r = \frac{5}{2}$

The terms are $\frac{2}{5}, 1, \frac{5}{2}$.

Check sum: $\frac{2}{5} + 1 + \frac{5}{2} = \frac{4}{10} + \frac{10}{10} + \frac{25}{10} = \frac{4+10+25}{10} = \frac{39}{10}$. (Correct)

Check product: $\frac{2}{5} \times 1 \times \frac{5}{2} = 1$. (Correct)

---

Both cases satisfy the given conditions.

---

Answer:

The common ratio is $\frac{2}{5}$ or $\frac{5}{2}$.

The terms of the G.P. are $\frac{5}{2}, 1, \frac{2}{5}$ (when $r = \frac{2}{5}$) or $\frac{2}{5}, 1, \frac{5}{2}$ (when $r = \frac{5}{2}$).

Given:

The geometric progression (G.P.) is 3, $3^2$, $3^3$, ... which is 3, 9, 27, ...

The sum of the first n terms is $S_n = 120$.

---

To Find:

The number of terms, $n$, needed to give the sum 120.

---

Solution:

The given sequence is 3, 9, 27, ...

This is a G.P. with the first term $a = 3$.

The common ratio, $r$, is found by dividing any term by its preceding term:

$r = \frac{9}{3} = 3$.

The common ratio is $r = 3$. Since $r \neq 1$, the formula for the sum of the first $n$ terms of a G.P. is $S_n = \frac{a(r^n - 1)}{r - 1}$.

We are given $S_n = 120$. Substitute the values of $a$, $r$, and $S_n$ into the formula:

$120 = \frac{3(3^n - 1)}{3 - 1}$

Simplify the denominator:

$3 - 1 = 2$

So the equation becomes:

$120 = \frac{3(3^n - 1)}{2}$

Multiply both sides by 2:

$120 \times 2 = 3(3^n - 1)$

$240 = 3(3^n - 1)$

Divide both sides by 3:

$\frac{240}{3} = 3^n - 1$

$80 = 3^n - 1$

Add 1 to both sides to isolate the term with $n$:

$80 + 1 = 3^n$

$81 = 3^n$

We need to express 81 as a power of 3:

$81 = 3 \times 3 \times 3 \times 3 = 3^4$.

So, we have:

$3^4 = 3^n$

Since the bases are equal, the exponents must be equal:

$n = 4$

---

Answer:

4 terms of the G.P. are needed to give the sum 120.

Given:

The sum of the first three terms of a G.P. is $S_3 = 16$.

The sum of the next three terms (i.e., the 4^th, 5^th, and 6^th terms) is 128.

---

To Find:

The first term ($a$), the common ratio ($r$), and the sum to n terms ($S_n$) of the G.P.

---

Solution:

Let the first term of the G.P. be $a$ and the common ratio be $r$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

The formula for the sum of the first n terms of a G.P. is $S_n = \frac{a(r^n - 1)}{r - 1}$ (if $r \neq 1$).

---

The sum of the first three terms is $S_3 = a_1 + a_2 + a_3 = a + ar + ar^2$.

Using the sum formula:

---

The sum of the next three terms (4^th, 5^th, and 6^th) is $a_4 + a_5 + a_6$.

$a_4 = ar^{4-1} = ar^3$

$a_5 = ar^{5-1} = ar^4$

$a_6 = ar^{6-1} = ar^5$

The sum is $ar^3 + ar^4 + ar^5 = 128$.

Factor out $ar^3$ from this sum:

$ar^3(1 + r + r^2) = 128$.

Alternatively, we can express the sum of the next three terms as the sum of the first six terms minus the sum of the first three terms:

Sum of next three terms = $S_6 - S_3 = 128$.

The sum of the first six terms is $S_6 = \frac{a(r^6 - 1)}{r - 1}$.

So, $S_6 - S_3 = \frac{a(r^6 - 1)}{r - 1} - \frac{a(r^3 - 1)}{r - 1} = 128$

$\frac{a(r^6 - 1 - (r^3 - 1))}{r - 1} = 128$

$\frac{a(r^6 - r^3)}{r - 1} = 128$

Factor out $r^3$ from the numerator:

---

Now we have a system of two equations (1) and (2):

Equation (1): $\frac{a(r^3 - 1)}{r - 1} = 16$

Equation (2): $\frac{ar^3(r^3 - 1)}{r - 1} = 128$

Substitute the value of $\frac{a(r^3 - 1)}{r - 1}$ from Equation (1) into Equation (2):

$r^3 \left(\frac{a(r^3 - 1)}{r - 1}\right) = 128$

$r^3 (16) = 128$

Solve for $r^3$:

$r^3 = \frac{128}{16}$

$r^3 = 8$

Take the cube root of both sides to find $r$:

$r = \sqrt[3]{8}$

$r = 2$

---

Now that we have the common ratio $r=2$, substitute this value back into Equation (1) to find the first term $a$:

$\frac{a(2^3 - 1)}{2 - 1} = 16$

$\frac{a(8 - 1)}{1} = 16$

$\frac{a(7)}{1} = 16$

$7a = 16$

Solve for $a$:

$a = \frac{16}{7}$

---

The first term is $a = \frac{16}{7}$ and the common ratio is $r = 2$.

Finally, find the sum to n terms ($S_n$) using the formula $S_n = \frac{a(r^n - 1)}{r - 1}$:

$S_n = \frac{\frac{16}{7}(2^n - 1)}{2 - 1}$

$S_n = \frac{\frac{16}{7}(2^n - 1)}{1}$

$S_n = \frac{16}{7}(2^n - 1)$

---

Answer:

The first term of the G.P. is $a = \frac{16}{7}$.

The common ratio of the G.P. is $r = 2$.

The sum to n terms of the G.P. is $S_n = \frac{16}{7}(2^n - 1)$.

Given:

The first term of the G.P. is $a = 729$.

The 7^th term of the G.P. is $a_7 = 64$.

---

To Find:

The sum of the first 7 terms of the G.P., $S_7$.

---

Solution:

Let the first term of the G.P. be $a$ and the common ratio be $r$.

We are given $a = 729$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Using the given 7^th term ($n=7$):

$a_7 = ar^{7-1} = ar^6$

Substitute the values of $a_7$ and $a$:

$64 = 729 r^6$

Solve for $r^6$:

$r^6 = \frac{64}{729}$

To find $r$, take the 6^th root of both sides. Note that both 64 and 729 are perfect 6^th powers. $64 = 2^6$ and $729 = 3^6$.

$r^6 = \frac{2^6}{3^6} = \left(\frac{2}{3}\right)^6$

Taking the 6^th root, the common ratio can be positive or negative:

$r = \pm \frac{2}{3}$

---

We need to find the sum of the first 7 terms ($S_7$). The number of terms is $n=7$. The first term is $a=729$. The common ratio is $r$.

The formula for the sum of the first n terms of a G.P. is $S_n = \frac{a(1 - r^n)}{1 - r}$ (if $r \neq 1$).

Since both $r = \frac{2}{3}$ and $r = -\frac{2}{3}$ are not equal to 1, we use this formula.

---

Case 1: Common ratio $r = \frac{2}{3}$

$S_7 = \frac{729 \left(1 - \left(\frac{2}{3}\right)^7\right)}{1 - \frac{2}{3}}$

$S_7 = \frac{729 \left(1 - \frac{2^7}{3^7}\right)}{\frac{1}{3}}$

Calculate $2^7 = 128$ and $3^7 = 2187$:

$S_7 = \frac{729 \left(1 - \frac{128}{2187}\right)}{\frac{1}{3}}$

$S_7 = 729 \times 3 \left(\frac{2187}{2187} - \frac{128}{2187}\right)$

$S_7 = 2187 \left(\frac{2187 - 128}{2187}\right)$

$S_7 = 2187 \left(\frac{2059}{2187}\right)$

$S_7 = 2059$

---

Case 2: Common ratio $r = -\frac{2}{3}$

$S_7 = \frac{729 \left(1 - \left(-\frac{2}{3}\right)^7\right)}{1 - \left(-\frac{2}{3}\right)}$

Calculate $\left(-\frac{2}{3}\right)^7 = \frac{(-2)^7}{3^7} = \frac{-128}{2187} = -\frac{128}{2187}$.

The numerator is $1 - \left(-\frac{128}{2187}\right) = 1 + \frac{128}{2187} = \frac{2187 + 128}{2187} = \frac{2315}{2187}$.

The denominator is $1 - \left(-\frac{2}{3}\right) = 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3}$.

$S_7 = \frac{729 \left(\frac{2315}{2187}\right)}{\frac{5}{3}}$

$S_7 = 729 \times \frac{2315}{2187} \times \frac{3}{5}$

Note that $2187 = 3^7$ and $729 = 3^6$.

$S_7 = 3^6 \times \frac{2315}{3^7} \times \frac{3}{5}$

$S_7 = \frac{3^6 \times 2315 \times 3}{3^7 \times 5}$

$S_7 = \frac{3^7 \times 2315}{3^7 \times 5}$

$S_7 = \frac{2315}{5}$

$S_7 = 463$

---

The problem does not specify that the terms are positive, so both values of the common ratio are possible, leading to two possible sums for the first 7 terms.

---

Answer:

The common ratio is $\frac{2}{3}$ or $-\frac{2}{3}$.

The sum of the first 7 terms is 2059 (if $r = \frac{2}{3}$) or 463 (if $r = -\frac{2}{3}$).

Given:

The sum of the first two terms of a G.P. is $-4$.

The fifth term of the G.P. is 4 times the third term.

---

To Find:

The geometric progression (G.P.), which means finding the first term and the common ratio.

---

Solution:

Let the first term of the G.P. be $a$ and the common ratio be $r$.

The terms of the G.P. are $a, ar, ar^2, ar^3, ar^4, \dots, ar^{n-1}, \dots$

The formula for the n^th term is $a_n = ar^{n-1}$.

---

According to the first condition, the sum of the first two terms is $-4$:

$a_1 + a_2 = -4$

$a + ar = -4$

---

According to the second condition, the fifth term is 4 times the third term:

$a_5 = 4 \times a_3$

Using the formula for the n^th term:

$ar^{5-1} = 4 \times ar^{3-1}$

$ar^4 = 4ar^2$

Assuming the G.P. is non-trivial ($a \neq 0$ and $r \neq 0$), we can divide both sides by $ar^2$:

$\frac{ar^4}{ar^2} = \frac{4ar^2}{ar^2}$

$r^{4-2} = 4$

Taking the square root of both sides of Equation (2):

$r = \pm \sqrt{4}$

$r = \pm 2$

So, there are two possible values for the common ratio: $r = 2$ or $r = -2$.

---

Now, substitute these values of $r$ back into Equation (1) to find the corresponding value(s) of $a$.

Equation (1) is $a(1 + r) = -4$.

---

Case 1: Common ratio $r = 2$

Substitute $r=2$ into Equation (1):

$a(1 + 2) = -4$

$a(3) = -4$

$a = -\frac{4}{3}$

In this case, the first term is $a = -\frac{4}{3}$ and the common ratio is $r = 2$. The G.P. starts as $-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots$

---

Case 2: Common ratio $r = -2$

Substitute $r=-2$ into Equation (1):

$a(1 + (-2)) = -4$

$a(-1) = -4$

$a = 4$

In this case, the first term is $a = 4$ and the common ratio is $r = -2$. The G.P. starts as $4, -8, 16, \dots$

---

Both cases satisfy the given conditions. The problem asks for "a G.P.", implying at least one solution, and we found two.

---

Answer:

There are two possible geometric progressions that satisfy the given conditions:

1. First term $a = -\frac{4}{3}$ and common ratio $r = 2$. The G.P. is $-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, -\frac{32}{3}, -\frac{64}{3}, \dots$

2. First term $a = 4$ and common ratio $r = -2$. The G.P. is $4, -8, 16, -32, 64, \dots$

Given:

The 4^th term of a G.P. is $a_4 = x$.

The 10^th term of the same G.P. is $a_{10} = y$.

The 16^th term of the same G.P. is $a_{16} = z$.

---

To Prove:

x, y, z are in G.P.

---

Proof:

Let the first term of the G.P. be $a$ and the common ratio be $r$.

The formula for the n^th term of a G.P. is $a_n = ar^{n-1}$.

Using this formula, we can express the given terms:

---

For three numbers x, y, and z to be in G.P., the ratio of consecutive terms must be the same. That is, $\frac{y}{x} = \frac{z}{y}$, or equivalently, $y^2 = xz$ (assuming $x \neq 0$ and $z \neq 0$).

---

Consider the square of the middle term, $y^2$, from Equation (2):

$y = ar^9$

$y^2 = (ar^9)^2$

$y^2 = a^2 (r^9)^2$

$y^2 = a^2 r^{18}$

---

Consider the product of the first and third terms, $xz$, from Equations (1) and (3):

$xz = (ar^3)(ar^{15})$

$xz = a \cdot a \cdot r^3 \cdot r^{15}$

$xz = a^{1+1} r^{3+15}$

---

Comparing Equation (4) and Equation (5), we see that $y^2 = a^2 r^{18}$ and $xz = a^2 r^{18}$.

Therefore, $y^2 = xz$.

This is the condition for three numbers x, y, and z to be in G.P.

(Assuming $a \neq 0$ and $r \neq 0$, which are usually implicit for a G.P. unless otherwise specified. If $a=0$, all terms are 0, and $0, 0, 0$ is a G.P. If $r=1$, the sequence is $a, a, a$. If $a \neq 0$, this is a G.P. with $r=1$. If $r \neq 0, 1$ and $a \neq 0$, $x, y, z$ are non-zero, and $y^2=xz$ implies $\frac{y}{x} = \frac{z}{y}$).

---

Answer:

Hence, it is proved that x, y, z are in G.P.

Given:

The sequence is 8, 88, 888, 8888, ... to n terms.

---

To Find:

The sum of the first n terms of the given sequence.

---

Solution:

Let the sum of the first n terms be $S_n$.

The terms of the sequence are $a_1 = 8$, $a_2 = 88$, $a_3 = 888$, and so on.

The n^th term, $a_n$, is a number consisting of n digits, all equal to 8.

We can write each term by factoring out 8:

$a_1 = 8 = 8 \times 1$

$a_2 = 88 = 8 \times 11$

$a_3 = 888 = 8 \times 111$

$a_n = 8 \times \underbrace{11\dots1}_{\text{n times}}$

The number $\underbrace{11\dots1}_{\text{n times}}$ can be expressed as $\frac{10^n - 1}{9}$.

So, the n^th term is $a_n = 8 \times \frac{10^n - 1}{9} = \frac{8}{9}(10^n - 1)$.

---

The sum of the first n terms is $S_n = a_1 + a_2 + \dots + a_n = \sum_{k=1}^{n} a_k$.

$S_n = \sum\limits_{k=1}^{n} \frac{8}{9}(10^k - 1)$

$S_n = \frac{8}{9} \sum\limits_{k=1}^{n} (10^k - 1)$

Separate the summation:

$S_n = \frac{8}{9} \left( \sum\limits_{k=1}^{n} 10^k - \sum\limits_{k=1}^{n} 1 \right)$

---

The second summation is simply the sum of n ones:

$\sum\limits_{k=1}^{n} 1 = 1 + 1 + \dots + 1$ (n times) $= n$.

---

The first summation is the sum of a geometric series:

$\sum\limits_{k=1}^{n} 10^k = 10^1 + 10^2 + 10^3 + \dots + 10^n$.

This is a geometric progression with the first term $A = 10$, the common ratio $R = 10$, and the number of terms is $n$.

The sum of the first n terms of a G.P. is $S_n = \frac{A(R^n - 1)}{R - 1}$.

Sum of this G.P. $= \frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9}$.

---

Substitute these results back into the expression for $S_n$:

$S_n = \frac{8}{9} \left( \frac{10(10^n - 1)}{9} - n \right)$

This can be further simplified:

$S_n = \frac{8}{9} \times \frac{10(10^n - 1)}{9} - \frac{8}{9}n$

$S_n = \frac{80(10^n - 1)}{81} - \frac{8n}{9}$

---

Answer:

The sum of the first n terms of the sequence is $S_n = \frac{80(10^n - 1)}{81} - \frac{8n}{9}$.

Given:

Sequence 1: $A = \{2, 4, 8, 16, 32\}$

Sequence 2: $B = \{128, 32, 8, 2, \frac{1}{2}\}$

---

To Find:

The sum of the products of the corresponding terms of the two sequences.

---

Solution:

Let the terms of the first sequence be $a_k$ and the terms of the second sequence be $b_k$, for $k=1, 2, 3, 4, 5$.

The first sequence is 2, 4, 8, 16, 32.

This is a G.P. with first term $a = 2$ and common ratio $r_A = \frac{4}{2} = 2$.

The k^th term of the first sequence is $a_k = 2 \cdot 2^{k-1} = 2^k$.

---

The second sequence is 128, 32, 8, 2, $\frac{1}{2}$.

This is a G.P. with first term $b = 128$ and common ratio $r_B = \frac{32}{128} = \frac{1}{4}$.

The k^th term of the second sequence is $b_k = 128 \cdot \left(\frac{1}{4}\right)^{k-1}$.

---

We need to find the sum of the products of the corresponding terms, which is $\sum\limits_{k=1}^{5} a_k b_k$.

The k^th product term is $p_k = a_k \times b_k$.

$p_k = (2^k) \times \left(128 \cdot \left(\frac{1}{4}\right)^{k-1}\right)$

Express 128 as a power of 2: $128 = 2^7$.

Express $\frac{1}{4}$ as a power of 2: $\frac{1}{4} = 4^{-1} = (2^2)^{-1} = 2^{-2}$.

Substitute these into the expression for $p_k$:

$p_k = 2^k \times 2^7 \times (2^{-2})^{k-1}$

$p_k = 2^{k+7} \times 2^{-2(k-1)}$

$p_k = 2^{k+7} \times 2^{-2k+2}$

Using the exponent rule $x^m \cdot x^n = x^{m+n}$:

$p_k = 2^{(k+7) + (-2k+2)}$

$p_k = 2^{k+7-2k+2}$

$p_k = 2^{9-k}$

---

The sequence of products is $p_k = 2^{9-k}$ for $k=1, 2, 3, 4, 5$.

The terms of this product sequence are:

$p_1 = 2^{9-1} = 2^8 = 256$

$p_2 = 2^{9-2} = 2^7 = 128$

$p_3 = 2^{9-3} = 2^6 = 64$

$p_4 = 2^{9-4} = 2^5 = 32$

$p_5 = 2^{9-5} = 2^4 = 16$

The sequence of products is 256, 128, 64, 32, 16.

---

This sequence of products is a G.P. with first term $P = 256$ and common ratio $R = \frac{128}{256} = \frac{1}{2}$. The number of terms is $N=5$.

The sum of this G.P. is given by $S_N = \frac{P(1 - R^N)}{1 - R}$, since $|R| < 1$.

Sum $= \frac{256 \left(1 - \left(\frac{1}{2}\right)^5\right)}{1 - \frac{1}{2}}$

Sum $= \frac{256 \left(1 - \frac{1}{32}\right)}{\frac{1}{2}}$

Sum $= 256 \times 2 \times \left(\frac{32 - 1}{32}\right)$

Sum $= 512 \times \frac{31}{32}$

Cancel out the common factor:

Sum $= \frac{\cancel{512}^{16} \times 31}{\cancel{32}_{1}}$

Sum $= 16 \times 31$

Perform the multiplication:

Sum $= 496$

---

Answer:

The sum of the products of the corresponding terms is 496.

Given:

Two sequences which are geometric progressions:

Sequence 1: a, ar, ar², ..., ar^n-1

Sequence 2: A, AR, AR², ..., AR^n-1

---

To Show:

The sequence formed by the products of the corresponding terms is a G.P.

---

To Find:

The common ratio of the resulting G.P.

---

Proof:

Let the terms of the first sequence be $a_k$ for $k=1, 2, \dots, n$. The first term is $a$ and the common ratio is $r$. The k^th term is $a_k = ar^{k-1}$.

Let the terms of the second sequence be $b_k$ for $k=1, 2, \dots, n$. The first term is $A$ and the common ratio is $R$. The k^th term is $b_k = AR^{k-1}$.

---

Consider the sequence formed by the products of the corresponding terms. Let the terms of this new sequence be $c_k = a_k \times b_k$.

The k^th term of the product sequence is:

$c_k = a_k \times b_k = (ar^{k-1}) \times (AR^{k-1})$

$c_k = (aA) (rR)^{k-1}$

---

To show that this sequence is a G.P., we need to show that the ratio of any term to its preceding term is a constant value.

Consider the ratio of the (k+1)^th term to the k^th term:

$\frac{c_{k+1}}{c_k} = \frac{(aA)(rR)^{(k+1)-1}}{(aA)(rR)^{k-1}}$

$\frac{c_{k+1}}{c_k} = \frac{(aA)(rR)^k}{(aA)(rR)^{k-1}}$

Assuming $aA \neq 0$ (for a non-trivial product sequence), we can cancel $(aA)$ from the numerator and denominator:

$\frac{c_{k+1}}{c_k} = \frac{(rR)^k}{(rR)^{k-1}}$

Using the exponent rule $\frac{x^m}{x^n} = x^{m-n}$:

$\frac{c_{k+1}}{c_k} = (rR)^{k - (k-1)}$

$\frac{c_{k+1}}{c_k} = (rR)^{k - k + 1}$

$\frac{c_{k+1}}{c_k} = (rR)^1 = rR$

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The ratio of consecutive terms in the sequence $\{c_k\}$ is $rR$, which is a constant value (since $r$ and $R$ are constants). This means the sequence of products is a geometric progression.

The first term of this product G.P. is $c_1 = a_1 b_1 = aA$.

The common ratio of this product G.P. is the constant ratio we found, which is $rR$.

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Answer:

The sequence formed by the products of the corresponding terms is $(aA)(rR)^0, (aA)(rR)^1, \dots, (aA)(rR)^{n-1}$, which is a G.P.

The common ratio of this G.P. is $rR$.

Given:

We are looking for four numbers that form a geometric progression (G.P.).

Let the four terms of the G.P. be $a_1, a_2, a_3, a_4$. Let the first term be $a$ and the common ratio be $r$. The terms are $a, ar, ar^2, ar^3$.

Condition 1: The third term is greater than the first term by 9. ($a_3 = a_1 + 9$).

Condition 2: The second term is greater than the 4^th term by 18. ($a_2 = a_4 + 18$).

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To Find:

The four numbers forming the G.P.

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Solution:

Using the terms in terms of $a$ and $r$, we write the given conditions as equations:

Condition 1: $a_3 = a_1 + 9$

$ar^2 = a + 9$

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Condition 2: $a_2 = a_4 + 18$

$ar = ar^3 + 18$

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We have a system of two equations with two variables $a$ and $r$:

(1) $a(r^2 - 1) = 9$

(2) $ar(1 - r^2) = 18$

From Equation (1), we can write $a(r^2 - 1) = 9$.

From Equation (2), we can rewrite $ar(1 - r^2) = ar(-(r^2 - 1)) = -ar(r^2 - 1) = 18$.

So, $-ar(r^2 - 1) = 18$.

Substitute the value of $a(r^2 - 1)$ from Equation (1) into this equation:

$-r \times [a(r^2 - 1)] = 18$

$-r \times [9] = 18$

$-9r = 18$

Divide both sides by -9:

$r = \frac{18}{-9}$

$r = -2$

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Now that we have the common ratio $r = -2$, substitute this value back into Equation (1) to find the first term $a$:

$a(r^2 - 1) = 9$

$a((-2)^2 - 1) = 9$

$a(4 - 1) = 9$

$a(3) = 9$

Divide both sides by 3:

$a = 3$

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The first term is $a = 3$ and the common ratio is $r = -2$.

The four numbers forming the G.P. are:

$a_1 = a = 3$

$a_2 = ar = 3 \times (-2) = -6$

$a_3 = ar^2 = 3 \times (-2)^2 = 3 \times 4 = 12$

$a_4 = ar^3 = 3 \times (-2)^3 = 3 \times (-8) = -24$

The four numbers are 3, -6, 12, -24.

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Verify the given conditions:

Condition 1: The third term is greater than the first term by 9.

$a_3 - a_1 = 12 - 3 = 9$. (Correct)

Condition 2: The second term is greater than the 4^th term by 18.

$a_2 - a_4 = -6 - (-24) = -6 + 24 = 18$. (Correct)

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Answer:

The four numbers forming the geometric progression are 3, -6, 12, and -24.

Let the first term of the G.P. be $A$ and the common ratio be $R$. The general term of a G.P. is given by $T_n = AR^{n-1}$.

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Given:

The p-th term of the G.P. is a.

The q-th term of the G.P. is b.

The r-th term of the G.P. is c.

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To Prove:

$a^{q – r} b^{r – p} c^{p – q} = 1$

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Proof:

Consider the Left Hand Side (L.H.S.) of the equation to be proved:

L.H.S. $= a^{q-r} b^{r-p} c^{p-q}$

Substitute the expressions for a, b, and c from equations (i), (ii), and (iii) into the L.H.S.:

L.H.S. $= (AR^{p-1})^{q-r} (AR^{q-1})^{r-p} (AR^{r-1})^{p-q}$

Using the exponent rules $(xy)^n = x^n y^n$ and $(x^m)^n = x^{mn}$:

L.H.S. $= A^{q-r} (R^{p-1})^{(q-r)} \cdot A^{r-p} (R^{q-1})^{(r-p)} \cdot A^{p-q} (R^{r-1})^{(p-q)}$

L.H.S. $= A^{q-r} R^{(p-1)(q-r)} \cdot A^{r-p} R^{(q-1)(r-p)} \cdot A^{p-q} R^{(r-1)(p-q)}$

Group the terms with the same base using the rule $x^m x^n = x^{m+n}$:

L.H.S. $= A^{(q-r) + (r-p) + (p-q)} \cdot R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}$

Let's simplify the exponent of A:

Exponent of A $= (q-r) + (r-p) + (p-q)$

$= q - r + r - p + p - q$

$= (q - q) + (-r + r) + (-p + p) = 0$

Thus, the term with base A is $A^0$.

Now, let's simplify the exponent of R:

Exponent of R $= (p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)$

Expanding the products:

$(p-1)(q-r) = pq - pr - q + r$

$(q-1)(r-p) = qr - qp - r + p$

$(r-1)(p-q) = pr - rq - p + q$

Sum of exponents of R $= (pq - pr - q + r) + (qr - pq - r + p) + (pr - qr - p + q)$

$= pq - pr - q + r + qr - pq - r + p + pr - qr - p + q$

$= (pq - pq) + (-pr + pr) + (-q + q) + (r - r) + (qr - qr) + (p - p) = 0$

Thus, the term with base R is $R^0$.

Substituting the simplified exponents back into the expression for L.H.S.:

L.H.S. $= A^0 \cdot R^0$

Assuming A and R are non-zero (as a, b, c are terms of a G.P., which implies the first term and common ratio are typically non-zero for distinct values), any non-zero number raised to the power of 0 is 1.

L.H.S. $= 1 \cdot 1 = 1$

L.H.S. = 1

This is equal to the Right Hand Side (R.H.S.).

L.H.S. = R.H.S.

Therefore, $a^{q – r} b^{r – p} c^{p – q} = 1$.

Hence Proved.

Let the first term of the G.P. be $A$ and the common ratio be $R$.

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Given:

First term of the G.P. is $a$. So, $A = a$.

The n-th term of the G.P. is $b$. The formula for the n-th term of a G.P. is $T_n = AR^{n-1}$.

Substitute $A=a$ into equation (i):

P is the product of the first n terms. The terms are $a, ar, ar^2, \dots, ar^{n-1}$.

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To Prove:

$P^2 = (ab)^n$

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Proof:

The product of the first n terms P is:

$P = a \cdot (ar) \cdot (ar^2) \cdot \dots \cdot (ar^{n-1})$

Group the terms with the same base:

$P = (a \cdot a \cdot \dots \cdot a) \cdot (r^0 \cdot r^1 \cdot r^2 \cdot \dots \cdot r^{n-1})$

There are n terms, so 'a' is multiplied n times. The exponents of 'r' form an arithmetic progression $0, 1, 2, \dots, n-1$.

Sum of exponents of r $= 0 + 1 + 2 + \dots + (n-1)$

This is the sum of the first $n-1$ non-negative integers, which is the same as the sum of the first $n-1$ positive integers, given by the formula $\frac{k(k+1)}{2}$ where $k = n-1$.

Sum of exponents of r $= \frac{(n-1)((n-1)+1)}{2} = \frac{(n-1)n}{2}$

So, the expression for P becomes:

Now, consider the square of P, i.e., $P^2$:

$P^2 = \left(a^n \cdot r^{\frac{n(n-1)}{2}}\right)^2$

$P^2 = (a^n)^2 \cdot \left(r^{\frac{n(n-1)}{2}}\right)^2$

$P^2 = a^{2n} \cdot r^{2 \cdot \frac{n(n-1)}{2}}$

Next, consider the Right Hand Side (R.H.S.) of the equation to be proved, which is $(ab)^n$.

From equation (ii), we know $b = ar^{n-1}$. Substitute this into $(ab)^n$:

$(ab)^n = (a \cdot (ar^{n-1}))^n$

$(ab)^n = (a^2 r^{n-1})^n$

$(ab)^n = (a^2)^n \cdot (r^{n-1})^n$

Comparing equation (iv) and equation (v), we see that:

$P^2 = a^{2n} r^{n(n-1)}$

$(ab)^n = a^{2n} r^{n(n-1)}$

Therefore, $P^2 = (ab)^n$.

L.H.S. = R.H.S.

Hence Proved.

Let the first term of the G.P. be $a$ and the common ratio be $r$.

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Given:

First term of the G.P. is $a$.

Common ratio of the G.P. is $r$.

The sum of the first $n$ terms of a G.P. is given by the formula:

If $r=1$, the sum is $S_n = na$.

The terms from the $(n+1)$th to the $(2n)$th term are $T_{n+1}, T_{n+2}, \dots, T_{2n}$.

The general term of a G.P. is $T_k = ar^{k-1}$.

So, $T_{n+1} = ar^{(n+1)-1} = ar^n$.

$T_{n+2} = ar^{(n+2)-1} = ar^{n+1}$.

...

$T_{2n} = ar^{2n-1}$.

The sequence of terms $T_{n+1}, T_{n+2}, \dots, T_{2n}$ is also a G.P.

Its first term is $a' = T_{n+1} = ar^n$.

Its common ratio is $r' = \frac{T_{n+2}}{T_{n+1}} = \frac{ar^{n+1}}{ar^n} = r$.

The number of terms in this sequence is $(2n) - (n+1) + 1 = n$.

The sum of these $n$ terms, let's denote it by $S_{n+1 \text{ to } 2n}$, is given by the sum formula for a G.P. with first term $a'$, common ratio $r'$, and $n$ terms:

If $r=1$, the sum is $S_{n+1 \text{ to } 2n} = n \cdot a' = n \cdot a(1)^n = na$.

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To Prove:

The ratio $\frac{S_n}{S_{n+1 \text{ to } 2n}} = \frac{1}{r^n}$.

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Proof:

Case 1: $r \neq 1$

The ratio of the sum of the first $n$ terms ($S_n$) to the sum of the terms from $(n+1)$th to $(2n)$th term ($S_{n+1 \text{ to } 2n}$) is:

Ratio $= \frac{S_n}{S_{n+1 \text{ to } 2n}}$

Substitute the formulas from (i) and (ii):

Ratio $= \frac{\frac{a(r^n - 1)}{r-1}}{\frac{ar^n(r^n - 1)}{r-1}}$

Assuming $a \neq 0$ (for a non-trivial G.P.) and $r^n \neq 1$ (which implies $r \neq 1$ and $r \neq -1$ if n is even), we can cancel the common factors $(r^n - 1)$ and $(r-1)$:

Ratio $= \frac{a}{ar^n}$

Cancel the common factor $a$:

Ratio $= \frac{1}{r^n}$

Case 2: $r = 1$

If $r = 1$, the G.P. is $a, a, a, \dots$.

The sum of the first $n$ terms is $S_n = na$.

The terms from the $(n+1)$th to the $(2n)$th term are also $n$ terms, each equal to $a$.

The sum of these terms is $S_{n+1 \text{ to } 2n} = na$.

The ratio is $\frac{S_n}{S_{n+1 \text{ to } 2n}} = \frac{na}{na}$.

Assuming $a \neq 0$, the ratio is $\frac{na}{na} = 1$.

The required ratio is $\frac{1}{r^n} = \frac{1}{1^n} = \frac{1}{1} = 1$.

In both cases ($r \neq 1$ and $r = 1$), the ratio is $\frac{1}{r^n}$.

Therefore, the ratio of the sum of first n terms of a G.P. to the sum of terms from $(n+1)$th to $(2n)$th term is $\frac{1}{r^{n}}$.

Hence Proved.

If a, b, c, and d are in Geometric Progression (G.P.), it means that the ratio between consecutive terms is constant. Let this common ratio be $r$.

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Given:

a, b, c, d are in G.P.

This implies:

$\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = r$

From this, we can express b, c, and d in terms of a and r:

$b = ar$

$c = br = (ar)r = ar^2$

$d = cr = (ar^2)r = ar^3$

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To Prove:

$(a^2 + b^2 + c^2 ) (b^2 + c^2 + d^2 ) = (ab + bc + cd)^2$

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Proof:

Consider the Left Hand Side (L.H.S.):

L.H.S. $= (a^2 + b^2 + c^2 ) (b^2 + c^2 + d^2 )$

Substitute the expressions for b, c, and d in terms of a and r into the first bracket:

$a^2 + b^2 + c^2 = a^2 + (ar)^2 + (ar^2)^2$

$= a^2 + a^2r^2 + a^2r^4$

$= a^2(1 + r^2 + r^4)$

Substitute the expressions for b, c, and d in terms of a and r into the second bracket:

$b^2 + c^2 + d^2 = (ar)^2 + (ar^2)^2 + (ar^3)^2$

$= a^2r^2 + a^2r^4 + a^2r^6$

$= a^2r^2(1 + r^2 + r^4)$

Now, multiply the two simplified brackets:

L.H.S. $= [a^2(1 + r^2 + r^4)] \cdot [a^2r^2(1 + r^2 + r^4)]$

L.H.S. $= a^2 \cdot a^2r^2 \cdot (1 + r^2 + r^4) \cdot (1 + r^2 + r^4)$

L.H.S. $= a^{2+2} r^2 (1 + r^2 + r^4)^2$

L.H.S. $= a^4 r^2 (1 + r^2 + r^4)^2$

Now, consider the Right Hand Side (R.H.S.):

R.H.S. $= (ab + bc + cd)^2$

Substitute the expressions for b, c, and d in terms of a and r into the bracket:

$ab + bc + cd = a(ar) + (ar)(ar^2) + (ar^2)(ar^3)$

$= a^2r + a^2r^3 + a^2r^5$

Factor out the common term $a^2r$:

$= a^2r(1 + r^2 + r^4)$

Now, square the expression:

R.H.S. $= [a^2r(1 + r^2 + r^4)]^2$

R.H.S. $= (a^2r)^2 \cdot (1 + r^2 + r^4)^2$

R.H.S. $= (a^2)^2 (r)^2 (1 + r^2 + r^4)^2$

R.H.S. $= a^4 r^2 (1 + r^2 + r^4)^2$

Comparing the simplified expressions for L.H.S. and R.H.S., we have:

L.H.S. $= a^4 r^2 (1 + r^2 + r^4)^2$

R.H.S. $= a^4 r^2 (1 + r^2 + r^4)^2$

Thus, L.H.S. = R.H.S.

Hence Proved.

We need to insert two numbers between 3 and 81 such that the sequence forms a Geometric Progression (G.P.).

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Given:

The first term of the resulting G.P. is $a = 3$.

The last term (which is the 4th term after inserting two numbers) is $T_4 = 81$.

Let the two inserted numbers be $x$ and $y$. The sequence is $3, x, y, 81$.

This is a G.P. with $n=4$ terms.

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To Find:

The two numbers to be inserted between 3 and 81.

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Solution:

Let the common ratio of the G.P. be $r$.

The terms of the G.P. are $a, ar, ar^2, ar^3, \dots$.

The first term is $a = 3$.

The fourth term is $T_4 = ar^{4-1} = ar^3$.

We are given that $T_4 = 81$.

Substitute the value of $a=3$ into equation (i):

$3r^3 = 81$

Divide both sides by 3:

$r^3 = \frac{81}{3}$

$r^3 = 27$

Taking the cube root of both sides:

$r = \sqrt[3]{27}$

$r = 3$

The two numbers to be inserted are the second and third terms of the G.P.

Second term ($x$) $= ar = 3 \times 3 = 9$.

Third term ($y$) $= ar^2 = 3 \times 3^2 = 3 \times 9 = 27$.

The resulting G.P. is 3, 9, 27, 81.

The two numbers inserted between 3 and 81 are 9 and 27.

The geometric mean (G.M.) between two positive numbers $a$ and $b$ is $\sqrt{ab}$.

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Given:

The expression $\frac{a^{n+1}\;+\;b^{n+1}}{a^{n}\;+\;b^{n}}$ is the geometric mean between $a$ and $b$.

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To Find:

The value of $n$.

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Solution:

Set the given expression equal to the geometric mean:

We can write $\sqrt{ab}$ as $a^{1/2}b^{1/2}$.

Substitute this into equation (i):

$\frac{a^{n+1}\;+\;b^{n+1}}{a^{n}\;+\;b^{n}} = a^{1/2}b^{1/2}$

Cross-multiply:

$a^{n+1}\;+\;b^{n+1} = a^{1/2}b^{1/2} (a^{n}\;+\;b^{n})$

Distribute $a^{1/2}b^{1/2}$ on the right side:

$a^{n+1}\;+\;b^{n+1} = a^{1/2}b^{1/2}a^{n}\;+\;a^{1/2}b^{1/2}b^{n}$

Use the exponent rule $x^m x^p = x^{m+p}$:

$a^{n+1}\;+\;b^{n+1} = a^{n+\frac{1}{2}}b^{\frac{1}{2}}\;+\;a^{\frac{1}{2}}b^{n+\frac{1}{2}}$

Rearrange the terms to group terms with $a^{n+1}$ and $b^{n+1}$:

$a^{n+1} - a^{n+\frac{1}{2}}b^{\frac{1}{2}} = a^{\frac{1}{2}}b^{n+\frac{1}{2}} - b^{n+1}$

Factor out the common terms on both sides:

$a^{n+\frac{1}{2}}(a^{1/2} - b^{1/2}) = b^{n+\frac{1}{2}}(a^{1/2} - b^{1/2})$

Move all terms to one side:

$a^{n+\frac{1}{2}}(a^{1/2} - b^{1/2}) - b^{n+\frac{1}{2}}(a^{1/2} - b^{1/2}) = 0$

Factor out the common bracket $(a^{1/2} - b^{1/2})$:

For equation (ii) to be true, either $a^{1/2} - b^{1/2} = 0$ or $a^{n+\frac{1}{2}} - b^{n+\frac{1}{2}} = 0$.

Case 1: $a^{1/2} - b^{1/2} = 0$

$a^{1/2} = b^{1/2}$

$\sqrt{a} = \sqrt{b}$

$a = b$

If $a=b$, the expression $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ becomes $\frac{a^{n+1}+a^{n+1}}{a^n+a^n} = \frac{2a^{n+1}}{2a^n} = a$.

The geometric mean between $a$ and $b$ (when $a=b$) is $\sqrt{a \cdot a} = \sqrt{a^2} = a$ (assuming $a>0$).

So, if $a=b$, the equality holds for any value of $n$. However, the question implies we need to find a specific value of $n$. This usually means $a \neq b$.

Case 2: $a^{n+\frac{1}{2}} - b^{n+\frac{1}{2}} = 0$

Assuming $a \neq b$:

$a^{n+\frac{1}{2}} = b^{n+\frac{1}{2}}$

Divide both sides by $b^{n+\frac{1}{2}}$ (assuming $b \neq 0$):

$\frac{a^{n+\frac{1}{2}}}{b^{n+\frac{1}{2}}} = 1$

Use the exponent rule $\frac{x^m}{y^m} = \left(\frac{x}{y}\right)^m$:

$\left(\frac{a}{b}\right)^{n+\frac{1}{2}} = 1$

For $(\frac{a}{b})^{n+\frac{1}{2}}$ to be equal to 1, given that $a \neq b$, the base $\frac{a}{b}$ is not equal to 1.

The only way a non-one base raised to a power equals 1 is if the power is 0.

$n + \frac{1}{2} = 0$

Solve for $n$:

$n = -\frac{1}{2}$

Thus, for the given expression to be the geometric mean between $a$ and $b$ (assuming $a \neq b$), the value of $n$ must be $-\frac{1}{2}$.

Note: The question implies a unique value for $n$, which corresponds to the case where $a \neq b$. If $a=b$, any value of $n$ works.

Let the two numbers be $x$ and $y$. Assume $x > 0$ and $y > 0$.

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Given:

The sum of the two numbers is 6 times their geometric mean.

The sum of the numbers is $x + y$.

The geometric mean of the numbers is $\sqrt{xy}$.

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To Show:

The ratio of the numbers is $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$. This means we need to show that $\frac{x}{y} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$ or $\frac{y}{x} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$.

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Proof:

Start with the given equation (i):

$x + y = 6\sqrt{xy}$

Divide both sides by $\sqrt{xy}$ (since $x, y > 0$, $\sqrt{xy} \neq 0$):

$\frac{x + y}{\sqrt{xy}} = 6$

Separate the terms on the left side:

$\frac{x}{\sqrt{xy}} + \frac{y}{\sqrt{xy}} = 6$

Simplify the terms using exponent rules ($x = \sqrt{x}\sqrt{x}$, $y = \sqrt{y}\sqrt{y}$):

$\frac{\sqrt{x}\sqrt{x}}{\sqrt{x}\sqrt{y}} + \frac{\sqrt{y}\sqrt{y}}{\sqrt{x}\sqrt{y}} = 6$

$\frac{\sqrt{x}}{\sqrt{y}} + \frac{\sqrt{y}}{\sqrt{x}} = 6$

Let $t = \frac{\sqrt{x}}{\sqrt{y}}$. Note that $\frac{\sqrt{y}}{\sqrt{x}} = \frac{1}{t}$.

The equation becomes:

$t + \frac{1}{t} = 6$

Multiply the equation by $t$ (since $t = \frac{\sqrt{x}}{\sqrt{y}} \neq 0$):

$t^2 + 1 = 6t$

Rearrange into a quadratic equation:

$t^2 - 6t + 1 = 0$

Solve for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$t = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(1)}}{2(1)}$

$t = \frac{6 \pm \sqrt{36 - 4}}{2}$

$t = \frac{6 \pm \sqrt{32}}{2}$

Simplify the square root: $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$.

$t = \frac{6 \pm 4\sqrt{2}}{2}$

$t = \frac{2(3 \pm 2\sqrt{2})}{2}$

$t = 3 \pm 2\sqrt{2}$

So we have two possible values for $t = \frac{\sqrt{x}}{\sqrt{y}}$:

$\frac{\sqrt{x}}{\sqrt{y}} = 3 + 2\sqrt{2} \quad \text{or} \quad \frac{\sqrt{x}}{\sqrt{y}} = 3 - 2\sqrt{2}$

The ratio of the numbers is $\frac{x}{y}$. Square both sides of the equations for $\frac{\sqrt{x}}{\sqrt{y}}$:

$\frac{x}{y} = (3 + 2\sqrt{2})^2 \quad \text{or} \quad \frac{x}{y} = (3 - 2\sqrt{2})^2$

Let's calculate $(3 + 2\sqrt{2})^2$:

$(3 + 2\sqrt{2})^2 = 3^2 + 2(3)(2\sqrt{2}) + (2\sqrt{2})^2$

$= 9 + 12\sqrt{2} + (4 \times 2)$

$= 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2}$

Now, let's calculate $(3 - 2\sqrt{2})^2$:

$(3 - 2\sqrt{2})^2 = 3^2 - 2(3)(2\sqrt{2}) + (2\sqrt{2})^2$

$= 9 - 12\sqrt{2} + (4 \times 2)$

$= 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2}$

So, the ratio $\frac{x}{y}$ is either $17 + 12\sqrt{2}$ or $17 - 12\sqrt{2}$.

We need to show that this is equivalent to the ratio $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$. Let's evaluate the ratio $\frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$ by rationalizing the denominator:

$\frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{(3 + 2\sqrt{2})(3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})}$

$= \frac{(3 + 2\sqrt{2})^2}{3^2 - (2\sqrt{2})^2}$

$= \frac{17 + 12\sqrt{2}}{9 - 8}$

$= \frac{17 + 12\sqrt{2}}{1} = 17 + 12\sqrt{2}$

Similarly, the reciprocal ratio $\frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}}$ is:

$\frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} = \frac{(3 - 2\sqrt{2})(3 - 2\sqrt{2})}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})}$

$= \frac{(3 - 2\sqrt{2})^2}{3^2 - (2\sqrt{2})^2}$

$= \frac{17 - 12\sqrt{2}}{9 - 8}$

$= \frac{17 - 12\sqrt{2}}{1} = 17 - 12\sqrt{2}$

We found that the ratio $\frac{x}{y}$ is either $17 + 12\sqrt{2}$ or $17 - 12\sqrt{2}$.

Since $17 + 12\sqrt{2} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$ and $17 - 12\sqrt{2} = \frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}}$, the ratio of the two numbers $\frac{x}{y}$ is either $\frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$ or $\frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}}$.

This means the numbers $x$ and $y$ are in the ratio $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$ or $(3 - 2\sqrt{2}) : (3 + 2\sqrt{2})$. Both express the same ratio between the two numbers, just ordered differently.

Thus, the numbers are in the ratio $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$.

Hence Showed.

Let the two positive numbers be $x$ and $y$.

---

Given:

A is the Arithmetic Mean (A.M.) between $x$ and $y$.

G is the Geometric Mean (G.M.) between $x$ and $y$.

From the definition of A.M., we get:

From the definition of G.M., we get:

$G^2 = xy$

---

To Prove:

The two numbers are $A + \sqrt{(A + G) (A - G)}$ and $A - \sqrt{(A + G) (A - G)}$.

---

Proof:

Consider a quadratic equation whose roots are the two numbers $x$ and $y$.

The standard form of a quadratic equation with roots $\alpha$ and $\beta$ is $t^2 - (\alpha + \beta)t + \alpha\beta = 0$.

In this case, the roots are $x$ and $y$. So, the quadratic equation is:

$t^2 - (x+y)t + xy = 0$

Substitute the values of $(x+y)$ from equation (i) and $xy$ from equation (ii) into the quadratic equation:

$t^2 - (2A)t + G^2 = 0$

This is a quadratic equation in $t$. We can find the roots ($x$ and $y$) using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a=1$, $b=-2A$, and $c=G^2$.

$t = \frac{-(-2A) \pm \sqrt{(-2A)^2 - 4(1)(G^2)}}{2(1)}$

$t = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2}$

$t = \frac{2A \pm \sqrt{4(A^2 - G^2)}}{2}$

$t = \frac{2A \pm 2\sqrt{A^2 - G^2}}{2}$

Cancel the factor of 2 from the numerator and denominator:

$t = A \pm \sqrt{A^2 - G^2}$

The roots of this equation are the two numbers $x$ and $y$. So, the two numbers are $A + \sqrt{A^2 - G^2}$ and $A - \sqrt{A^2 - G^2}$.

Now, we need to show that $\sqrt{A^2 - G^2}$ is equal to $\sqrt{(A + G) (A - G)}$.

Using the difference of squares formula, $A^2 - G^2 = (A + G)(A - G)$.

Therefore, $\sqrt{A^2 - G^2} = \sqrt{(A + G) (A - G)}$.

Substituting this back into the expression for the numbers, we get:

The numbers are $A \pm \sqrt{(A + G) (A - G)}$.

This means the two numbers are $A + \sqrt{(A + G) (A - G)}$ and $A - \sqrt{(A + G) (A - G)}$.

Hence Proved.

Let the number of bacteria present at the start (originally, at time $t=0$) be the initial amount.

The number of bacteria doubles every hour. This indicates that the number of bacteria at the end of each successive hour forms a Geometric Progression (G.P.).

---

Given:

Initial number of bacteria (at $t=0$) = 30. This can be considered the first term if we think of the count at the start of each hour as the term. However, the question asks for the count at the end of the hour.

Let $N(t)$ be the number of bacteria present at the end of $t$ hours.

$N(0)$ = 30 (originally present).

The number doubles every hour, so the common ratio is $r = 2$.

At the end of 1st hour, the number of bacteria will be $30 \times 2^1$.

At the end of 2nd hour, the number of bacteria will be $(30 \times 2^1) \times 2 = 30 \times 2^2$.

At the end of 3rd hour, the number of bacteria will be $(30 \times 2^2) \times 2 = 30 \times 2^3$.

In general, the number of bacteria at the end of the $t$-th hour is given by:

$N(t) = N(0) \times r^t = 30 \times 2^t$

---

To Find:

The number of bacteria at the end of the 2nd hour.

The number of bacteria at the end of the 4th hour.

The number of bacteria at the end of the n-th hour.

---

Solution:

Using the formula $N(t) = 30 \times 2^t$:

Number of bacteria at the end of the 2nd hour ($t=2$):

$N(2) = 30 \times 2^2$

$N(2) = 30 \times 4$

$N(2) = 120$

Number of bacteria at the end of the 4th hour ($t=4$):

$N(4) = 30 \times 2^4$

$N(4) = 30 \times 16$

$N(4) = 480$

Number of bacteria at the end of the n-th hour ($t=n$):

$N(n) = 30 \times 2^n$

So,

At the end of the 2nd hour, there will be 120 bacteria.

At the end of the 4th hour, there will be 480 bacteria.

At the end of the n-th hour, there will be $30 \cdot 2^n$ bacteria.

This problem involves compound interest, where the interest is calculated on the initial principal and also on the accumulated interest of previous periods.

---

Given:

Principal amount, $P = \textsf{₹}500$.

Annual interest rate, $r = 10\% = \frac{10}{100} = 0.10$.

Time period, $n = 10$ years.

Interest is compounded annually.

---

To Find:

The amount ($A$) after 10 years.

---

Solution:

The formula for the amount ($A$) when interest is compounded annually is:

$A = P(1 + r)^n$

Substitute the given values into the formula:

$A = 500(1 + 0.10)^{10}$

$A = 500(1.10)^{10}$

Now, we need to calculate $(1.10)^{10}$.

$(1.10)^{10} \approx 2.59374246$

Substitute this value back into the formula for A:

$A = 500 \times 2.59374246$

$A \approx 1296.87123$

Rounding the amount to two decimal places (for currency):

$A \approx 1296.87$

The amount will be approximately $\textsf{₹}1296.87$ after 10 years.

Let the roots of the quadratic equation be $\alpha$ and $\beta$.

---

Given:

The Arithmetic Mean (A.M.) of the roots is 8.

A.M. $= \frac{\alpha + \beta}{2}$

So,

This gives the sum of the roots:

The Geometric Mean (G.M.) of the roots is 5.

G.M. $= \sqrt{\alpha \beta}$

So,

Squaring both sides gives the product of the roots:

---

To Find:

The quadratic equation whose roots are $\alpha$ and $\beta$.

---

Solution:

A quadratic equation with roots $\alpha$ and $\beta$ can be written in the form:

$x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0$

Substitute the sum of roots from (i) and the product of roots from (ii) into this general form:

$x^2 - (\alpha + \beta)x + (\alpha \beta) = 0$

$x^2 - (16)x + (25) = 0$

$x^2 - 16x + 25 = 0$

This is the required quadratic equation.

Let the given series be denoted by $S_n$. The terms of the series are $a_1 = 5$, $a_2 = 11$, $a_3 = 19$, $a_4 = 29$, $a_5 = 41$, ...

Let's find the differences between consecutive terms:

$a_2 - a_1 = 11 - 5 = 6$

$a_3 - a_2 = 19 - 11 = 8$

$a_4 - a_3 = 29 - 19 = 10$

$a_5 - a_4 = 41 - 29 = 12$

The sequence of these differences is $6, 8, 10, 12, \dots$.

These differences form an Arithmetic Progression (AP) with the first term $A' = 6$ and the common difference $D = 8 - 6 = 2$.

---

Let $a_n$ be the n-th term of the given series. The difference between consecutive terms is $a_k - a_{k-1}$.

The n-th term of the original series can be written as the sum of the first term and the sum of the first $(n-1)$ differences:

$a_n = a_1 + (a_2 - a_1) + (a_3 - a_2) + \dots + (a_n - a_{n-1})$

$a_n = a_1 + \sum_{k=1}^{n-1} (a_{k+1} - a_k)$

The differences $(a_{k+1} - a_k)$ form the AP $6, 8, 10, \dots$. The k-th term of this difference AP is $d_k = 6 + (k-1)2 = 6 + 2k - 2 = 2k + 4$.

So, for $n > 1$:

$a_n = 5 + \sum_{k=1}^{n-1} (2k + 4)$

$a_n = 5 + \sum_{k=1}^{n-1} 2k + \sum_{k=1}^{n-1} 4$

$a_n = 5 + 2 \sum_{k=1}^{n-1} k + 4 \sum_{k=1}^{n-1} 1$

Using the sum formulas $\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$ and $\sum_{k=1}^{m} 1 = m$, with $m = n-1$:

$a_n = 5 + 2 \left( \frac{(n-1)((n-1)+1)}{2} \right) + 4(n-1)$

$a_n = 5 + 2 \left( \frac{(n-1)n}{2} \right) + 4(n-1)$

$a_n = 5 + n(n-1) + 4(n-1)$

$a_n = 5 + n^2 - n + 4n - 4$

$a_n = n^2 + 3n + 1$

This formula also works for $n=1$: $a_1 = 1^2 + 3(1) + 1 = 1 + 3 + 1 = 5$.

---

Now, we need to find the sum of the first n terms, $S_n = \sum_{k=1}^{n} a_k$.

$S_n = \sum_{k=1}^{n} (k^2 + 3k + 1)$

$S_n = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} 3k + \sum_{k=1}^{n} 1$

$S_n = \sum_{k=1}^{n} k^2 + 3 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1$

Using the standard sum formulas: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$, $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$, and $\sum_{k=1}^{n} 1 = n$.

$S_n = \frac{n(n+1)(2n+1)}{6} + 3 \frac{n(n+1)}{2} + n$

Find a common denominator, which is 6:

$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{9n(n+1)}{6} + \frac{6n}{6}$

$S_n = \frac{n[(n+1)(2n+1) + 9(n+1) + 6]}{6}$

Expand and simplify the expression inside the square brackets:

$(n+1)(2n+1) = 2n^2 + n + 2n + 1 = 2n^2 + 3n + 1$

$9(n+1) = 9n + 9$

$[2n^2 + 3n + 1 + 9n + 9 + 6] = 2n^2 + 12n + 16$

Substitute this back into the expression for $S_n$:

$S_n = \frac{n(2n^2 + 12n + 16)}{6}$

Factor out 2 from the bracket in the numerator:

$S_n = \frac{n \cdot 2(n^2 + 6n + 8)}{6}$

$S_n = \frac{n(n^2 + 6n + 8)}{3}$

Factor the quadratic $n^2 + 6n + 8$:

$n^2 + 6n + 8 = (n+2)(n+4)$

So, the sum to n terms is:

$S_n = \frac{n(n+2)(n+4)}{3}$

Let the n-th term of the series be $a_n$.

---

Given:

The n-th term of the series is $a_n = n(n+3)$.

Expand the expression for $a_n$:

$a_n = n^2 + 3n$

---

To Find:

The sum to n terms of the series, denoted by $S_n = \sum_{k=1}^{n} a_k$.

---

Solution:

The sum of the first n terms is given by:

$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (k^2 + 3k)$

Using the properties of summation, we can write:

$S_n = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} 3k$

$S_n = \sum_{k=1}^{n} k^2 + 3 \sum_{k=1}^{n} k$

Now, we use the standard formulas for the sum of the first n squares and the sum of the first n natural numbers:

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

Substitute these formulas into the expression for $S_n$:

$S_n = \frac{n(n+1)(2n+1)}{6} + 3 \frac{n(n+1)}{2}$

To combine these terms, find a common denominator, which is 6:

$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{3 \cdot 3n(n+1)}{6}$

$S_n = \frac{n(n+1)(2n+1) + 9n(n+1)}{6}$

Factor out the common term $n(n+1)$ from the numerator:

$S_n = \frac{n(n+1) [(2n+1) + 9]}{6}$

Simplify the expression inside the square brackets:

$S_n = \frac{n(n+1) [2n + 10]}{6}$

Factor out 2 from the term in the square brackets:

$S_n = \frac{n(n+1) \cdot 2(n + 5)}{6}$

Cancel the common factor of 2 in the numerator and denominator:

$S_n = \frac{n(n+1)(n+5)}{3}$

The sum to n terms of the series is $\frac{n(n+1)(n+5)}{3}$.

Let the given series be denoted by $S_n$.

The terms of the series are $a_1 = 1 \times 2 = 2$, $a_2 = 2 \times 3 = 6$, $a_3 = 3 \times 4 = 12$, $a_4 = 4 \times 5 = 20$, ...

---

To Find:

The sum to n terms of the series, $S_n$.

---

Solution:

Observe the pattern in the terms. The k-th term of the series, $a_k$, is the product of $k$ and $(k+1)$.

$a_k = k(k+1) = k^2 + k$

The sum of the first n terms, $S_n$, is given by:

$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (k^2 + k)$

Using the properties of summation:

$S_n = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$

Now, use the standard formulas for the sum of the first n squares and the sum of the first n natural numbers:

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

Substitute these formulas into the expression for $S_n$:

$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$

To combine these terms, find a common denominator, which is 6:

$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}$

Factor out the common term $n(n+1)$ from the numerator:

$S_n = \frac{n(n+1)[(2n+1) + 3]}{6}$

Simplify the expression inside the square brackets:

$S_n = \frac{n(n+1)[2n + 4]}{6}$

Factor out 2 from the term in the square brackets:

$S_n = \frac{n(n+1) \cdot 2(n + 2)}{6}$

Cancel the common factor of 2 in the numerator and denominator:

$S_n = \frac{n(n+1)(n+2)}{3}$

The sum to n terms of the series is $\frac{n(n+1)(n+2)}{3}$.

Let the given series be denoted by $S_n$.

The terms of the series are $a_1 = 1 \times 2 \times 3 = 6$, $a_2 = 2 \times 3 \times 4 = 24$, $a_3 = 3 \times 4 \times 5 = 60$, ...

---

To Find:

The sum to n terms of the series, $S_n$.

---

Solution:

Observe the pattern in the terms. The k-th term of the series, $a_k$, is the product of three consecutive integers starting from $k$.

$a_k = k(k+1)(k+2)$

Expand the expression for $a_k$:

$a_k = k(k^2 + 2k + k + 2) = k(k^2 + 3k + 2)$

$a_k = k^3 + 3k^2 + 2k$

The sum of the first n terms, $S_n$, is given by:

$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (k^3 + 3k^2 + 2k)$

Using the properties of summation:

$S_n = \sum_{k=1}^{n} k^3 + \sum_{k=1}^{n} 3k^2 + \sum_{k=1}^{n} 2k$

$S_n = \sum_{k=1}^{n} k^3 + 3 \sum_{k=1}^{n} k^2 + 2 \sum_{k=1}^{n} k$

Now, use the standard formulas for the sum of the first n cubes, squares, and natural numbers:

$\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{4}$

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

Substitute these formulas into the expression for $S_n$:

$S_n = \frac{n^2(n+1)^2}{4} + 3 \left(\frac{n(n+1)(2n+1)}{6}\right) + 2 \left(\frac{n(n+1)}{2}\right)$

$S_n = \frac{n^2(n+1)^2}{4} + \frac{3n(n+1)(2n+1)}{6} + \frac{2n(n+1)}{2}$

$S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + n(n+1)$

To combine these terms, find a common denominator, which is 4:

$S_n = \frac{n^2(n+1)^2}{4} + \frac{2n(n+1)(2n+1)}{4} + \frac{4n(n+1)}{4}$

Factor out the common term $n(n+1)$ from the numerator:

$S_n = \frac{n(n+1) [n(n+1) + 2(2n+1) + 4]}{4}$

Simplify the expression inside the square brackets:

$n(n+1) + 2(2n+1) + 4 = n^2 + n + 4n + 2 + 4$

$= n^2 + 5n + 6$

Factor the quadratic inside the brackets:

$n^2 + 5n + 6 = (n+2)(n+3)$

Substitute this back into the expression for $S_n$:

$S_n = \frac{n(n+1)(n+2)(n+3)}{4}$

The sum to n terms of the series is $\frac{n(n+1)(n+2)(n+3)}{4}$.

Let the given series be denoted by $S_n$.

The k-th term of the series involves the product of two parts:

1. The first part is $3, 5, 7, \dots$. This is an Arithmetic Progression (A.P.) with the first term $a_1' = 3$ and the common difference $d = 5 - 3 = 2$. The k-th term of this A.P. is $a_k' = a_1' + (k-1)d = 3 + (k-1)2 = 3 + 2k - 2 = 2k + 1$.

2. The second part is $1^2, 2^2, 3^2, \dots$. This is the square of the term number, which is $k^2$ for the k-th term.

So, the k-th term of the given series, $a_k$, is the product of these two parts:

$a_k = (2k+1)k^2$

Expand the expression for $a_k$:

$a_k = 2k^3 + k^2$

---

To Find:

The sum to n terms of the series, $S_n = \sum_{k=1}^{n} a_k$.

---

Solution:

The sum of the first n terms is given by:

$S_n = \sum_{k=1}^{n} (2k^3 + k^2)$

Using the properties of summation:

$S_n = \sum_{k=1}^{n} 2k^3 + \sum_{k=1}^{n} k^2$

$S_n = 2 \sum_{k=1}^{n} k^3 + \sum_{k=1}^{n} k^2$

Now, use the standard formulas for the sum of the first n cubes and the sum of the first n squares:

$\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{4}$

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

Substitute these formulas into the expression for $S_n$:

$S_n = 2 \left(\frac{n^2(n+1)^2}{4}\right) + \frac{n(n+1)(2n+1)}{6}$

$S_n = \frac{n^2(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{6}$

To combine these terms, find a common denominator, which is 6:

$S_n = \frac{3n^2(n+1)^2}{6} + \frac{n(n+1)(2n+1)}{6}$

Factor out the common term $\frac{n(n+1)}{6}$ from the numerator:

$S_n = \frac{n(n+1)}{6} [3n(n+1) + (2n+1)]$

Simplify the expression inside the square brackets:

$3n(n+1) + (2n+1) = 3n^2 + 3n + 2n + 1$

$= 3n^2 + 5n + 1$

Substitute this back into the expression for $S_n$:

$S_n = \frac{n(n+1)(3n^2 + 5n + 1)}{6}$

The sum to n terms of the series is $\frac{n(n+1)(3n^2 + 5n + 1)}{6}$.

Let the given series be denoted by $S_n$. The terms of the series are $a_1 = \frac{1}{1 \times 2}$, $a_2 = \frac{1}{2 \times 3}$, $a_3 = \frac{1}{3 \times 4}$, ...

---

To Find:

The sum to n terms of the series, $S_n$.

---

Solution:

Observe the pattern in the terms. The k-th term of the series, $a_k$, is given by:

$a_k = \frac{1}{k(k+1)}$

We can decompose the k-th term using partial fractions:

$\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}$

Multiply both sides by $k(k+1)$:

$1 = A(k+1) + Bk$

Setting $k=0$: $1 = A(0+1) + B(0) \implies 1 = A$.

Setting $k=-1$: $1 = A(-1+1) + B(-1) \implies 1 = -B \implies B = -1$.

So, the k-th term can be written as:

$a_k = \frac{1}{k} - \frac{1}{k+1}$

The sum of the first n terms, $S_n$, is given by $\sum_{k=1}^{n} a_k$:

$S_n = \sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k+1}\right)$

Write out the first few terms of the sum:

$S_n = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n-1} - \frac{1}{n}\right) + \left(\frac{1}{n} - \frac{1}{n+1}\right)$

This is a telescoping sum. The intermediate terms cancel each other out:

$S_n = 1 - \cancel{\frac{1}{2}} + \cancel{\frac{1}{2}} - \cancel{\frac{1}{3}} + \cancel{\frac{1}{3}} - \cancel{\frac{1}{4}} + \dots + \cancel{\frac{1}{n}} - \frac{1}{n+1}$

$S_n = 1 - \frac{1}{n+1}$

Combine the terms on the right side:

$S_n = \frac{n+1}{n+1} - \frac{1}{n+1}$

$S_n = \frac{n+1 - 1}{n+1}$

$S_n = \frac{n}{n+1}$

The sum to n terms of the series is $\frac{n}{n+1}$.

Let the given series be denoted by $S$.

The series is the sum of the squares of integers from 5 to 20.

$S = 5^2 + 6^2 + 7^2 + \dots + 20^2$

---

To Find:

The sum of the given series.

---

Solution:

The sum can be written in summation notation as:

$S = \sum_{k=5}^{20} k^2$

We can express this sum as the difference between the sum of the first 20 squares and the sum of the first 4 squares:

$S = \sum_{k=1}^{20} k^2 - \sum_{k=1}^{4} k^2$

The formula for the sum of the first n squares is:

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

Calculate the sum of the first 20 squares ($n=20$):

$\sum_{k=1}^{20} k^2 = \frac{20(20+1)(2 \times 20 + 1)}{6}$

$= \frac{20 \times 21 \times 41}{6}$

$= \frac{\cancel{20}^{10} \times \cancel{21}^{7} \times 41}{\cancel{6}^{1}}$

$= 10 \times 7 \times 41$

$= 70 \times 41$

$= 2870$

Calculate the sum of the first 4 squares ($n=4$):

$\sum_{k=1}^{4} k^2 = \frac{4(4+1)(2 \times 4 + 1)}{6}$

$= \frac{4 \times 5 \times 9}{6}$

$= \frac{\cancel{4}^{2} \times 5 \times \cancel{9}^{3}}{\cancel{6}^{1}}$

$= 2 \times 5 \times 3$

$= 30$

Now, subtract the sum of the first 4 squares from the sum of the first 20 squares:

$S = \sum_{k=1}^{20} k^2 - \sum_{k=1}^{4} k^2$

$S = 2870 - 30$

$S = 2840$

The sum of the series 5² + 6² + 7² + ... + 20² is 2840.

Let the given series be denoted by $S_n$.

The terms of the series are $a_1 = 3 \times 8$, $a_2 = 6 \times 11$, $a_3 = 9 \times 14$, ...

---

To Find:

The sum to n terms of the series, $S_n$.

---

Solution:

Observe the pattern in the terms. The k-th term of the series, $a_k$, is the product of two factors.

1. The first factor is $3, 6, 9, \dots$. This is an Arithmetic Progression (A.P.) with the first term $a_1' = 3$ and the common difference $d_1 = 6 - 3 = 3$. The k-th term of this A.P. is $a_k' = a_1' + (k-1)d_1 = 3 + (k-1)3 = 3 + 3k - 3 = 3k$.

2. The second factor is $8, 11, 14, \dots$. This is an Arithmetic Progression (A.P.) with the first term $a_1'' = 8$ and the common difference $d_2 = 11 - 8 = 3$. The k-th term of this A.P. is $a_k'' = a_1'' + (k-1)d_2 = 8 + (k-1)3 = 8 + 3k - 3 = 3k + 5$.

So, the k-th term of the given series, $a_k$, is the product of these two factors:

$a_k = (3k)(3k + 5)$

Expand the expression for $a_k$:

$a_k = 9k^2 + 15k$

The sum of the first n terms, $S_n$, is given by:

$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (9k^2 + 15k)$

Using the properties of summation:

$S_n = \sum_{k=1}^{n} 9k^2 + \sum_{k=1}^{n} 15k$

$S_n = 9 \sum_{k=1}^{n} k^2 + 15 \sum_{k=1}^{n} k$

Now, use the standard formulas for the sum of the first n squares and the sum of the first n natural numbers:

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

Substitute these formulas into the expression for $S_n$:

$S_n = 9 \left(\frac{n(n+1)(2n+1)}{6}\right) + 15 \left(\frac{n(n+1)}{2}\right)$

$S_n = \frac{9n(n+1)(2n+1)}{6} + \frac{15n(n+1)}{2}$

$S_n = \frac{3n(n+1)(2n+1)}{2} + \frac{15n(n+1)}{2}$

The terms now have a common denominator, 2. Combine the numerators:

$S_n = \frac{3n(n+1)(2n+1) + 15n(n+1)}{2}$

Factor out the common term $3n(n+1)$ from the numerator:

$S_n = \frac{3n(n+1) [(2n+1) + 5]}{2}$

Simplify the expression inside the square brackets:

$S_n = \frac{3n(n+1) [2n + 6]}{2}$

Factor out 2 from the term in the square brackets:

$S_n = \frac{3n(n+1) \cdot 2(n + 3)}{2}$

Cancel the common factor of 2 in the numerator and denominator:

$S_n = 3n(n+1)(n+3)$

The sum to n terms of the series is $3n(n+1)(n+3)$.

Let the given series be denoted by $S_n$. The terms of the series are:

$a_1 = 1^2$

$a_2 = 1^2 + 2^2$

$a_3 = 1^2 + 2^2 + 3^2$

In general, the k-th term of the series, $a_k$, is the sum of the squares of the first k natural numbers.

$a_k = 1^2 + 2^2 + 3^2 + \dots + k^2 = \sum_{i=1}^{k} i^2$

Using the formula for the sum of the first k squares:

$\sum_{i=1}^{k} i^2 = \frac{k(k+1)(2k+1)}{6}$

So, the k-th term is:

$a_k = \frac{k(k+1)(2k+1)}{6}$

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To Find:

The sum to n terms of the series, $S_n = \sum_{k=1}^{n} a_k$.

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Solution:

Substitute the expression for $a_k$ into the summation for $S_n$:

$S_n = \sum_{k=1}^{n} \frac{k(k+1)(2k+1)}{6}$

Factor out the constant $\frac{1}{6}$:

$S_n = \frac{1}{6} \sum_{k=1}^{n} k(k+1)(2k+1)$

Expand the expression inside the summation:

$k(k+1)(2k+1) = (k^2+k)(2k+1) = 2k^3 + k^2 + 2k^2 + k = 2k^3 + 3k^2 + k$

So,

$S_n = \frac{1}{6} \sum_{k=1}^{n} (2k^3 + 3k^2 + k)$

Use the properties of summation:

$S_n = \frac{1}{6} \left( \sum_{k=1}^{n} 2k^3 + \sum_{k=1}^{n} 3k^2 + \sum_{k=1}^{n} k \right)$

$S_n = \frac{1}{6} \left( 2 \sum_{k=1}^{n} k^3 + 3 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \right)$

Now, use the standard formulas for the sum of the first n cubes, squares, and natural numbers:

$\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

Substitute these formulas into the expression for $S_n$:

$S_n = \frac{1}{6} \left( 2 \cdot \left(\frac{n(n+1)}{2}\right)^2 + 3 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right)$

$S_n = \frac{1}{6} \left( 2 \cdot \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + \frac{n(n+1)}{2} \right)$

$S_n = \frac{1}{6} \left( \frac{n^2(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{2} + \frac{n(n+1)}{2} \right)$

Combine the terms inside the bracket, which have a common denominator of 2:

$S_n = \frac{1}{6} \left( \frac{n^2(n+1)^2 + n(n+1)(2n+1) + n(n+1)}{2} \right)$

$S_n = \frac{n^2(n+1)^2 + n(n+1)(2n+1) + n(n+1)}{12}$

Factor out the common term $n(n+1)$ from the numerator:

$S_n = \frac{n(n+1) [n(n+1) + (2n+1) + 1]}{12}$

Simplify the expression inside the square brackets:

$n(n+1) + 2n + 1 + 1 = n^2 + n + 2n + 2 = n^2 + 3n + 2$

Factor the quadratic $n^2 + 3n + 2$:

$n^2 + 3n + 2 = (n+1)(n+2)$

Substitute this back into the expression for $S_n$:

$S_n = \frac{n(n+1) (n+1)(n+2)}{12}$

$S_n = \frac{n(n+1)^2(n+2)}{12}$

The sum to n terms of the series is $\frac{n(n+1)^2(n+2)}{12}$.

Let the n-th term of the series be $a_n$.

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Given:

The n-th term of the series is $a_n = n(n+1)(n+4)$.

To find the sum, we first express the k-th term:

$a_k = k(k+1)(k+4)$

Expand the expression for $a_k$:

$a_k = (k^2 + k)(k + 4)$

$a_k = k^2(k+4) + k(k+4)$

$a_k = k^3 + 4k^2 + k^2 + 4k$

$a_k = k^3 + 5k^2 + 4k$

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To Find:

The sum to n terms of the series, denoted by $S_n = \sum_{k=1}^{n} a_k$.

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Solution:

The sum of the first n terms is given by:

$S_n = \sum_{k=1}^{n} (k^3 + 5k^2 + 4k)$

Using the properties of summation:

$S_n = \sum_{k=1}^{n} k^3 + \sum_{k=1}^{n} 5k^2 + \sum_{k=1}^{n} 4k$

$S_n = \sum_{k=1}^{n} k^3 + 5 \sum_{k=1}^{n} k^2 + 4 \sum_{k=1}^{n} k$

Now, use the standard formulas for the sum of the first n cubes, squares, and natural numbers:

$\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

Substitute these formulas into the expression for $S_n$:

$S_n = \frac{n^2(n+1)^2}{4} + 5 \left(\frac{n(n+1)(2n+1)}{6}\right) + 4 \left(\frac{n(n+1)}{2}\right)$

$S_n = \frac{n^2(n+1)^2}{4} + \frac{5n(n+1)(2n+1)}{6} + 2n(n+1)$

To combine these terms, find a common denominator, which is 12:

$S_n = \frac{3n^2(n+1)^2}{12} + \frac{10n(n+1)(2n+1)}{12} + \frac{24n(n+1)}{12}$

Combine the terms with the common denominator:

$S_n = \frac{3n^2(n+1)^2 + 10n(n+1)(2n+1) + 24n(n+1)}{12}$

Factor out the common term $n(n+1)$ from the numerator:

$S_n = \frac{n(n+1) [3n(n+1) + 10(2n+1) + 24]}{12}$

Simplify the expression inside the square brackets:

$3n(n+1) + 10(2n+1) + 24 = 3n^2 + 3n + 20n + 10 + 24$

$= 3n^2 + 23n + 34$

Substitute the simplified bracket back into the expression for $S_n$:

$S_n = \frac{n(n+1)(3n^2 + 23n + 34)}{12}$

Factor the quadratic $3n^2 + 23n + 34$. We look for two numbers that multiply to $3 \times 34 = 102$ and add up to 23. These numbers are 6 and 17.

$3n^2 + 23n + 34 = 3n^2 + 6n + 17n + 34$

$= 3n(n+2) + 17(n+2)$

$= (3n+17)(n+2)$

Substitute the factored quadratic back into the expression for $S_n$:

$S_n = \frac{n(n+1)(n+2)(3n+17)}{12}$

The sum to n terms of the series is $\frac{n(n+1)(n+2)(3n+17)}{12}$.

Let the n-th term of the series be $a_n$.

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Given:

The n-th term of the series is $a_n = n^2 + 2^n$.

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To Find:

The sum to n terms of the series, denoted by $S_n = \sum_{k=1}^{n} a_k$.

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Solution:

The sum of the first n terms is given by:

$S_n = \sum_{k=1}^{n} (k^2 + 2^k)$

Using the properties of summation, we can split this sum into two parts:

$S_n = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} 2^k$

The first part is the sum of the squares of the first n natural numbers:

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

The second part is the sum of the first n terms of a Geometric Progression (G.P.) with the first term $a=2^1=2$ and the common ratio $r=2$.

$\sum_{k=1}^{n} 2^k = 2^1 + 2^2 + 2^3 + \dots + 2^n$

The sum of the first n terms of a G.P. with first term A and common ratio R is $S_n' = A \frac{R^n - 1}{R-1}$.

Here, $A=2$ and $R=2$. So,

$\sum_{k=1}^{n} 2^k = 2 \frac{2^n - 1}{2-1} = 2 \frac{2^n - 1}{1} = 2(2^n - 1) = 2^{n+1} - 2$

Now, combine the two parts to find $S_n$:

$S_n = \left(\sum_{k=1}^{n} k^2\right) + \left(\sum_{k=1}^{n} 2^k\right)$

$S_n = \frac{n(n+1)(2n+1)}{6} + (2^{n+1} - 2)$

The sum to n terms of the series is $\frac{n(n+1)(2n+1)}{6} + 2^{n+1} - 2$.

Let the n-th term of the series be $a_n$.

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Given:

The n-th term of the series is $a_n = (2n-1)^2$.

To find the sum, we first express the k-th term:

$a_k = (2k-1)^2$

Expand the expression for $a_k$:

$a_k = (2k)^2 - 2(2k)(1) + 1^2$

$a_k = 4k^2 - 4k + 1$

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To Find:

The sum to n terms of the series, denoted by $S_n = \sum_{k=1}^{n} a_k$.

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Solution:

The sum of the first n terms is given by:

$S_n = \sum_{k=1}^{n} (4k^2 - 4k + 1)$

Using the properties of summation:

$S_n = \sum_{k=1}^{n} 4k^2 - \sum_{k=1}^{n} 4k + \sum_{k=1}^{n} 1$

$S_n = 4 \sum_{k=1}^{n} k^2 - 4 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1$

Now, use the standard formulas for the sum of the first n squares, natural numbers, and the sum of a constant:

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

$\sum_{k=1}^{n} 1 = n$

Substitute these formulas into the expression for $S_n$:

$S_n = 4 \left(\frac{n(n+1)(2n+1)}{6}\right) - 4 \left(\frac{n(n+1)}{2}\right) + n$

$S_n = \frac{4n(n+1)(2n+1)}{6} - \frac{4n(n+1)}{2} + n$

$S_n = \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n$

To combine these terms, find a common denominator, which is 3:

$S_n = \frac{2n(n+1)(2n+1)}{3} - \frac{6n(n+1)}{3} + \frac{3n}{3}$

Combine the numerators:

$S_n = \frac{2n(n+1)(2n+1) - 6n(n+1) + 3n}{3}$

Factor out the common term $n$ from the numerator:

$S_n = \frac{n [2(n+1)(2n+1) - 6(n+1) + 3]}{3}$

Simplify the expression inside the square brackets:

$2(n+1)(2n+1) = 2(2n^2 + n + 2n + 1) = 2(2n^2 + 3n + 1) = 4n^2 + 6n + 2$

$6(n+1) = 6n + 6$

Expression in brackets $= (4n^2 + 6n + 2) - (6n + 6) + 3$

$= 4n^2 + 6n + 2 - 6n - 6 + 3$

$= 4n^2 - 1$

Substitute the simplified bracket back into the expression for $S_n$:

$S_n = \frac{n(4n^2 - 1)}{3}$

Alternatively, the term $(4n^2 - 1)$ can be factored as a difference of squares: $4n^2 - 1 = (2n)^2 - 1^2 = (2n-1)(2n+1)$.

So, the sum to n terms is:

$S_n = \frac{n(2n-1)(2n+1)}{3}$

The sum to n terms of the series is $\frac{n(2n-1)(2n+1)}{3}$.

Let the first term of the Arithmetic Progression (A.P.) be $\textsf{A}_1$ and the common difference be $D$.

The k-th term of the A.P. is given by $T_k = \textsf{A}_1 + (k-1)D$.

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Given:

The p-th, q-th, r-th, and s-th terms of the A.P., $T_p, T_q, T_r, T_s$, are in Geometric Progression (G.P.).

The terms are:

$T_p = \textsf{A}_1 + (p-1)D$

$T_q = \textsf{A}_1 + (q-1)D$

$T_r = \textsf{A}_1 + (r-1)D$

$T_s = \textsf{A}_1 + (s-1)D$

Since $T_p, T_q, T_r, T_s$ are in G.P., there exists a common ratio, say $R$, such that:

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To Show:

The terms $(p-q), (q-r), (r-s)$ are in G.P.

This means we need to show that $\frac{q-r}{p-q} = \frac{r-s}{q-r}$.

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Proof:

From the properties of an A.P., the difference between terms is related to the difference in their indices:

$T_q - T_p = (\textsf{A}_1 + (q-1)D) - (\textsf{A}_1 + (p-1)D) = (q-1 - p+1)D = (q-p)D$

$T_r - T_q = (\textsf{A}_1 + (r-1)D) - (\textsf{A}_1 + (q-1)D) = (r-1 - q+1)D = (r-q)D$

$T_s - T_r = (\textsf{A}_1 + (s-1)D) - (\textsf{A}_1 + (r-1)D) = (s-1 - r+1)D = (s-r)D$

Since $T_p, T_q, T_r, T_s$ are in G.P. with common ratio $R$:

$T_q = T_p R$

$T_r = T_q R$

$T_s = T_r R$

Consider the differences $T_p - T_q$, $T_q - T_r$, $T_r - T_s$:

$T_p - T_q = T_p - T_p R = T_p(1-R)$

$T_q - T_r = T_q - T_q R = T_q(1-R)$

$T_r - T_s = T_r - T_r R = T_r(1-R)$

From the A.P. differences, we also have:

Equating the two expressions for each difference:

We want to show that $(p-q), (q-r), (r-s)$ are in G.P., i.e., $\frac{q-r}{p-q} = \frac{r-s}{q-r}$.

Case 1: $D \neq 0$ and $R \neq 1$.

In this case, $T_p, T_q, T_r, T_s$ must be non-zero. We can divide equations (B) by (A) and (C) by (B).

Consider the ratio $\frac{q-r}{p-q}$:

Since $1-R \neq 0$, we can cancel it:

From the given information, $T_p, T_q, T_r, T_s$ are in G.P., so $\frac{T_q}{T_p} = R$.

Consider the ratio $\frac{r-s}{q-r}$:

Since $1-R \neq 0$, we can cancel it:

From the given information, $T_p, T_q, T_r, T_s$ are in G.P., so $\frac{T_r}{T_q} = R$.

Comparing (X) and (Y), we have:

This shows that the terms $(p-q), (q-r), (r-s)$ have a common ratio $R$, and thus are in G.P.

Case 2: $D = 0$.

If $D=0$, the A.P. is constant: $T_k = \textsf{A}_1$ for all $k$.

The terms $T_p, T_q, T_r, T_s$ are all equal to $\textsf{A}_1$.

For these equal terms to be in G.P., we must have $\textsf{A}_1 \neq 0$ (as G.P. terms are typically non-zero). In this case, the common ratio of the G.P. is $R=1$.

The differences are $(p-q)D = (p-q) \cdot 0 = 0$, $(q-r)D = (q-r) \cdot 0 = 0$, and $(r-s)D = (r-s) \cdot 0 = 0$.

The sequence of differences is $0, 0, 0$. This sequence is in G.P. with any non-zero common ratio (or common ratio 1 if we allow it).

Case 3: $R = 1$.

If $R=1$, then $T_p = T_q = T_r = T_s$. Since the terms are from an A.P., this equality implies $\textsf{A}_1 + (p-1)D = \textsf{A}_1 + (q-1)D = \dots$. If the indices $p, q, r, s$ are distinct, this equality is only possible if the common difference of the A.P. is $D=0$. This reduces to Case 2.

In all valid scenarios where the p-th, q-th, r-th, and s-th terms of an A.P. form a G.P., the differences $(p-q), (q-r), (r-s)$ are also in G.P.

Hence Showed.

Let a, b, c be positive numbers (as they are terms of a G.P. and raised to powers, implying they are usually positive).

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Given:

1. a, b, c are in G.P.

This means the ratio of consecutive terms is constant:

Cross-multiplying gives the property of G.P.:

2. $a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}}$

Let this common value be equal to some constant $K$.

Assuming $K \neq 1$ (otherwise $a=b=c=1$, which is a trivial case of G.P. and A.P.), and $a,b,c$ are positive.

From (ii), we can write:

$a^{\frac{1}{x}} = K \implies a = K^x$

$b^{\frac{1}{y}} = K \implies b = K^y$

$c^{\frac{1}{z}} = K \implies c = K^z$

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To Prove:

x, y, z are in A.P.

This means the difference between consecutive terms is constant:

$y - x = z - y$

Rearranging this gives the property of A.P.:

$2y = x + z$

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Proof:

Substitute the expressions for a, b, and c in terms of K and x, y, z into the G.P. property $b^2 = ac$ (equation (i)):

$(K^y)^2 = (K^x)(K^z)$

Using the exponent rules $(m^p)^q = m^{pq}$ and $m^p m^q = m^{p+q}$:

$K^{2y} = K^{x+z}$

Since the bases are equal and $K \neq 1$, the exponents must be equal:

$2y = x + z$

Rearranging this equation gives:

$y - x = z - y$

This shows that the difference between the second and first term ($y-x$) is equal to the difference between the third and second term ($z-y$).

By the definition of an A.P., this proves that x, y, and z are in A.P.

Hence Proved.

Let the given inequality be:

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Given:

a, b, c, d, and p are different real numbers, and the inequality (i) holds.

The inequality (i) is a quadratic in the variable $p$. Let this quadratic function be $f(p)$.

$f(p) = (a^2 + b^2 + c^2 )p^2 – 2(ab + bc + cd) p + (b^2 + c^2 + d^2 )$

The coefficient of $p^2$ is $(a^2 + b^2 + c^2)$. Since a, b, c are real numbers, $a^2 \geq 0$, $b^2 \geq 0$, $c^2 \geq 0$. If a, b, c are different real numbers, at least one must be non-zero, so $(a^2 + b^2 + c^2) > 0$.

Thus, $f(p)$ represents an upward-opening parabola.

An upward-opening parabola $f(p)$ is less than or equal to zero for some real value of $p$ if and only if it touches the p-axis at exactly one point. This occurs when the discriminant of the quadratic is equal to zero.

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To Show:

a, b, c, and d are in G.P.

This means we need to show that $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$, which is equivalent to showing $b^2 = ac$ and $c^2 = bd$, assuming $a,b,c,d \neq 0$. Since the numbers are different, none of them can be zero in a non-trivial G.P. sequence.

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Proof:

The discriminant ($\Delta$) of the quadratic $Ap^2 + Bp + C$ is $B^2 - 4AC$.

Here, $A = (a^2 + b^2 + c^2)$, $B = -2(ab + bc + cd)$, and $C = (b^2 + c^2 + d^2)$.

Set the discriminant to zero:

$\Delta = [-2(ab + bc + cd)]^2 - 4(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = 0$

$4(ab + bc + cd)^2 - 4(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = 0$

Divide by 4:

Expand both sides of equation (ii):

L.H.S. $= (ab)^2 + (bc)^2 + (cd)^2 + 2(ab)(bc) + 2(ab)(cd) + 2(bc)(cd)$

$= a^2b^2 + b^2c^2 + c^2d^2 + 2ab^2c + 2abcd + 2bc^2d$

R.H.S. $= a^2(b^2 + c^2 + d^2) + b^2(b^2 + c^2 + d^2) + c^2(b^2 + c^2 + d^2)$

$= a^2b^2 + a^2c^2 + a^2d^2 + b^4 + b^2c^2 + b^2d^2 + c^2b^2 + c^4 + c^2d^2$

$= a^2b^2 + a^2c^2 + a^2d^2 + b^4 + 2b^2c^2 + b^2d^2 + c^4 + c^2d^2$

Equating L.H.S. and R.H.S.:

$a^2b^2 + b^2c^2 + c^2d^2 + 2ab^2c + 2abcd + 2bc^2d = a^2b^2 + a^2c^2 + a^2d^2 + b^4 + 2b^2c^2 + b^2d^2 + c^4 + c^2d^2$

Move all terms to one side and simplify:

$0 = a^2b^2 + a^2c^2 + a^2d^2 + b^4 + 2b^2c^2 + b^2d^2 + c^4 + c^2d^2 - (a^2b^2 + b^2c^2 + c^2d^2 + 2ab^2c + 2abcd + 2bc^2d)$

$0 = a^2b^2 + a^2c^2 + a^2d^2 + b^4 + 2b^2c^2 + b^2d^2 + c^4 + c^2d^2 - a^2b^2 - b^2c^2 - c^2d^2 - 2ab^2c - 2abcd - 2bc^2d$

$0 = a^2c^2 + a^2d^2 + b^4 + b^2c^2 + b^2d^2 + c^4 - 2ab^2c - 2abcd - 2bc^2d$

Rearrange and group terms to form perfect squares:

$0 = (b^4 - 2ab^2c + a^2c^2) + (c^4 - 2bc^2d + b^2d^2) + (a^2d^2 - 2abcd + b^2c^2)$

Rewrite the grouped terms as squares:

$0 = (b^2 - ac)^2 + (c^2 - bd)^2 + (ad - bc)^2$

Since a, b, c, and d are real numbers, the squares of real numbers are non-negative. The sum of non-negative terms is zero if and only if each term is zero.

$(b^2 - ac)^2 = 0 \implies b^2 - ac = 0 \implies b^2 = ac$

$(c^2 - bd)^2 = 0 \implies c^2 - bd = 0 \implies c^2 = bd$

$(ad - bc)^2 = 0 \implies ad - bc = 0 \implies ad = bc$

The condition $b^2 = ac$ implies $\frac{b}{a} = \frac{c}{b}$ (if $a, b, c \neq 0$).

The condition $c^2 = bd$ implies $\frac{c}{b} = \frac{d}{c}$ (if $b, c, d \neq 0$).

If a, b, c, d are different real numbers, none of them can be zero for the concepts of ratios to be well-defined in a non-trivial G.P.

From $b^2 = ac$ and $c^2 = bd$, since $a, b, c, d \neq 0$, we get $\frac{b}{a} = \frac{c}{b}$ and $\frac{c}{b} = \frac{d}{c}$.

Therefore, $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.

This is the definition of a Geometric Progression. Thus, a, b, c, and d are in G.P.

The condition $ad=bc$ is consistent with $a, ar, ar^2, ar^3$ being in G.P., as $a(ar^3) = ar(ar^2) = a^2r^3$.

Hence Showed.

Let the common root of the two quadratic equations be $\alpha$.

---

Given:

1. p, q, r are in G.P.

This means the square of the middle term is equal to the product of the other two terms:

Since p, q, r are coefficients of a quadratic equation, we assume $p \neq 0$. For $\frac{e}{q}$ and $\frac{f}{r}$ to be defined later, we also assume $q \neq 0$ and $r \neq 0$. If $q=0$, then from (i), $pr=0$. Since $p \neq 0$, $r$ must be 0. The G.P. is p, 0, 0, where $p \neq 0$. The first equation becomes $px^2=0$, which has the root $x=0$. If 0 is a common root, then $d(0)^2 + 2e(0) + f = 0 \implies f=0$. The terms $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ become $\frac{d}{p}, \frac{e}{0}, \frac{0}{0}$, which are not all defined in the standard sense. We proceed assuming $p, q, r \neq 0$.

2. $\alpha$ is a root of $px^2 + 2qx + r = 0$.

3. $\alpha$ is a root of $dx^2 + 2ex + f = 0$.

---

To Show:

$\frac{d}{p}$, $\frac{e}{q}$, $\frac{f}{r}$ are in A.P.

This means we need to show that the difference between consecutive terms is constant:

$\frac{e}{q} - \frac{d}{p} = \frac{f}{r} - \frac{e}{q}$

Or equivalently, the middle term is the average of the other two:

$2 \frac{e}{q} = \frac{d}{p} + \frac{f}{r}$

---

Proof:

Consider the discriminant of the first quadratic equation $px^2 + 2qx + r = 0$.

Discriminant $\Delta = (2q)^2 - 4pr = 4q^2 - 4pr$

Using equation (i), $q^2 = pr$:

$\Delta = 4pr - 4pr = 0$

Since the discriminant is 0, the quadratic equation $px^2 + 2qx + r = 0$ has exactly one distinct real root (a repeated root). This root is given by $\alpha = \frac{-2q}{2p} = -\frac{q}{p}$.

Since $\alpha$ is the common root, it must be this value:

Substitute the value of $\alpha$ from (iv) into equation (iii):

$d\left(-\frac{q}{p}\right)^2 + 2e\left(-\frac{q}{p}\right) + f = 0$

$d \frac{q^2}{p^2} - \frac{2eq}{p} + f = 0$

Now, use the G.P. property $q^2 = pr$ (equation (i)) to substitute for $q^2$:

$d \frac{pr}{p^2} - \frac{2eq}{p} + f = 0$

Assuming $p \neq 0$, we can cancel one $p$ from the first term:

$d \frac{r}{p} - \frac{2eq}{p} + f = 0$

Multiply the entire equation by $\frac{1}{r}$ (assuming $r \neq 0$):

$\left(d \frac{r}{p}\right) \frac{1}{r} - \left(\frac{2eq}{p}\right) \frac{1}{r} + f \frac{1}{r} = 0$

$\frac{d}{p} - \frac{2eq}{pr} + \frac{f}{r} = 0$

Again, use the G.P. property $pr = q^2$ (equation (i)) to substitute for $pr$ in the denominator of the second term:

$\frac{d}{p} - \frac{2eq}{q^2} + \frac{f}{r} = 0$

Assuming $q \neq 0$, we can cancel one $q$ from the second term:

$\frac{d}{p} - \frac{2e}{q} + \frac{f}{r} = 0$

Rearrange the terms:

$\frac{d}{p} + \frac{f}{r} = \frac{2e}{q}$

This equation shows that $\frac{e}{q}$ is the arithmetic mean of $\frac{d}{p}$ and $\frac{f}{r}$.

By the definition of an A.P., this proves that $\frac{d}{p}$, $\frac{e}{q}$, and $\frac{f}{r}$ are in A.P.

Hence Showed.

Given:

An arithmetic progression (A.P.).

---

To Prove:

The sum of the $(m+n)$-th term and the $(m-n)$-th term is equal to twice the $m$-th term.

That is, $a_{m+n} + a_{m-n} = 2 \cdot a_m$, where $a_k$ represents the $k$-th term of the A.P.

---

Proof:

Let the first term of the A.P. be $a$ and the common difference be $d$.

The formula for the $k$-th term of an A.P. is given by:

Using this formula, we can write the expressions for the required terms:

The $(m+n)$-th term is obtained by substituting $k = m+n$ in equation (1):

The $(m-n)$-th term is obtained by substituting $k = m-n$ in equation (1). (Assume $m > n$ for the $(m-n)$-th term to be a positive indexed term. The formula holds even if $m \le n$, representing a term before the first term, which is valid in theory.):

The $m$-th term is obtained by substituting $k = m$ in equation (1):

Now, consider the sum of the $(m+n)$-th term and the $(m-n)$-th term by adding equations (2) and (3):

$a_{m+n} + a_{m-n} = [a + ((m+n)-1)d] + [a + ((m-n)-1)d]$

Remove the brackets and group terms:

$a_{m+n} + a_{m-n} = a + (m+n-1)d + a + (m-n-1)d$

$a_{m+n} + a_{m-n} = (a + a) + (m+n-1)d + (m-n-1)d$

Factor out the common difference $d$ from the terms containing $d$:

$a_{m+n} + a_{m-n} = 2a + d[(m+n-1) + (m-n-1)]$

Simplify the expression inside the square brackets:

$a_{m+n} + a_{m-n} = 2a + d[m+n-1 + m-n-1]$

$a_{m+n} + a_{m-n} = 2a + d[(m+m) + (n-n) + (-1-1)]$

$a_{m+n} + a_{m-n} = 2a + d[2m + 0 - 2]$

$a_{m+n} + a_{m-n} = 2a + d[2m - 2]$

Factor out 2 from the term containing $d$:

$a_{m+n} + a_{m-n} = 2a + 2d(m - 1)$

Now, factor out the common term 2 from the entire expression:

$a_{m+n} + a_{m-n} = 2[a + (m - 1)d]$

From equation (4), we know that $a + (m-1)d$ is the $m$-th term ($a_m$).

Substituting $a_m$ into the expression:

This result shows that the sum of the $(m+n)$-th term and the $(m-n)$-th term of an A.P. is equal to twice the $m$-th term.

---

Hence Proved.

Solution:

---

Let the three numbers in Arithmetic Progression (A.P.) be $a - d$, $a$, and $a + d$, where $a$ is the middle term and $d$ is the common difference.

---

According to the problem statement, the sum of the three numbers is 24.

$(a - d) + a + (a + d) = 24$

Simplify the equation:

$3a = 24$

Divide by 3:

---

According to the problem statement, the product of the three numbers is 440.

$(a - d) \cdot a \cdot (a + d) = 440$

Rearrange the terms and use the difference of squares formula $(x-y)(x+y) = x^2 - y^2$:

$a(a^2 - d^2) = 440$

---

Now, substitute the value of $a = 8$ into the product equation:

$8(8^2 - d^2) = 440$

$8(64 - d^2) = 440$

Divide both sides by 8:

$64 - d^2 = \frac{440}{8}$

Rearrange the equation to solve for $d^2$:

$d^2 = 64 - 55$

$d^2 = 9$

Take the square root of both sides to find the value of $d$:

$d = \pm \sqrt{9}$

$d = \pm 3$

---

We have two possible values for the common difference $d$:

Case 1: $d = 3$

The three numbers are:

$a - d = 8 - 3 = 5$

$a = 8$

$a + d = 8 + 3 = 11$

The numbers are 5, 8, and 11.

---

Case 2: $d = -3$

The three numbers are:

$a - d = 8 - (-3) = 8 + 3 = 11$

$a = 8$

$a + d = 8 + (-3) = 8 - 3 = 5$

The numbers are 11, 8, and 5.

---

In both cases, the set of numbers is the same {5, 8, 11}.

The sum of these numbers is $5 + 8 + 11 = 24$.

The product of these numbers is $5 \times 8 \times 11 = 40 \times 11 = 440$.

Both conditions given in the problem are satisfied.

---

The numbers are 5, 8, and 11.

Given:

An Arithmetic Progression (A.P.).

$S_1$ = Sum of the first $n$ terms.

$S_2$ = Sum of the first $2n$ terms.

$S_3$ = Sum of the first $3n$ terms.

---

To Show:

$S_3 = 3(S_2 - S_1)$

---

Proof:

Let the first term of the A.P. be $a$ and the common difference be $d$.

The formula for the sum of the first $k$ terms of an A.P. is given by:

Using this formula, we can write the expressions for $S_1$, $S_2$, and $S_3$:

$S_1$ is the sum of the first $n$ terms ($k=n$):

$S_2$ is the sum of the first $2n$ terms ($k=2n$):

$S_3$ is the sum of the first $3n$ terms ($k=3n$):

Now, consider the expression $S_2 - S_1$. Subtract equation (2) from equation (3):

$S_2 - S_1 = n[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d]$

$S_2 - S_1 = \frac{n}{2} \left[ 2 \cdot (2a + (2n-1)d) - (2a + (n-1)d) \right]$

$S_2 - S_1 = \frac{n}{2} \left[ 4a + 2(2n-1)d - 2a - (n-1)d \right]$

$S_2 - S_1 = \frac{n}{2} \left[ (4a - 2a) + (4n-2)d - (n-1)d \right]$

$S_2 - S_1 = \frac{n}{2} \left[ 2a + \{(4n-2) - (n-1)\}d \right]$

$S_2 - S_1 = \frac{n}{2} \left[ 2a + (4n - 2 - n + 1)d \right]$

$S_2 - S_1 = \frac{n}{2} \left[ 2a + (3n - 1)d \right]$

Now, consider the expression $3(S_2 - S_1)$. Multiply equation (5) by 3:

$3(S_2 - S_1) = 3 \cdot \frac{n}{2} [2a + (3n - 1)d]$

Compare equation (6) with equation (4). We observe that the right-hand sides are identical.

Therefore, from equations (4) and (6), we can conclude that:

---

Hence Shown.

Solution:

---

We need to find the sum of all numbers between 200 and 400 that are divisible by 7.

These numbers form an Arithmetic Progression (A.P.) with a common difference $d = 7$.

---

Finding the first term ($a$):

The first number greater than 200 that is divisible by 7.

Divide 200 by 7: $200 = 7 \times 28 + 4$.

This means 200 is 4 more than a multiple of 7 ($7 \times 28 = 196$).

The next multiple of 7 will be $196 + 7 = 203$.

So, the first term $a = 203$.

---

Finding the last term ($l$ or $a_n$):

The last number less than 400 that is divisible by 7.

Divide 400 by 7: $400 = 7 \times 57 + 1$.

This means 400 is 1 more than a multiple of 7 ($7 \times 57 = 399$).

The last multiple of 7 before 400 is 399.

So, the last term $l = a_n = 399$.

---

Finding the number of terms ($n$):

We use the formula for the $n$-th term of an A.P.: $a_n = a + (n-1)d$.

We have $a_n = 399$, $a = 203$, and $d = 7$.

$399 = 203 + (n-1)7$

Subtract 203 from both sides:

$399 - 203 = (n-1)7$

$196 = (n-1)7$

Divide both sides by 7:

$\frac{196}{7} = n-1$

Add 1 to both sides:

$n = 28 + 1$

---

Finding the sum ($S_n$):

The sum of the first $n$ terms of an A.P. is given by the formula: $S_n = \frac{n}{2}(a + l)$.

Substitute the values $n = 29$, $a = 203$, and $l = 399$ into the formula:

$S_{29} = \frac{29}{2}(203 + 399)$

$S_{29} = \frac{29}{2}(602)$

Divide 602 by 2:

$S_{29} = 29 \times 301$

Multiply 29 by 301:

$S_{29} = 8729$

---

The sum of all numbers between 200 and 400 which are divisible by 7 is 8729.

Solution:

---

We want to find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Using the Principle of Inclusion-Exclusion, the required sum is:

Sum (divisible by 2 or 5) = Sum (divisible by 2) + Sum (divisible by 5) - Sum (divisible by both 2 and 5)

Numbers divisible by both 2 and 5 are those divisible by the Least Common Multiple (LCM) of 2 and 5, which is 10.

So, the required sum is: $S_{\text{by 2 or 5}} = S_{\text{by 2}} + S_{\text{by 5}} - S_{\text{by 10}}$.

---

1. Sum of integers from 1 to 100 divisible by 2:

These are 2, 4, 6, ..., 100.

This is an Arithmetic Progression (A.P.) with first term $a = 2$, common difference $d = 2$, and last term $l = 100$.

To find the number of terms ($n_2$), we use the formula $l = a + (n-1)d$:

$100 = 2 + (n_2-1)2$

$98 = (n_2-1)2$

$\frac{98}{2} = n_2-1$

$49 = n_2-1$

The sum ($S_{\text{by 2}}$) is given by $S_n = \frac{n}{2}(a + l)$:

$S_{\text{by 2}} = \frac{50}{2}(2 + 100)$

$S_{\text{by 2}} = 25 \times 102$

---

2. Sum of integers from 1 to 100 divisible by 5:

These are 5, 10, 15, ..., 100.

This is an A.P. with first term $a = 5$, common difference $d = 5$, and last term $l = 100$.

To find the number of terms ($n_5$), we use the formula $l = a + (n-1)d$:

$100 = 5 + (n_5-1)5$

$95 = (n_5-1)5$

$\frac{95}{5} = n_5-1$

$19 = n_5-1$

The sum ($S_{\text{by 5}}$) is given by $S_n = \frac{n}{2}(a + l)$:

$S_{\text{by 5}} = \frac{20}{2}(5 + 100)$

$S_{\text{by 5}} = 10 \times 105$

---

3. Sum of integers from 1 to 100 divisible by 10:

These are 10, 20, 30, ..., 100.

This is an A.P. with first term $a = 10$, common difference $d = 10$, and last term $l = 100$.

To find the number of terms ($n_{10}$), we use the formula $l = a + (n-1)d$:

$100 = 10 + (n_{10}-1)10$

$90 = (n_{10}-1)10$

$\frac{90}{10} = n_{10}-1$

$9 = n_{10}-1$

The sum ($S_{\text{by 10}}$) is given by $S_n = \frac{n}{2}(a + l)$:

$S_{\text{by 10}} = \frac{10}{2}(10 + 100)$

$S_{\text{by 10}} = 5 \times 110$

---

4. Calculate the required sum:

Using the Inclusion-Exclusion Principle and the sums calculated in (i), (ii), and (iii):

$S_{\text{by 2 or 5}} = S_{\text{by 2}} + S_{\text{by 5}} - S_{\text{by 10}}$

$S_{\text{by 2 or 5}} = 2550 + 1050 - 550$

$S_{\text{by 2 or 5}} = 3600 - 550$

$S_{\text{by 2 or 5}} = 3050$

---

The sum of integers from 1 to 100 that are divisible by 2 or 5 is 3050.

Solution:

---

We are looking for two-digit numbers that leave a remainder of 1 when divided by 4.

A number $x$ leaves a remainder of 1 when divided by 4 can be written in the form $x = 4k + 1$, where $k$ is an integer.

The two-digit numbers range from 10 to 99, inclusive.

---

Finding the first term:

We need to find the smallest integer $k$ such that $4k + 1 \ge 10$.

$4k \ge 10 - 1$

$4k \ge 9$

$k \ge \frac{9}{4}$

$k \ge 2.25$

Since $k$ must be an integer, the smallest possible value for $k$ is 3.

The first number in the sequence is $4(3) + 1 = 12 + 1 = 13$.

So, the first term $a = 13$.

---

Finding the last term:

We need to find the largest integer $k$ such that $4k + 1 \le 99$.

$4k \le 99 - 1$

$4k \le 98$

$k \le \frac{98}{4}$

$k \le 24.5$

Since $k$ must be an integer, the largest possible value for $k$ is 24.

The last number in the sequence is $4(24) + 1 = 96 + 1 = 97$.

So, the last term $l = 97$.

---

The sequence of numbers is 13, 17, 21, ..., 97.

This is an Arithmetic Progression (A.P.) with the first term $a = 13$ and the common difference $d = 4$ (since each number is of the form $4k+1$, the difference between consecutive terms is $(4(k+1)+1) - (4k+1) = 4$).

---

Finding the number of terms ($n$):

We use the formula for the $n$-th term of an A.P.: $l = a + (n-1)d$.

Substitute the values $l = 97$, $a = 13$, and $d = 4$:

$97 = 13 + (n-1)4$

Subtract 13 from both sides:

$97 - 13 = (n-1)4$

$84 = (n-1)4$

Divide both sides by 4:

$\frac{84}{4} = n-1$

Add 1 to both sides:

$n = 21 + 1$

---

Finding the sum ($S_n$):

The sum of the first $n$ terms of an A.P. is given by the formula: $S_n = \frac{n}{2}(a + l)$.

Substitute the values $n = 22$, $a = 13$, and $l = 97$ into the formula:

$S_{22} = \frac{22}{2}(13 + 97)$

$S_{22} = 11(110)$

$S_{22} = 1210$

---

The sum of all two-digit numbers which when divided by 4, yields 1 as remainder is 1210.

Solution:

---

The given functional equation is $f(x + y) = f(x) f(y)$ for all $x, y \in \mathbb{N}$.

This type of functional equation suggests that the function $f(x)$ is of the form $f(x) = a^x$ for some base $a$.

We are given that $f(1) = 3$. Substitute $x=1$ into equation (1):

$f(1) = a^1 = a$

Since $f(1) = 3$, we have:

Therefore, the function is:

---

We are also given that $\sum\limits_{x=1}^{n} f(x) = 120$.

Substitute $f(x) = 3^x$ from equation (2) into the summation:

The summation on the left side of equation (3) is $3^1 + 3^2 + 3^3 + \dots + 3^n$.

This is a geometric progression (G.P.) with:

First term, $A = 3^1 = 3$.

Common ratio, $R = \frac{3^2}{3^1} = 3$.

Number of terms = $n$.

The sum of the first $n$ terms of a G.P. with first term $A$ and common ratio $R$ (where $R \neq 1$) is given by the formula:

Substitute the values $S_n = 120$, $A = 3$, and $R = 3$ into equation (4):

$120 = 3 \frac{3^n - 1}{3 - 1}$

Now, we need to solve equation (5) for $n$.

Multiply both sides by 2:

$120 \times 2 = 3 (3^n - 1)$

$240 = 3 (3^n - 1)$

Divide both sides by 3:

$\frac{240}{3} = 3^n - 1$

Add 1 to both sides of equation (6):

$80 + 1 = 3^n$

To solve for $n$, express 81 as a power of 3:

$81 = 3 \times 27 = 3 \times 3^3 = 3^4$

Substitute this into equation (7):

Since the bases are equal, the exponents must be equal:

$n = 4$

---

The value of n is 4.

Solution:

---

Given:

Sum of terms of a G.P., $S_n = 315$.

First term, $a = 5$.

Common ratio, $r = 2$.

---

To Find:

The number of terms, $n$.

The last term, $a_n$.

---

Solution:

The sum of the first $n$ terms of a Geometric Progression (G.P.) with first term $a$ and common ratio $r$ is given by the formula:

Substitute the given values $S_n = 315$, $a = 5$, and $r = 2$ into equation (1):

$315 = 5 \frac{2^n - 1}{2 - 1}$

$315 = 5 \frac{2^n - 1}{1}$

$315 = 5 (2^n - 1)$

Divide both sides by 5:

$\frac{315}{5} = 2^n - 1$

Add 1 to both sides of equation (2):

$63 + 1 = 2^n$

Express 64 as a power of 2:

$64 = 2 \times 32 = 2 \times 2 \times 16 = 2 \times 2 \times 2 \times 8 = 2^4 \times 4 = 2^4 \times 2^2 = 2^6$

Substitute $64 = 2^6$ into equation (3):

Since the bases are equal in equation (4), the exponents must be equal:

---

Now, we need to find the last term, which is the $n$-th term ($a_n$).

The formula for the $n$-th term of a G.P. is:

Substitute the values $a = 5$, $r = 2$, and $n = 6$ into equation (5) to find the 6th term ($a_6$):

$a_6 = 5 \times 2^{6-1}$

$a_6 = 5 \times 2^5$

Calculate $2^5$:

$2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$

Substitute $2^5 = 32$ back into the equation for $a_6$:

Calculate the product:

---

The number of terms is 6 and the last term is 160.

Solution:

---

Given:

The first term of a Geometric Progression (G.P.), $a = 1$.

The sum of the third term ($a_3$) and the fifth term ($a_5$) is 90.

$a_3 + a_5 = 90$

---

To Find:

The common ratio, $r$, of the G.P.

---

The formula for the $n$-th term of a Geometric Progression (G.P.) with first term $a$ and common ratio $r$ is given by:

Using this formula, we can express the third term ($a_3$) and the fifth term ($a_5$) in terms of $a$ and $r$:

For the third term, substitute $n=3$ into equation (1):

For the fifth term, substitute $n=5$ into equation (1):

We are given that the first term $a = 1$. Substitute $a=1$ into equations (2) and (3):

$a_3 = (1)r^2 = r^2$

$a_5 = (1)r^4 = r^4$

We are also given that the sum of the third and fifth terms is 90:

Substitute the expressions for $a_3$ and $a_5$ into this equation:

Rearrange equation (4) into a standard quadratic form in terms of $r^2$:

$r^4 + r^2 - 90 = 0$

Let $x = r^2$. The equation becomes a quadratic equation in $x$:

We can solve this quadratic equation by factoring. We look for two numbers that multiply to -90 and add up to 1. These numbers are 10 and -9.

So, equation (5) can be factored as:

$(x + 10)(x - 9) = 0$

This gives two possible values for $x$:

$x + 10 = 0$ or $x - 9 = 0$

$x = -10$ $x = 9$

Now, substitute back $x = r^2$:

$r^2 = -10$ or $r^2 = 9$

For real values of $r$, $r^2$ must be non-negative. Thus, $r^2 = -10$ has no real solutions for $r$.

Consider the second possibility:

$r^2 = 9$

Taking the square root of both sides:

$r = \pm \sqrt{9}$

$r = \pm 3$

So, the possible real values for the common ratio are 3 and -3.

---

The common ratio of the G.P. can be 3 or -3.

Solution:

---

Given:

Three numbers are in Geometric Progression (G.P.).

The sum of these three numbers is 56.

When 1, 7, and 21 are subtracted from these numbers respectively, the new numbers are in Arithmetic Progression (A.P.).

---

To Find:

The three numbers in G.P.

---

Let the three numbers in G.P. be $\frac{a}{r}$, $a$, and $ar$, where $a$ is the middle term and $r$ is the common ratio.

---

Condition 1: Sum of the numbers in G.P. is 56.

$\frac{a}{r} + a + ar = 56$

Factor out $a$:

---

Condition 2: After subtracting 1, 7, 21, the new numbers are in A.P.

The new numbers are: $(\frac{a}{r} - 1)$, $(a - 7)$, and $(ar - 21)$.

For these three numbers to be in A.P., the difference between consecutive terms must be constant. That is, the common difference must be the same.

$(a - 7) - (\frac{a}{r} - 1) = (ar - 21) - (a - 7)$

Simplify both sides:

$a - 7 - \frac{a}{r} + 1 = ar - 21 - a + 7$

$a - \frac{a}{r} - 6 = ar - a - 14$

Gather terms involving $a$ on one side and constants on the other:

$a - \frac{a}{r} - ar + a = -14 + 6$

$2a - \frac{a}{r} - ar = -8$

Multiply by $r$ to clear the denominator (assuming $r \neq 0$. If $r=0$, the GP is $a, 0, 0, ...$, which sums to $a$. $a=56$, giving $56,0,0$. Subtracting 1,7,21 gives $55,-7,-21$. This is not an AP. So $r \neq 0$.):

$2ar - a - ar^2 = -8r$

Multiply by -1 to make the $ar^2$ term positive:

$ar^2 - 2ar + a = 8r$

Factor out $a$ from the left side:

Recognize that the term in the parenthesis is a perfect square: $(r-1)^2$.

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From equation (1), we can express $a$ in terms of $r$ (assuming $1/r + 1 + r \neq 0$):

$a = \frac{56}{\frac{1}{r} + 1 + r} = \frac{56}{\frac{1 + r + r^2}{r}} = \frac{56r}{1 + r + r^2}$

Substitute this expression for $a$ into equation (3):

$\frac{56r}{1 + r + r^2} (r - 1)^2 = 8r$

We can divide both sides by $r$, assuming $r \neq 0$ (as established earlier):

$\frac{56}{1 + r + r^2} (r - 1)^2 = 8$

Divide both sides by 8:

$\frac{7}{1 + r + r^2} (r - 1)^2 = 1$

$7(r - 1)^2 = 1 + r + r^2$

Expand the left side: $7(r^2 - 2r + 1) = 1 + r + r^2$

$7r^2 - 14r + 7 = 1 + r + r^2$

Move all terms to one side to form a quadratic equation:

$7r^2 - r^2 - 14r - r + 7 - 1 = 0$

Divide the quadratic equation by 3 to simplify:

$2r^2 - 5r + 2 = 0$

Solve this quadratic equation for $r$. We can factor it. We look for two numbers that multiply to $2 \times 2 = 4$ and add up to -5. These numbers are -4 and -1.

$2r^2 - 4r - r + 2 = 0$

Factor by grouping:

$2r(r - 2) - 1(r - 2) = 0$

$(2r - 1)(r - 2) = 0$

This gives two possible values for $r$:

$2r - 1 = 0$ or $r - 2 = 0$

$2r = 1$ $r = 2$

$r = \frac{1}{2}$

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Now find the value of $a$ for each value of $r$ using equation (3): $a(r - 1)^2 = 8r$.

Case 1: $r = 2$

$a(2 - 1)^2 = 8(2)$

$a(1)^2 = 16$

The three numbers in G.P. are $\frac{a}{r}$, $a$, $ar$:

$\frac{16}{2} = 8$

$16$

$16 \times 2 = 32$

The numbers are 8, 16, 32.

Check the conditions:

Sum: $8 + 16 + 32 = 24 + 32 = 56$. (Matches the given sum)

Subtracting 1, 7, 21: $8-1=7$, $16-7=9$, $32-21=11$. The numbers are 7, 9, 11.

Check if they are in A.P.: $9 - 7 = 2$, $11 - 9 = 2$. Yes, they are in A.P. with a common difference of 2.

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Case 2: $r = \frac{1}{2}$

$a\left(\frac{1}{2} - 1\right)^2 = 8\left(\frac{1}{2}\right)$

$a\left(-\frac{1}{2}\right)^2 = 4$

$a\left(\frac{1}{4}\right) = 4$

$a = 4 \times 4$

The three numbers in G.P. are $\frac{a}{r}$, $a$, $ar$:

$\frac{16}{1/2} = 16 \times 2 = 32$

$16$

$16 \times \frac{1}{2} = 8$

The numbers are 32, 16, 8.

Check the conditions:

Sum: $32 + 16 + 8 = 48 + 8 = 56$. (Matches the given sum)

Subtracting 1, 7, 21: $32-1=31$, $16-7=9$, $8-21=-13$. The numbers are 31, 9, -13.

Check if they are in A.P.: $9 - 31 = -22$, $-13 - 9 = -22$. Yes, they are in A.P. with a common difference of -22.

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Both sets of numbers (8, 16, 32) and (32, 16, 8) satisfy the given conditions. They are the same set of numbers, just in reverse order.

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The three numbers are 8, 16, and 32.

Solution:

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Given:

A Geometric Progression (G.P.) consists of an even number of terms. Let the number of terms be $2n$, where $n \in \mathbb{N}$ and $n \ge 1$.

Let the first term of the G.P. be $a$ and the common ratio be $r$. The terms of the G.P. are $a, ar, ar^2, ar^3, \dots, ar^{2n-1}$.

The sum of all $2n$ terms is $S_{2n}$.

The sum of terms occupying odd places is $S_{odd}$.

The given condition is:

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To Find:

The common ratio, $r$, of the G.P.

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Solution:

The formula for the sum of the first $k$ terms of a G.P. with first term $A$ and common ratio $R$ is given by:

The sum of all $2n$ terms ($S_{2n}$) of the given G.P. has first term $a$, common ratio $r$, and number of terms $2n$. Using formula (1) (assuming $r \neq 1$):

The terms occupying odd places are the 1st, 3rd, 5th, ..., $(2n-1)$-th terms of the original G.P.

These terms are $a_1, a_3, a_5, \dots, a_{2n-1}$, which are $a, ar^2, ar^4, \dots, ar^{2n-2}$.

This sequence itself forms a G.P. with:

First term, $A_{odd} = a$.

Common ratio, $R_{odd} = \frac{ar^2}{a} = r^2$.

Number of terms, $N_{odd}$. Since there are $2n$ terms in the original G.P., and we are taking terms at odd positions (1, 3, 5, ..., 2n-1), there are $n$ such terms.

The sum of terms in odd places ($S_{odd}$) is calculated using formula (1) with $A=a$, $R=r^2$, and $k=n$ (assuming $r^2 \neq 1$, i.e., $r \neq \pm 1$):

Now, substitute the expressions for $S_{2n}$ from equation (2) and $S_{odd}$ from equation (3) into the given condition $S_{2n} = 5 \cdot S_{odd}$:

We are given that the G.P. consists of an even number of terms ($2n \ge 2$, so $n \ge 1$). Assume the G.P. is non-trivial, i.e., $a \neq 0$.

Consider the cases for $r$:

If $r=1$, the terms are $a, a, \dots, a$ ($2n$ times). $S_{2n} = 2na$. The terms in odd places are $a, a, \dots, a$ ($n$ times). $S_{odd} = na$. The condition $S_{2n} = 5 S_{odd}$ becomes $2na = 5na$. If $a \neq 0$ and $n \ge 1$, then $2 = 5$, which is false. So $r \neq 1$.

If $r=-1$, the terms are $a, -a, a, -a, \dots, a, -a$ ($2n$ terms). $S_{2n} = (a-a) + (a-a) + \dots + (a-a) = 0$. The terms in odd places are $a, a, \dots, a$ ($n$ terms). $S_{odd} = na$. The condition $S_{2n} = 5 S_{odd}$ becomes $0 = 5na$. Since $n \ge 1$, this implies $a=0$. If $a=0$, all terms are 0, $S_{2n}=0$, $S_{odd}=0$, and $0=5 \cdot 0$ holds, but the common ratio is undefined or could be anything in a GP of zeros. We usually consider a non-zero first term. So, assuming $a \neq 0$, $r \neq -1$.

Since $r \neq \pm 1$, we have $r-1 \neq 0$ and $r^2-1 \neq 0$. Also, $r^{2n}-1 \neq 0$ because $r^{2n} \neq 1$ for $r \neq \pm 1$ when $2n$ is even.

We can safely divide both sides of equation (4) by $a$ and $(r^{2n} - 1)$:

$\frac{1}{r - 1} = \frac{5}{r^2 - 1}$

Factor the denominator on the right side using the difference of squares formula $r^2 - 1 = (r-1)(r+1)$:

Multiply both sides of equation (5) by $(r - 1)(r + 1)$. Since $r \neq 1$, $r-1 \neq 0$.

$(r - 1)(r + 1) \cdot \frac{1}{r - 1} = (r - 1)(r + 1) \cdot \frac{5}{(r - 1)(r + 1)}$

Cancel the common term $(r - 1)$ on the left side and $(r - 1)(r + 1)$ on the right side:

$r + 1 = 5$

Subtract 1 from both sides:

$r = 5 - 1$

This value $r=4$ satisfies the assumption $r \neq \pm 1$.

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The common ratio of the G.P. is 4.

Solution:

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Given:

An Arithmetic Progression (A.P.) with first term $a = 11$.

Sum of the first four terms = 56.

Sum of the last four terms = 112.

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To Find:

The number of terms, $n$, in the A.P.

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Solution:

Let the common difference of the A.P. be $d$. The terms of the A.P. are denoted by $a_k$, where $k$ is the term number.

The formula for the $k$-th term of an A.P. is:

We are given the first term $a_1 = a = 11$.

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The first four terms of the A.P. are $a_1, a_2, a_3, a_4$.

Using formula (1) with $a=11$:

$a_1 = 11 + (1-1)d = 11 + 0d = 11$

$a_2 = 11 + (2-1)d = 11 + d$

$a_3 = 11 + (3-1)d = 11 + 2d$

$a_4 = 11 + (4-1)d = 11 + 3d$

The sum of the first four terms is given as 56:

Substitute the expressions for the terms:

$(11) + (11 + d) + (11 + 2d) + (11 + 3d) = 56$

Combine the terms with 11 and the terms with $d$:

$(11 + 11 + 11 + 11) + (d + 2d + 3d) = 56$

$44 + 6d = 56$

Subtract 44 from both sides of the equation:

$6d = 56 - 44$

$6d = 12$

Divide both sides by 6 to find the common difference:

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Now we consider the last four terms of the A.P. If the total number of terms is $n$, the last four terms are $a_{n-3}, a_{n-2}, a_{n-1}, a_n$.

Using formula (1) with $a=11$ and $d=2$:

$a_{n-3} = 11 + (n-3-1)2 = 11 + (n-4)2$

$a_{n-2} = 11 + (n-2-1)2 = 11 + (n-3)2$

$a_{n-1} = 11 + (n-1-1)2 = 11 + (n-2)2$

$a_n = 11 + (n-1)2$

The sum of the last four terms is given as 112:

Substitute the expressions for the terms:

$[11 + (n-4)2] + [11 + (n-3)2] + [11 + (n-2)2] + [11 + (n-1)2] = 112$

Combine the terms with 11 and the terms with 2:

$(11 + 11 + 11 + 11) + [2(n-4) + 2(n-3) + 2(n-2) + 2(n-1)] = 112$

$44 + 2[(n-4) + (n-3) + (n-2) + (n-1)] = 112$

$44 + 2[n - 4 + n - 3 + n - 2 + n - 1] = 112$

$44 + 2[(n + n + n + n) + (-4 - 3 - 2 - 1)] = 112$

$44 + 2[4n - 10] = 112$

Distribute the 2 on the left side:

$44 + 8n - 20 = 112$

Combine the constant terms on the left side:

$(44 - 20) + 8n = 112$

$24 + 8n = 112$

Subtract 24 from both sides:

$8n = 112 - 24$

$8n = 88$

Divide both sides by 8 to find the value of $n$:

$n = \frac{88}{8}$

Since the number of terms is 11, which is greater than 4, the last four terms are well-defined.

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The number of terms in the A.P. is 11.

Given:

The equality: $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$, where $x \neq 0$.

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To Show:

That $a, b, c,$ and $d$ are in Geometric Progression (G.P.).

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Proof:

For $a, b, c, d$ to be in G.P., the ratio of consecutive terms must be constant, i.e., $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$. This is equivalent to showing that $b^2 = ac$ and $c^2 = bd$ (assuming $a, b, c, d$ are non-zero for a non-degenerate G.P.).

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Consider the first equality from the given relation:

Cross-multiply equation (1):

$(a + bx)(b - cx) = (b + cx)(a - bx)$

Expand both sides of the equation:

$ab - acx + b^2x - bcx^2 = ab - b^2x + acx - bcx^2$

Subtract $ab$ from both sides and add $bcx^2$ to both sides:

$- acx + b^2x = - b^2x + acx$

Move the terms involving $b^2x$ to one side and the terms involving $acx$ to the other side:

$b^2x + b^2x = acx + acx$

$2b^2x = 2acx$

Since we are given that $x \neq 0$, we can divide both sides by $2x$:

Equation (2), $b^2 = ac$, is a necessary and sufficient condition for $a, b, c$ to be in G.P., assuming $a, b, c$ are non-zero. It implies that $\frac{b}{a} = \frac{c}{b}$.

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Now, consider the second equality from the given relation:

Cross-multiply equation (3):

$(b + cx)(c - dx) = (c + dx)(b - cx)$

Expand both sides of the equation:

$bc - bdx + c^2x - cdx^2 = bc - c^2x + bdx - cdx^2$

Subtract $bc$ from both sides and add $cdx^2$ to both sides:

$- bdx + c^2x = - c^2x + bdx$

Move the terms involving $c^2x$ to one side and the terms involving $bdx$ to the other side:

$c^2x + c^2x = bdx + bdx$

$2c^2x = 2bdx$

Since $x \neq 0$, we can divide both sides by $2x$:

Equation (4), $c^2 = bd$, is a necessary and sufficient condition for $b, c, d$ to be in G.P., assuming $b, c, d$ are non-zero. It implies that $\frac{c}{b} = \frac{d}{c}$.

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From equation (2), we have $\frac{b}{a} = \frac{c}{b}$ (assuming $a, b$ are non-zero).
From equation (4), we have $\frac{c}{b} = \frac{d}{c}$ (assuming $b, c$ are non-zero).

Combining these two results, we get:

This shows that the ratio of consecutive terms is constant.

Thus, $a, b, c,$ and $d$ are in G.P.

Note: This proof assumes that the initial denominators $(a-bx, b-cx, c-dx)$ and the terms $a, b, c, d$ are non-zero for the ratios and G.P. to be well-defined in the standard sense.

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Hence Shown.

Given:

A Geometric Progression (G.P.) with $n$ terms.

$S$ = Sum of the $n$ terms.

$P$ = Product of the $n$ terms.

$R$ = Sum of the reciprocals of the $n$ terms.

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To Prove:

$P^2 R^n = S^n$

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Proof:

Let the first term of the G.P. be $a$ and the common ratio be $r$. The $n$ terms of the G.P. are $a, ar, ar^2, \dots, ar^{n-1}$.

We assume that $a \neq 0$ and $r \neq 0$, which is standard for a non-degenerate G.P.

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1. Sum of the terms (S):

The sum of the first $n$ terms of a G.P. is given by:

Case 1: If $r = 1$

Case 2: If $r \neq 1$

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2. Product of the terms (P):

The product of the $n$ terms is:

$P = a \cdot (ar) \cdot (ar^2) \cdot \dots \cdot (ar^{n-1})$

$P = a^n \cdot r^{0+1+2+\dots+(n-1)}$

The sum of the exponents of $r$ is the sum of the first $n-1$ non-negative integers: $0+1+2+\dots+(n-1) = \frac{(n-1)n}{2}$.

This formula holds for both $r=1$ (where $r^{\frac{n(n-1)}{2}} = 1$) and $r \neq 1$.

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3. Sum of the reciprocals (R):

The reciprocals of the terms are $\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \dots, \frac{1}{ar^{n-1}}$.

This sequence is also a G.P. with first term $A' = \frac{1}{a}$ and common ratio $R' = \frac{1}{r}$. The number of terms is $n$.

The sum of these reciprocals is $R$.

Case 1: If $r = 1$ (which implies $R' = 1$)

Case 2: If $r \neq 1$ (which implies $R' = \frac{1}{r} \neq 1$)

Using the sum formula for the reciprocal G.P. with first term $\frac{1}{a}$ and common ratio $\frac{1}{r}$:

$R = \frac{1}{a} \frac{(1/r)^n - 1}{(1/r) - 1} = \frac{1}{a} \frac{\frac{1 - r^n}{r^n}}{\frac{1 - r}{r}} = \frac{1}{a} \frac{1 - r^n}{r^n} \cdot \frac{r}{1 - r} = \frac{1}{a} \frac{-(r^n - 1)}{r^{n-1}(-(r - 1))}$

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Now, let's evaluate $P^2 R^n$ and $S^n$ for both cases.

Case 1: $r = 1$

From (1a), $S = na$.

From (2), $P = a^n (1)^{\frac{n(n-1)}{2}} = a^n$.

From (3a), $R = \frac{n}{a}$.

Calculate $P^2 R^n$:

$P^2 R^n = (a^n)^2 \left( \frac{n}{a} \right)^n = a^{2n} \cdot \frac{n^n}{a^n} = a^{2n - n} n^n = a^n n^n = (na)^n$

Calculate $S^n$:

$S^n = (na)^n$

Thus, $P^2 R^n = S^n$ holds for $r=1$.

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Case 2: $r \neq 1$

From (1b), $S = a \frac{r^n - 1}{r - 1}$.

From (2), $P = a^n r^{\frac{n(n-1)}{2}}$.

From (3b), $R = \frac{1}{a} \frac{r^n - 1}{r^{n-1}(r - 1)}$.

Calculate $P^2 R^n$:

$P^2 = \left( a^n r^{\frac{n(n-1)}{2}} \right)^2 = a^{2n} \left( r^{\frac{n(n-1)}{2}} \right)^2 = a^{2n} r^{n(n-1)}$

$R^n = \left( \frac{1}{a} \frac{r^n - 1}{r^{n-1}(r - 1)} \right)^n = \left( \frac{1}{a} \right)^n \left( \frac{r^n - 1}{r^{n-1}(r - 1)} \right)^n = \frac{1}{a^n} \frac{(r^n - 1)^n}{(r^{n-1})^n (r - 1)^n} = \frac{1}{a^n} \frac{(r^n - 1)^n}{r^{n(n-1)} (r - 1)^n}$

$P^2 R^n = \left( a^{2n} r^{n(n-1)} \right) \cdot \left( \frac{1}{a^n} \frac{(r^n - 1)^n}{r^{n(n-1)} (r - 1)^n} \right)$

$P^2 R^n = a^{2n} \cdot \frac{1}{a^n} \cdot r^{n(n-1)} \cdot \frac{1}{r^{n(n-1)}} \cdot \frac{(r^n - 1)^n}{(r - 1)^n}$

$P^2 R^n = a^{2n - n} \cdot r^{n(n-1) - n(n-1)} \cdot \left( \frac{r^n - 1}{r - 1} \right)^n$

$P^2 R^n = a^n \cdot r^0 \cdot \left( \frac{r^n - 1}{r - 1} \right)^n$

$P^2 R^n = a^n \cdot 1 \cdot \left( \frac{r^n - 1}{r - 1} \right)^n$

Calculate $S^n$:

From (1b), $S = a \frac{r^n - 1}{r - 1}$.

Comparing equation (4) and equation (5), we see that $P^2 R^n = S^n$.

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The result holds for both $r=1$ and $r \neq 1$.

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Hence Proved.

Given:

An Arithmetic Progression (A.P.).

The p-th term of the A.P. is $a$.

The q-th term of the A.P. is $b$.

The r-th term of the A.P. is $c$.

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To Show:

$(q – r)a + (r – p)b + (p – q)c = 0$

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Proof:

Let the first term of the A.P. be $A$ and the common difference be $D$.

The formula for the $k$-th term of an A.P. is $T_k = A + (k-1)D$.

Based on the given information, we can write the expressions for the p-th, q-th, and r-th terms:

Consider the expression $(q – r)a + (r – p)b + (p – q)c$. Substitute the expressions for $a$, $b$, and $c$ from equations (1), (2), and (3):

$(q – r)[A + (p-1)D] + (r – p)[A + (q-1)D] + (p – q)[A + (r-1)D]$

Expand each term:

$(q – r)A + (q – r)(p-1)D + (r – p)A + (r – p)(q-1)D + (p – q)A + (p – q)(r-1)D$

Group the terms containing $A$ and the terms containing $D$:

$[ (q – r)A + (r – p)A + (p – q)A ] + [ (q – r)(p-1)D + (r – p)(q-1)D + (p – q)(r-1)D ]$

Factor out $A$ from the first group and $D$ from the second group:

$A[(q – r) + (r – p) + (p – q)] + D[(q – r)(p-1) + (r – p)(q-1) + (p – q)(r-1)]$

Simplify the expression inside the square brackets for the term with $A$:

$(q – r) + (r – p) + (p – q) = q - r + r - p + p - q = (q - q) + (-r + r) + (-p + p) = 0$

So, the first part of the expression is $A[0] = 0$.

Now, simplify the expression inside the square brackets for the term with $D$. Expand each product:

$(q – r)(p-1) = qp - q - rp + r$

$(r – p)(q-1) = rq - r - pq + p$

$(p – q)(r-1) = pr - p - qr + q$

Summing these three expanded terms:

$(qp - q - rp + r) + (rq - r - pq + p) + (pr - p - qr + q)$

Combine like terms:

$ (qp - pq) + (rq - qr) + (pr - rp) + (-q + q) + (-r + r) + (-p + p) $

$ 0 + 0 + 0 + 0 + 0 + 0 = 0$

So, the second part of the expression is $D[0] = 0$.

Therefore, the entire expression is:

$(q – r)a + (r – p)b + (p – q)c = 0 + 0 = 0$

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Hence Showed.

Given:

$a\left( \frac{1}{b}+\frac{1}{c} \right)$, $b\left( \frac{1}{c}+\frac{1}{a} \right)$, and $c\left( \frac{1}{a}+\frac{1}{b} \right)$ are in A.P.

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To Prove:

$a$, $b$, $c$ are in A.P.

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Proof:

Let the given terms be $T_1$, $T_2$, and $T_3$ respectively.

$T_1 = a\left( \frac{1}{b}+\frac{1}{c} \right)$

$T_2 = b\left( \frac{1}{c}+\frac{1}{a} \right)$

$T_3 = c\left( \frac{1}{a}+\frac{1}{b} \right)$

Since $T_1$, $T_2$, $T_3$ are in A.P., the common difference is constant. Thus, $T_2 - T_1 = T_3 - T_2$, which implies $2T_2 = T_1 + T_3$.

Alternatively, we know that if terms are in A.P., adding or subtracting a constant from each term results in a new sequence that is also in A.P.

Consider adding $1$ to each term:

New term 1: $T_1 + 1 = a\left( \frac{1}{b}+\frac{1}{c} \right) + 1 = \frac{a}{b} + \frac{a}{c} + 1 = \frac{ac + ab + bc}{bc}$

New term 2: $T_2 + 1 = b\left( \frac{1}{c}+\frac{1}{a} \right) + 1 = \frac{b}{c} + \frac{b}{a} + 1 = \frac{ab + bc + ac}{ac}$

New term 3: $T_3 + 1 = c\left( \frac{1}{a}+\frac{1}{b} \right) + 1 = \frac{c}{a} + \frac{c}{b} + 1 = \frac{bc + ac + ab}{ab}$

Since the original terms are in A.P., the new terms are also in A.P. Let the new terms be $T'_1, T'_2, T'_3$.

$T'_1 = \frac{ab+bc+ca}{bc}$

$T'_2 = \frac{ab+bc+ca}{ac}$

$T'_3 = \frac{ab+bc+ca}{ab}$

Since $T'_1, T'_2, T'_3$ are in A.P., the common difference is the same:

$T'_2 - T'_1 = T'_3 - T'_2$

$\frac{ab+bc+ca}{ac} - \frac{ab+bc+ca}{bc} = \frac{ab+bc+ca}{ab} - \frac{ab+bc+ca}{ac}$

Assuming $ab+bc+ca \neq 0$, we can divide both sides by $(ab+bc+ca)$:

$\frac{1}{ac} - \frac{1}{bc} = \frac{1}{ab} - \frac{1}{ac}$

Rearranging the terms:

$\frac{1}{ac} + \frac{1}{ac} = \frac{1}{ab} + \frac{1}{bc}$

$\frac{2}{ac} = \frac{c+a}{abc}$

Multiplying both sides by $abc$ (assuming $a, b, c \neq 0$):

$2b = c+a$

or

$a + c = 2b$

This is the condition for $a$, $b$, and $c$ to be in A.P.

Therefore, $a$, $b$, $c$ are in A.P.

Given:

$a, b, c, d$ are in G.P.

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To Prove:

$(a^n + b^n)$, $(b^n + c^n)$, $(c^n + d^n)$ are in G.P.

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Proof:

Since $a, b, c, d$ are in G.P., there exists a common ratio $r$ such that:

$b = ar$

$c = br = (ar)r = ar^2$

$d = cr = (ar^2)r = ar^3$

We need to show that $(a^n + b^n)$, $(b^n + c^n)$, $(c^n + d^n)$ are in G.P.

This is equivalent to showing that the square of the middle term is equal to the product of the first and third terms, i.e., $(b^n + c^n)^2 = (a^n + b^n)(c^n + d^n)$.

Consider the left-hand side (LHS):

LHS = $(b^n + c^n)^2$

Substitute $b = ar$ and $c = ar^2$:

LHS = $((ar)^n + (ar^2)^n)^2$

LHS = $(a^n r^n + a^n r^{2n})^2$

LHS = $(a^n (r^n + r^{2n}))^2$

LHS = $(a^n r^n (1 + r^n))^2$

LHS = $a^{2n} r^{2n} (1 + r^n)^2$

Now consider the right-hand side (RHS):

RHS = $(a^n + b^n)(c^n + d^n)$

Substitute $b = ar$, $c = ar^2$, and $d = ar^3$:

RHS = $(a^n + (ar)^n)((ar^2)^n + (ar^3)^n)$

RHS = $(a^n + a^n r^n)(a^n r^{2n} + a^n r^{3n})$

RHS = $a^n (1 + r^n) \cdot a^n (r^{2n} + r^{3n})$

RHS = $a^{2n} (1 + r^n) r^{2n} (1 + r^n)$

RHS = $a^{2n} r^{2n} (1 + r^n)^2$

Comparing LHS and RHS:

LHS = $a^{2n} r^{2n} (1 + r^n)^2$

RHS = $a^{2n} r^{2n} (1 + r^n)^2$

Thus, LHS = RHS.

This shows that $(b^n + c^n)^2 = (a^n + b^n)(c^n + d^n)$, which is the condition for $(a^n + b^n)$, $(b^n + c^n)$, and $(c^n + d^n)$ to be in G.P.

Therefore, $(a^n + b^n)$, $(b^n + c^n)$, $(c^n + d^n)$ are in G.P.

Given:

$a$ and $b$ are roots of $x^2 - 3x + p = 0$.

$c$ and $d$ are roots of $x^2 - 12x + q = 0$.

$a, b, c, d$ form a G.P.

---

To Prove:

$(q + p) : (q - p) = 17:15$.

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Proof:

Since $a$ and $b$ are the roots of the quadratic equation $x^2 - 3x + p = 0$, by Vieta's formulas, we have:

Since $c$ and $d$ are the roots of the quadratic equation $x^2 - 12x + q = 0$, by Vieta's formulas, we have:

Since $a, b, c, d$ are in G.P., let the common ratio be $r$. Then we can write:

$b = ar$

$c = ar^2$

$d = ar^3$

Substitute these into equations (i) and (iii):

From (i): $a + ar = 3$

From (iii): $ar^2 + ar^3 = 12$

Divide equation (vi) by equation (v) (assuming $a \neq 0$ and $1+r \neq 0$):

$r^2 = 4$

Now, substitute the G.P. terms into equations (ii) and (iv):

From (ii): $p = ab = a(ar) = a^2 r$

From (iv): $q = cd = (ar^2)(ar^3) = a^2 r^5$

We need to find the ratio $(q+p) : (q-p)$. Consider the expression $\frac{q+p}{q-p}$.

Substitute the values of $p$ and $q$ from (viii) and (ix):

Factor out $a^2 r$ from the numerator and denominator (since $a \neq 0$ and $r \neq 0$, $a^2 r \neq 0$):

From equation (vii), $r^2 = 4$. Squaring both sides gives $r^4 = (r^2)^2 = 4^2 = 16$.

Substitute $r^4 = 16$ into equation (x):

Thus, $(q+p) : (q-p) = 17:15$.

This completes the proof.

Given:

$a$ and $b$ are two positive numbers.

Ratio of A.M. to G.M. is $m:n$.

A.M. $= \frac{a+b}{2}$

G.M. $= \sqrt{ab}$

---

To Show:

$a : b = \left( m+\sqrt{m^{2}-n^{2}} \right) : \left( m-\sqrt{m^{2}-n^{2}} \right)$

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Proof:

According to the given information, the ratio of the A.M. and G.M. of $a$ and $b$ is $m:n$.

Since $a$ and $b$ are positive and $a \neq b$ (otherwise A.M. = G.M., $m=n$, and the required ratio would involve $\sqrt{0}$), A.M. > G.M., which implies $m > n$. Thus, $m^2 - n^2 > 0$.

From equation (i), squaring both sides gives:

We know that $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$. Adding $4ab$ to $(a-b)^2$ gives $(a-b)^2 + 4ab = a^2 - 2ab + b^2 + 4ab = a^2 + 2ab + b^2 = (a+b)^2$. So, $(a+b)^2 - 4ab = (a-b)^2$.

From $\frac{(a+b)^2}{4ab} = \frac{m^2}{n^2}$, subtract 1 from both sides:

$\frac{(a+b)^2}{4ab} - 1 = \frac{m^2}{n^2} - 1$

$\frac{(a+b)^2 - 4ab}{4ab} = \frac{m^2 - n^2}{n^2}$

Substitute $(a+b)^2 - 4ab = (a-b)^2$:

Taking the square root of both sides (since $a,b$ are positive, $\sqrt{ab}$ is positive, and assuming $a>b$, $a-b$ is positive, so we take the positive root on the RHS):

Now, divide equation (i) by equation (ii):

Now, apply Componendo and Dividendo to $\frac{a+b}{a-b} = \frac{m}{\sqrt{m^2 - n^2}}$:

Simplifying the left-hand side:

$\frac{a+b+a-b}{a+b-a+b} = \frac{2a}{2b} = \frac{a}{b}$

So, we get:

This can be written in ratio form as:

$a : b = \left( m+\sqrt{m^{2}-n^{2}} \right) : \left( m-\sqrt{m^{2}-n^{2}} \right)$

Hence, proved.

Given:

1. $a, b, c$ are in A.P.

2. $b, c, d$ are in G.P.

3. $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in A.P.

---

To Prove:

$a, c, e$ are in G.P.

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Proof:

Since $a, b, c$ are in A.P., we have:

Since $b, c, d$ are in G.P., we have:

Since $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in A.P., we have:

Taking the reciprocal of both sides (assuming $d, c, e, c+e$ are non-zero):

Multiplying by 2:

Substitute the value of $b$ from (i) into (ii): $b = \frac{a+c}{2}$.

Now substitute the value of $d$ from (iii) into (iv):

Simplify the right-hand side:

Assuming $c \neq 0$, divide both sides by $c$:

Multiply both sides by $(c+e)$ (assuming $c+e \neq 0$):

Expand both sides:

Subtract $ce$ from both sides:

The condition $c^2 = ae$ implies that $a, c, e$ are in G.P.

Therefore, $a, c, e$ are in G.P.

(i) 5 + 55 + 555 + … up to n terms

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To Find:

The sum of the series $5 + 55 + 555 + \dots$ up to $n$ terms.

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Solution:

The given series is $5 + 55 + 555 + \dots$

Let the $k$-th term of the series be $a_k$. The terms are formed by repeating the digit 5.

The $k$-th term can be written as:

$a_k = 5 \times \underbrace{11\dots1}_{k \text{ times}}$

The number $\underbrace{11\dots1}_{k \text{ times}}$ can be expressed as $\frac{10^k - 1}{9}$.

So, the $k$-th term is $a_k = 5 \left( \frac{10^k - 1}{9} \right) = \frac{5}{9}(10^k - 1)$.

The sum of the series up to $n$ terms, $S_n$, is the sum of the first $n$ terms:

$S_n = \sum_{k=1}^{n} a_k = \sum\limits_{k=1}^{n} \frac{5}{9}(10^k - 1)$

$S_n = \frac{5}{9} \sum\limits_{k=1}^{n} (10^k - 1)$

$S_n = \frac{5}{9} \left( \sum\limits_{k=1}^{n} 10^k - \sum\limits_{k=1}^{n} 1 \right)$

The first part, $\sum\limits_{k=1}^{n} 10^k$, is the sum of a geometric series $10^1 + 10^2 + \dots + 10^n$.

This G.P. has the first term $a = 10$, the common ratio $r = 10$, and $n$ terms.

The sum of a G.P. is given by the formula $\frac{a(r^n - 1)}{r-1}$.

So, $\sum\limits_{k=1}^{n} 10^k = \frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9}$.

The second part, $\sum\limits_{k=1}^{n} 1$, is the sum of $n$ terms, each equal to 1. This sum is $n$.

Substitute these values back into the expression for $S_n$:

$S_n = \frac{5}{9} \left( \frac{10(10^n - 1)}{9} - n \right)$

$S_n = \frac{5}{9} \left( \frac{10^{n+1} - 10}{9} - \frac{9n}{9} \right)$

$S_n = \frac{5}{9} \left( \frac{10^{n+1} - 10 - 9n}{9} \right)$

$S_n = \frac{5}{81} (10^{n+1} - 9n - 10)$

Thus, the sum of the series $5 + 55 + 555 + \dots$ up to $n$ terms is $\frac{5}{81} (10^{n+1} - 9n - 10)$.

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(ii) .6 + .66 + .666 + … up to n terms

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To Find:

The sum of the series $0.6 + 0.66 + 0.666 + \dots$ up to $n$ terms.

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Solution:

The given series is $0.6 + 0.66 + 0.666 + \dots$

Let the $k$-th term of the series be $a_k$. The terms are formed by repeating the digit 6 after the decimal point.

The $k$-th term can be written as:

$a_k = 0.\underbrace{66\dots6}_{k \text{ times}}$

We can express this term as:

$a_k = 6 \times 0.\underbrace{11\dots1}_{k \text{ times}}$

The number $0.\underbrace{11\dots1}_{k \text{ times}}$ is the sum of a geometric series:

$0.\underbrace{11\dots1}_{k \text{ times}} = \frac{1}{10} + \frac{1}{100} + \dots + \frac{1}{10^k}$

This is a G.P. with the first term $a = \frac{1}{10}$, the common ratio $r = \frac{1}{10}$, and $k$ terms.

The sum of this G.P. is $\frac{a(1-r^k)}{1-r}$.

Sum $= \frac{\frac{1}{10} \left( 1 - \left(\frac{1}{10}\right)^k \right)}{1 - \frac{1}{10}} = \frac{\frac{1}{10} (1 - 10^{-k})}{\frac{9}{10}} = \frac{1}{9} (1 - 10^{-k})$.

So, the $k$-th term is $a_k = 6 \times \frac{1}{9} (1 - 10^{-k}) = \frac{2}{3}(1 - 10^{-k})$.

The sum of the series up to $n$ terms, $S_n$, is the sum of the first $n$ terms:

$S_n = \sum_{k=1}^{n} a_k = \sum\limits_{k=1}^{n} \frac{2}{3}(1 - 10^{-k})$

$S_n = \frac{2}{3} \sum\limits_{k=1}^{n} (1 - 10^{-k})$

$S_n = \frac{2}{3} \left( \sum\limits_{k=1}^{n} 1 - \sum\limits_{k=1}^{n} 10^{-k} \right)$

The first part, $\sum\limits_{k=1}^{n} 1$, is the sum of $n$ terms, each equal to 1. This sum is $n$.

The second part, $\sum\limits_{k=1}^{n} 10^{-k}$, is the sum of a geometric series $10^{-1} + 10^{-2} + \dots + 10^{-n}$.

This G.P. has the first term $a = 10^{-1} = \frac{1}{10}$, the common ratio $r = 10^{-1} = \frac{1}{10}$, and $n$ terms.

The sum of this G.P. is $\frac{a(1-r^n)}{1-r}$.

$\sum\limits_{k=1}^{n} 10^{-k} = \frac{\frac{1}{10} \left( 1 - \left(\frac{1}{10}\right)^n \right)}{1 - \frac{1}{10}} = \frac{\frac{1}{10} (1 - 10^{-n})}{\frac{9}{10}} = \frac{1}{9} (1 - 10^{-n})$.

Substitute these values back into the expression for $S_n$:

$S_n = \frac{2}{3} \left( n - \frac{1}{9}(1 - 10^{-n}) \right)$

$S_n = \frac{2n}{3} - \frac{2}{27}(1 - 10^{-n})$

$S_n = \frac{2n}{3} - \frac{2}{27} + \frac{2}{27 \cdot 10^n}$

$S_n = \frac{18n}{27} - \frac{2}{27} + \frac{2}{27 \cdot 10^n}$

$S_n = \frac{18n - 2 + 2 \cdot 10^{-n}}{27}$

$S_n = \frac{2}{27} (9n - 1 + 10^{-n})$

Thus, the sum of the series $0.6 + 0.66 + 0.666 + \dots$ up to $n$ terms is $\frac{2}{3}n - \frac{2}{27}(1 - 10^{-n})$ or $\frac{2}{27} (9n - 1 + 10^{-n})$.

Given:

The series is $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$

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To Find:

The 20th term of the given series.

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Solution:

Let the given series be $S = 2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$

Let the $n$-th term of the series be denoted by $T_n$.

Observe the pattern of the first factor in each term:

The first factors are $2, 4, 6, \dots$

This sequence is an Arithmetic Progression (A.P.) with the first term $a_1 = 2$ and the common difference $d = 4 - 2 = 2$.

The $n$-th term of this A.P. is given by $a_n = a_1 + (n-1)d = 2 + (n-1)2 = 2 + 2n - 2 = 2n$.

Observe the pattern of the second factor in each term:

The second factors are $4, 6, 8, \dots$

This sequence is also an Arithmetic Progression (A.P.) with the first term $b_1 = 4$ and the common difference $d = 6 - 4 = 2$.

The $n$-th term of this A.P. is given by $b_n = b_1 + (n-1)d = 4 + (n-1)2 = 4 + 2n - 2 = 2n + 2$.

The $n$-th term of the given series is the product of the $n$-th term of the first sequence and the $n$-th term of the second sequence.

$T_n = (2n)(2n + 2)$

We need to find the 20th term of the series, so we need to find $T_{20}$.

Substitute $n=20$ into the expression for $T_n$:

$T_{20} = (2 \times 20)(2 \times 20 + 2)$

$T_{20} = (40)(40 + 2)$

$T_{20} = 40 \times 42$

Calculate the product:

$40 \times 42 = 1680$

Therefore, the 20th term of the series is 1680.

Given:

The series is $3 + 7 + 13 + 21 + 31 + \dots$

---

To Find:

The sum of the first $n$ terms of the series.

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Solution:

Let the terms of the series be $a_1, a_2, a_3, \dots, a_n$.

The series is $3, 7, 13, 21, 31, \dots$

Let's find the differences between consecutive terms:

$a_2 - a_1 = 7 - 3 = 4$

$a_3 - a_2 = 13 - 7 = 6$

$a_4 - a_3 = 21 - 13 = 8$

$a_5 - a_4 = 31 - 21 = 10$

The sequence of these first differences is $4, 6, 8, 10, \dots$ which is an A.P.

Now let's find the differences between the consecutive first differences:

$6 - 4 = 2$

$8 - 6 = 2$

$10 - 8 = 2$

Since the second differences are constant, the $n$-th term of the series is a quadratic expression in $n$. Let the $n$-th term be $a_n = An^2 + Bn + C$.

For $n=1$, $a_1 = A(1)^2 + B(1) + C = A + B + C = 3$

For $n=2$, $a_2 = A(2)^2 + B(2) + C = 4A + 2B + C = 7$

For $n=3$, $a_3 = A(3)^2 + B(3) + C = 9A + 3B + C = 13$

Subtracting the first equation from the second:

$(4A + 2B + C) - (A + B + C) = 7 - 3$

$3A + B = 4$

Subtracting the second equation from the third:

$(9A + 3B + C) - (4A + 2B + C) = 13 - 7$

$5A + B = 6$

Subtracting $(3A + B = 4)$ from $(5A + B = 6)$:

$(5A + B) - (3A + B) = 6 - 4$

$2A = 2$

$A = 1$

Substitute $A=1$ into $3A + B = 4$:

$3(1) + B = 4$

$3 + B = 4$

$B = 1$

Substitute $A=1$ and $B=1$ into $A + B + C = 3$:

$1 + 1 + C = 3$

$2 + C = 3$

$C = 1$

So, the $n$-th term of the series is $a_n = 1n^2 + 1n + 1 = n^2 + n + 1$.

The sum of the first $n$ terms, $S_n$, is given by $\sum\limits_{k=1}^{n} a_k$.

$S_n = \sum\limits_{k=1}^{n} (k^2 + k + 1)$

$S_n = \sum\limits_{k=1}^{n} k^2 + \sum\limits_{k=1}^{n} k + \sum\limits_{k=1}^{n} 1$

Using the standard summation formulas:

$\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$

$\sum\limits_{k=1}^{n} 1 = n$

Substitute these formulas into the expression for $S_n$:

$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + n$

Combine the terms with a common denominator of 6:

$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6} + \frac{6n}{6}$

$S_n = \frac{n[(n+1)(2n+1) + 3(n+1) + 6]}{6}$

Expand the terms inside the square brackets:

$(n+1)(2n+1) = 2n^2 + n + 2n + 1 = 2n^2 + 3n + 1$

$3(n+1) = 3n + 3$

$S_n = \frac{n[2n^2 + 3n + 1 + 3n + 3 + 6]}{6}$

$S_n = \frac{n[2n^2 + 6n + 10]}{6}$

Factor out 2 from the terms in the bracket:

$S_n = \frac{n \cdot 2(n^2 + 3n + 5)}{6}$

Cancel the common factor of 2 in the numerator and denominator:

$S_n = \frac{n(n^2 + 3n + 5)}{3}$

Thus, the sum of the first $n$ terms of the series is $\frac{n(n^2 + 3n + 5)}{3}$.

Given:

$S_1$ is the sum of the first $n$ natural numbers.

$S_2$ is the sum of the squares of the first $n$ natural numbers.

$S_3$ is the sum of the cubes of the first $n$ natural numbers.

---

To Show:

$9S_2^2 = S_3 (1 + 8S_1)$

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Proof:

The formulas for the sum of the first $n$ natural numbers, squares, and cubes are:

From (i) and (iii), we notice that $S_3 = S_1^2$.

Consider the Left Hand Side (LHS) of the equation we need to show:

LHS = $9S_2^2$

Substitute the formula for $S_2$ from (ii):

Now consider the Right Hand Side (RHS) of the equation:

RHS = $S_3 (1 + 8S_1)$

Substitute the formulas for $S_3$ from (iii) and $S_1$ from (i):

Recognize that $1 + 4n^2 + 4n$ is a perfect square, $(2n+1)^2$.

Comparing equation (iv) and equation (v), we see that LHS = RHS.

Thus, $9S_2^2 = S_3 (1 + 8S_1)$.

Hence, shown.

Given:

The series is $\frac{1^{3}}{1} + \frac{1^{3}\;+\;2^{3}}{1\;+\;3} + \frac{1^{3}\;+\;2^{3}\;+\;3^{3}}{1\;+\;3\;+\;5}+...$

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To Find:

The sum of the first $n$ terms of the series.

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Solution:

Let the $k$-th term of the series be denoted by $a_k$.

The numerator of the $k$-th term is the sum of the cubes of the first $k$ natural numbers:

Numerator $= 1^3 + 2^3 + \dots + k^3 = \sum\limits_{i=1}^{k} i^3$

Using the formula for the sum of the first $k$ cubes, $\sum\limits_{i=1}^{k} i^3 = \left(\frac{k(k+1)}{2}\right)^2$:

Numerator $= \left(\frac{k(k+1)}{2}\right)^2 = \frac{k^2(k+1)^2}{4}$

The denominator of the $k$-th term is the sum of the first $k$ odd natural numbers:

Denominator $= 1 + 3 + 5 + \dots$ (up to $k$ terms)

This is an A.P. with the first term $1$ and common difference $2$. The $k$-th term is $1 + (k-1)2 = 1 + 2k - 2 = 2k - 1$.

The sum of an A.P. is $\frac{\text{number of terms}}{2} \times (\text{first term} + \text{last term})$.

Denominator $= \frac{k}{2}(1 + (2k-1)) = \frac{k}{2}(2k) = k^2$.

Alternatively, the sum of the first $k$ odd numbers is $k^2$.

So, the $k$-th term $a_k$ is:

$a_k = \frac{\text{Numerator}}{\text{Denominator}} = \frac{\frac{k^2(k+1)^2}{4}}{k^2}$

Assuming $k \neq 0$, we can simplify this expression:

$a_k = \frac{k^2(k+1)^2}{4k^2} = \frac{(k+1)^2}{4}$

The sum of the first $n$ terms, $S_n$, is the sum of the first $n$ terms $a_k$:

$S_n = \sum\limits_{k=1}^{n} a_k = \sum\limits_{k=1}^{n} \frac{(k+1)^2}{4}$

$S_n = \frac{1}{4} \sum\limits_{k=1}^{n} (k+1)^2$

Expand $(k+1)^2 = k^2 + 2k + 1$:

$S_n = \frac{1}{4} \sum\limits_{k=1}^{n} (k^2 + 2k + 1)$

$S_n = \frac{1}{4} \left( \sum\limits_{k=1}^{n} k^2 + 2\sum\limits_{k=1}^{n} k + \sum\limits_{k=1}^{n} 1 \right)$

Use the standard summation formulas:

$\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$

$\sum\limits_{k=1}^{n} 1 = n$

Substitute these formulas into the expression for $S_n$:

$S_n = \frac{1}{4} \left( \frac{n(n+1)(2n+1)}{6} + 2\frac{n(n+1)}{2} + n \right)$

$S_n = \frac{1}{4} \left( \frac{n(n+1)(2n+1)}{6} + n(n+1) + n \right)$

Combine the terms inside the bracket with a common denominator of 6:

$S_n = \frac{1}{4} \left( \frac{n(n+1)(2n+1) + 6n(n+1) + 6n}{6} \right)$

$S_n = \frac{n[(n+1)(2n+1) + 6(n+1) + 6]}{24}$

Expand the terms in the bracket:

$(n+1)(2n+1) = 2n^2 + 2n + n + 1 = 2n^2 + 3n + 1$

$6(n+1) = 6n + 6$

$S_n = \frac{n[2n^2 + 3n + 1 + 6n + 6 + 6]}{24}$

$S_n = \frac{n[2n^2 + 9n + 13]}{24}$

Thus, the sum of the first $n$ terms of the series is $\frac{n(2n^2 + 9n + 13)}{24}$.

Given:

The series in the numerator is $1 \times 2^2 + 2 \times 3^2 + 3 \times 4^2 + \dots + n \times (n+1)^2$.

The series in the denominator is $1^2 \times 2 + 2^2 \times 3 + 3^2 \times 4 + \dots + n^2 \times (n+1)$.

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To Show:

$\frac{1 \times 2^{2}\;+\;2 \times 3^{2}\;+\;...\;+\; n \times (n+1)^{2}}{1^{2}\times 2 \;+\; 2^{2} \times 3\;+\;...\;+\; n^{2} \times (n + 1)} = \frac{3n \;+\; 5}{3n \;+\; 1}$

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Proof:

Let the numerator series be $N$ and the denominator series be $D$.

The $k$-th term of the numerator series is $T_N(k) = k \times (k+1)^2$.

$T_N(k) = k(k^2 + 2k + 1) = k^3 + 2k^2 + k$.

The sum of the numerator series up to $n$ terms is $S_N = \sum\limits_{k=1}^{n} T_N(k)$:

$S_N = \sum\limits_{k=1}^{n} (k^3 + 2k^2 + k)$

$S_N = \sum\limits_{k=1}^{n} k^3 + 2\sum\limits_{k=1}^{n} k^2 + \sum\limits_{k=1}^{n} k$

Using the standard formulas for the sum of powers of natural numbers:

$\sum\limits_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{4}$

$\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

$\sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}$

Substitute these into the expression for $S_N$:

$S_N = \frac{n^2(n+1)^2}{4} + 2 \left( \frac{n(n+1)(2n+1)}{6} \right) + \frac{n(n+1)}{2}$

$S_N = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$

Find a common denominator, which is 12:

$S_N = \frac{3n^2(n+1)^2}{12} + \frac{4n(n+1)(2n+1)}{12} + \frac{6n(n+1)}{12}$

Factor out $\frac{n(n+1)}{12}$:

$S_N = \frac{n(n+1)}{12} [3n(n+1) + 4(2n+1) + 6]$

$S_N = \frac{n(n+1)}{12} [3n^2 + 3n + 8n + 4 + 6]$

$S_N = \frac{n(n+1)}{12} [3n^2 + 11n + 10]$

Factor the quadratic $3n^2 + 11n + 10 = (3n+5)(n+2)$:

$S_N = \frac{n(n+1)(n+2)(3n+5)}{12}$

The $k$-th term of the denominator series is $T_D(k) = k^2 \times (k+1)$.

$T_D(k) = k^3 + k^2$.

The sum of the denominator series up to $n$ terms is $S_D = \sum\limits_{k=1}^{n} T_D(k)$:

$S_D = \sum\limits_{k=1}^{n} (k^3 + k^2)$

$S_D = \sum\limits_{k=1}^{n} k^3 + \sum\limits_{k=1}^{n} k^2$

Substitute the formulas for the sums of powers:

$S_D = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6}$

Find a common denominator, which is 12:

$S_D = \frac{3n^2(n+1)^2}{12} + \frac{2n(n+1)(2n+1)}{12}$

Factor out $\frac{n(n+1)}{12}$:

$S_D = \frac{n(n+1)}{12} [3n(n+1) + 2(2n+1)]$

$S_D = \frac{n(n+1)}{12} [3n^2 + 3n + 4n + 2]$

$S_D = \frac{n(n+1)}{12} [3n^2 + 7n + 2]$

Factor the quadratic $3n^2 + 7n + 2 = (3n+1)(n+2)$:

$S_D = \frac{n(n+1)(n+2)(3n+1)}{12}$

Now, consider the ratio $\frac{S_N}{S_D}$:

$\frac{S_N}{S_D} = \frac{\frac{n(n+1)(n+2)(3n+5)}{12}}{\frac{n(n+1)(n+2)(3n+1)}{12}}$

For $n \geq 1$, the term $\frac{n(n+1)(n+2)}{12}$ is non-zero. We can cancel it from the numerator and the denominator:

$\frac{S_N}{S_D} = \frac{(3n+5)}{(3n+1)}$

Thus, $\frac{1 \times 2^{2}\;+\;2 \times 3^{2}\;+\;...\;+\; n \times (n+1)^{2}}{1^{2}\times 2 \;+\; 2^{2} \times 3\;+\;...\;+\; n^{2} \times (n + 1)} = \frac{3n \;+\; 5}{3n \;+\; 1}$.

Hence, shown.

Given:

Cost of the tractor = $\textsf{₹ } 12000$

Cash payment = $\textsf{₹ } 6000$

Annual principal instalment = $\textsf{₹ } 500$

Interest rate = 12% per annum on the unpaid amount.

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To Find:

The total cost of the tractor (including cash payment and total interest).

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Solution:

The balance amount to be paid in instalments is:

Balance Amount = Cost of Tractor - Cash Payment

Balance Amount = $\textsf{₹ } 12000 - \textsf{₹ } 6000 = \textsf{₹ } 6000$.

The balance amount is paid in annual principal instalments of $\textsf{₹ } 500$.

Number of instalments = $\frac{\text{Balance Amount}}{\text{Annual Principal Instalment}}$

Number of instalments = $\frac{6000}{500} = 12$ instalments.

Interest is paid at 12% per annum on the unpaid amount at the beginning of each year.

The unpaid amounts at the beginning of each year (before the principal payment for that year) are:

Year 1: $\textsf{₹ } 6000$

Year 2: $\textsf{₹ } 6000 - \textsf{₹ } 500 = \textsf{₹ } 5500$

Year 3: $\textsf{₹ } 5500 - \textsf{₹ } 500 = \textsf{₹ } 5000$

$\dots$

Year 12: $\textsf{₹ } 500$ (after 11 payments of $\textsf{₹ } 500$ each, the unpaid amount is $6000 - 11 \times 500 = 6000 - 5500 = 500$).

The unpaid amounts form an Arithmetic Progression:

$6000, 5500, 5000, \dots, 500$.

This A.P. has the first term $a = 6000$, common difference $d = -500$, and the number of terms $n = 12$. The last term is $l = 500$.

The total interest paid is 12% of the sum of these unpaid amounts over the 12 years.

Sum of Unpaid Amounts ($S_{12}$) = $\frac{n}{2}(a+l)$

$S_{12} = \frac{12}{2}(6000 + 500) = 6(6500) = \textsf{₹ } 39000$.

Total Interest Paid = 12% of the Sum of Unpaid Amounts

Total Interest Paid = $\frac{12}{100} \times 39000$

Total Interest Paid = $12 \times 390 = \textsf{₹ } 4680$.

The total cost of the tractor is the sum of the cash payment, the balance principal paid, and the total interest paid.

Total Cost = Cash Payment + Balance Amount + Total Interest Paid

Total Cost = $\textsf{₹ } 6000 + \textsf{₹ } 6000 + \textsf{₹ } 4680$

Total Cost = $\textsf{₹ } 12000 + \textsf{₹ } 4680 = \textsf{₹ } 16680$.

The tractor will cost him $\textsf{₹ } 16680$.

Given:

Cost of the scooter = $\textsf{₹ } 22000$

Cash payment = $\textsf{₹ } 4000$

Annual principal instalment = $\textsf{₹ } 1000$

Interest rate = 10% per annum on the unpaid amount.

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To Find:

The total cost of the scooter (including cash payment and total interest).

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Solution:

The balance amount to be paid in instalments is:

Balance Amount = Cost of Scooter - Cash Payment

Balance Amount = $\textsf{₹ } 22000 - \textsf{₹ } 4000 = \textsf{₹ } 18000$.

The balance amount is paid in annual principal instalments of $\textsf{₹ } 1000$.

Number of instalments = $\frac{\text{Balance Amount}}{\text{Annual Principal Instalment}}$

Number of instalments = $\frac{18000}{1000} = 18$ instalments.

Interest is paid at 10% per annum on the unpaid amount at the beginning of each year.

The unpaid amounts at the beginning of each year (before the principal payment for that year) are:

Year 1: $\textsf{₹ } 18000$

Year 2: $\textsf{₹ } 18000 - \textsf{₹ } 1000 = \textsf{₹ } 17000$

Year 3: $\textsf{₹ } 17000 - \textsf{₹ } 1000 = \textsf{₹ } 16000$

$\dots$

Year 18: $\textsf{₹ } 18000 - (17 \times \textsf{₹ } 1000) = \textsf{₹ } 18000 - \textsf{₹ } 17000 = \textsf{₹ } 1000$.

The unpaid amounts form an Arithmetic Progression:

$18000, 17000, 16000, \dots, 1000$.

This A.P. has the first term $a = 18000$, common difference $d = -1000$, and the number of terms $n = 18$. The last term is $l = 1000$.

The total interest paid is 10% of the sum of these unpaid amounts over the 18 years.

Sum of Unpaid Amounts ($S_{18}$) = $\frac{n}{2}(a+l)$

$S_{18} = \frac{18}{2}(18000 + 1000) = 9(19000) = \textsf{₹ } 171000$.

Total Interest Paid = 10% of the Sum of Unpaid Amounts

Total Interest Paid = $\frac{10}{100} \times 171000$

Total Interest Paid = $\frac{1}{10} \times 171000 = \textsf{₹ } 17100$.

The total cost of the scooter is the sum of the cash payment, the balance principal paid, and the total interest paid.

Total Cost = Cash Payment + Balance Amount + Total Interest Paid

Total Cost = $\textsf{₹ } 4000 + \textsf{₹ } 18000 + \textsf{₹ } 17100$

Total Cost = $\textsf{₹ } 22000 + \textsf{₹ } 17100 = \textsf{₹ } 39100$.

The scooter will cost him $\textsf{₹ } 39100$.

Given:

Initial letters sent = 4

Each recipient sends letters to 4 new persons.

Cost per letter = 50 paise = $\textsf{₹ } 0.50$.

The chain is not broken.

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To Find:

The total amount spent on postage when the 8th set of letters is mailed.

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Solution:

The number of letters mailed at each set forms a geometric progression (G.P.).

In the 1st set, the initial person mails 4 letters.

Number of letters in Set 1: $4 = 4^1$

In the 2nd set, each of the 4 recipients mails 4 letters, so $4 \times 4 = 16$ letters are mailed.

Number of letters in Set 2: $16 = 4^2$

In the 3rd set, each of the 16 recipients mails 4 letters, so $16 \times 4 = 64$ letters are mailed.

Number of letters in Set 3: $64 = 4^3$

Continuing this pattern, the number of letters mailed in the $k$-th set is $4^k$.

We need to find the total amount spent on postage when the 8th set of letters is mailed. This implies finding the total cost of postage for all sets from the 1st up to and including the 8th set.

The total number of letters mailed up to the 8th set is the sum of the number of letters in each set from 1 to 8.

Total Letters = (Letters in Set 1) + (Letters in Set 2) + ... + (Letters in Set 8)

Total Letters = $4^1 + 4^2 + 4^3 + \dots + 4^8$

This is a geometric series with:

First term, $a = 4$

Common ratio, $r = 4$

Number of terms, $n = 8$

The sum of a G.P. is $S_n = \frac{a(r^n - 1)}{r-1}$.

Total Letters ($S_8$) = $\frac{4(4^8 - 1)}{4 - 1}$

$S_8 = \frac{4(4^8 - 1)}{3}$

Calculate $4^8$:

$4^8 = (4^4)^2 = (256)^2 = 65536$

$S_8 = \frac{4(65536 - 1)}{3}$

$S_8 = \frac{4 \times 65535}{3}$

Divide 65535 by 3:

$S_8 = 4 \times 21845 = 87380$ letters.

The total number of letters mailed up to the 8th set is 87380.

The cost of mailing one letter is 50 paise or $\textsf{₹ } 0.50$ or $\textsf{₹ } \frac{1}{2}$.

Total amount spent on postage = Total Letters $\times$ Cost per letter

Total amount spent = $87380 \times \textsf{₹ } 0.50$

Total amount spent = $87380 \times \frac{1}{2}$

Total amount spent = $43690$

The total amount spent on postage when the 8th set of letters is mailed is $\textsf{₹ } 43690$.

Note: If the question was interpreted as the cost of mailing only the 8th set of letters, the calculation would be $4^8 \times \textsf{₹ } 0.50 = 65536 \times \textsf{₹ } 0.50 = \textsf{₹ } 32768$. However, the phrase "when 8th set is mailed" typically includes all mailings leading up to and including the 8th set in such problems.

Given:

Principal amount (P) = $\textsf{₹ } 10000$

Rate of simple interest (R) = 5% per annum.

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To Find:

1. The amount in the 15th year (interpreted as the amount at the end of the 15th year).

2. The total amount after 20 years (interpreted as the amount at the end of the 20th year).

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Solution:

Simple interest is calculated on the original principal amount each year.

Annual Simple Interest = $\frac{P \times R \times T}{100}$ where T = 1 year.

This is the interest earned each year.

Amount in the 15th year:

The interest earned in 15 years is the annual interest multiplied by the number of years.

Total Interest after 15 years = Annual Interest $\times$ 15

Total Interest after 15 years = $\textsf{₹ } 500 \times 15 = \textsf{₹ } 7500$.

The amount at the end of the 15th year is the sum of the principal and the total interest earned up to that year.

Amount after 15 years = Principal + Total Interest after 15 years

Amount after 15 years = $\textsf{₹ } 10000 + \textsf{₹ } 7500 = \textsf{₹ } 17500$.

Total amount after 20 years:

The interest earned in 20 years is the annual interest multiplied by the number of years.

Total Interest after 20 years = Annual Interest $\times$ 20

Total Interest after 20 years = $\textsf{₹ } 500 \times 20 = \textsf{₹ } 10000$.

The total amount after 20 years is the sum of the principal and the total interest earned up to that year.

Amount after 20 years = Principal + Total Interest after 20 years

Amount after 20 years = $\textsf{₹ } 10000 + \textsf{₹ } 10000 = \textsf{₹ } 20000$.

The amount in the 15th year is $\textsf{₹ } 17500$.

The total amount after 20 years is $\textsf{₹ } 20000$.

Given:

Initial cost of the machine (Principal, P) = $\textsf{₹ } 15625$

Annual depreciation rate (r) = 20% = $\frac{20}{100} = 0.20$

Time period (n) = 5 years.

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To Find:

The estimated value of the machine at the end of 5 years.

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Solution:

The value of the machine depreciates each year by a fixed percentage of its value at the beginning of that year. This is a case of value decreasing over time, similar to compound interest but with a negative rate.

The formula for the value of an asset after $n$ years with an initial value $P$ and an annual depreciation rate $r$ is:

$V_n = P(1 - r)^n$

Here, $P = \textsf{₹ } 15625$, $r = 0.20$, and $n = 5$.

$V_5 = 15625 (1 - 0.20)^5$

$V_5 = 15625 (0.80)^5$

$V_5 = 15625 \left(\frac{80}{100}\right)^5$

$V_5 = 15625 \left(\frac{4}{5}\right)^5$

Calculate the powers:

$4^5 = 4 \times 4 \times 4 \times 4 \times 4 = 1024$

$5^5 = 5 \times 5 \times 5 \times 5 \times 5 = 3125$

Substitute these values into the expression for $V_5$:

$V_5 = 15625 \times \frac{1024}{3125}$

We notice that $15625 = 5 \times 3125$. So, $\frac{15625}{3125} = 5$.

$V_5 = 5 \times 1024$

$V_5 = 5120$

The estimated value of the machine at the end of 5 years is $\textsf{₹ } 5120$.