Chapter 5 Arithmetic Progressions

Given:

The Arithmetic Progression (AP) is: $\frac{3}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}, \dots$

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To Find:

The first term ($a$) and the common difference ($d$).

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Solution:

The first term ($a$) of an AP is the first number in the sequence.

In this AP, the first term is $a = \frac{3}{2}$.

The common difference ($d$) is the difference between any term and its preceding term.

We can find the common difference by subtracting the first term from the second term:

$d = a_2 - a_1 = \frac{1}{2} - \frac{3}{2}$

$d = \frac{1 - 3}{2}$

$d = \frac{-2}{2}$

$d = -1$

We can verify this by subtracting the second term from the third term:

$d = a_3 - a_2 = -\frac{1}{2} - \frac{1}{2}$

$d = \frac{-1 - 1}{2}$

$d = \frac{-2}{2}$

$d = -1$

The common difference is consistent.

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Final Answer:

For the given AP, the first term $a = \mathbf{\frac{3}{2}}$ and the common difference $d = \mathbf{-1}$.

General Concept: Arithmetic Progression (AP)

A list of numbers $a_1, a_2, a_3, \dots$ is an Arithmetic Progression (AP) if the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by $d$.

$d = a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = \dots$

If a list forms an AP, the next term ($a_{n+1}$) is found by adding the common difference ($d$) to the last known term ($a_n$).

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Solution (i): 4, 10, 16, 22, ...

Given list: 4, 10, 16, 22, ...

Let the terms be $a_1 = 4, a_2 = 10, a_3 = 16, a_4 = 22$.

Calculate the differences between consecutive terms:

$a_2 - a_1 = 10 - 4 = 6$

$a_3 - a_2 = 16 - 10 = 6$

$a_4 - a_3 = 22 - 16 = 6$

Since the difference between consecutive terms is constant (6), the list of numbers forms an AP.

The common difference $d = 6$.

The last given term is $a_4 = 22$.

The next term ($a_5$) = $a_4 + d = 22 + 6 = 28$.

The term after that ($a_6$) = $a_5 + d = 28 + 6 = 34$.

Conclusion: The list forms an AP with a common difference of 6. The next two terms are 28 and 34.

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Solution (ii): 1, -1, -3, -5, ...

Given list: 1, -1, -3, -5, ...

Let the terms be $a_1 = 1, a_2 = -1, a_3 = -3, a_4 = -5$.

Calculate the differences between consecutive terms:

$a_2 - a_1 = -1 - 1 = -2$

$a_3 - a_2 = -3 - (-1) = -3 + 1 = -2$

$a_4 - a_3 = -5 - (-3) = -5 + 3 = -2$

Since the difference between consecutive terms is constant (-2), the list of numbers forms an AP.

The common difference $d = -2$.

The last given term is $a_4 = -5$.

The next term ($a_5$) = $a_4 + d = -5 + (-2) = -7$.

The term after that ($a_6$) = $a_5 + d = -7 + (-2) = -9$.

Conclusion: The list forms an AP with a common difference of -2. The next two terms are -7 and -9.

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Solution (iii): -2, 2, -2, 2, -2, ...

Given list: -2, 2, -2, 2, -2, ...

Let the terms be $a_1 = -2, a_2 = 2, a_3 = -2, a_4 = 2$.

Calculate the differences between consecutive terms:

$a_2 - a_1 = 2 - (-2) = 2 + 2 = 4$

$a_3 - a_2 = -2 - 2 = -4$

Since the difference between the second and first term (4) is not equal to the difference between the third and second term (-4), the differences are not constant.

Conclusion: The list of numbers does not form an AP.

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Solution (iv): 1, 1, 1, 2, 2, 2, 3, 3, 3, ...

Given list: 1, 1, 1, 2, 2, 2, 3, 3, 3, ...

Let the terms be $a_1 = 1, a_2 = 1, a_3 = 1, a_4 = 2$.

Calculate the differences between consecutive terms:

$a_2 - a_1 = 1 - 1 = 0$

$a_3 - a_2 = 1 - 1 = 0$

$a_4 - a_3 = 2 - 1 = 1$

Since the difference between the fourth and third term (1) is not equal to the difference between the third and second term (0), the differences are not constant.

Conclusion: The list of numbers does not form an AP.

General Concept: Arithmetic Progression (AP)

An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number (the common difference) to the preceding term, except the first term.

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Solution (i): Taxi Fare

Let $a_n$ be the taxi fare after $n$ km.

Fare for the 1st km ($a_1$) = $\textsf{₹} 15$.

Fare for the first 2 km ($a_2$) = Fare for 1st km + Fare for 2nd km = $15 + 8 = \textsf{₹} 23$.

Fare for the first 3 km ($a_3$) = Fare for first 2 km + Fare for 3rd km = $23 + 8 = \textsf{₹} 31$.

Fare for the first 4 km ($a_4$) = Fare for first 3 km + Fare for 4th km = $31 + 8 = \textsf{₹} 39$.

The list of fares after each km is: 15, 23, 31, 39, ...

Check the difference between consecutive terms:

$a_2 - a_1 = 23 - 15 = 8$

$a_3 - a_2 = 31 - 23 = 8$

$a_4 - a_3 = 39 - 31 = 8$

Since the difference between consecutive terms is constant ($d = 8$), the list of numbers involved makes an arithmetic progression.

Why: Each subsequent term (fare after the next km) is obtained by adding a fixed amount (₹ 8) to the previous term.

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Solution (ii): Air in Cylinder

Let the initial amount of air present in the cylinder be $V$ units.

Amount of air after 1st removal ($a_1$) = $V - \frac{1}{4}V = \frac{3}{4}V$.

Amount of air after 2nd removal ($a_2$) = $\frac{3}{4}V - \frac{1}{4}\left(\frac{3}{4}V\right) = \frac{3}{4}V - \frac{3}{16}V = \left(\frac{12-3}{16}\right)V = \frac{9}{16}V$.

Amount of air after 3rd removal ($a_3$) = $\frac{9}{16}V - \frac{1}{4}\left(\frac{9}{16}V\right) = \frac{9}{16}V - \frac{9}{64}V = \left(\frac{36-9}{64}\right)V = \frac{27}{64}V$.

The list of amounts of air remaining is: $\frac{3}{4}V, \frac{9}{16}V, \frac{27}{64}V, \dots$ (assuming $a_1$ is after the first removal).

Check the difference between consecutive terms:

$a_2 - a_1 = \frac{9}{16}V - \frac{3}{4}V = \frac{9V - 12V}{16} = -\frac{3}{16}V$.

$a_3 - a_2 = \frac{27}{64}V - \frac{9}{16}V = \frac{27V - 36V}{64} = -\frac{9}{64}V$.

Since the difference between consecutive terms ($-\frac{3}{16}V$ and $-\frac{9}{64}V$) is not constant, the list of numbers involved does not make an arithmetic progression.

Why: Each subsequent term is obtained by multiplying the previous term by a fixed factor ($\frac{3}{4}$), not by adding a fixed amount.

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Solution (iii): Cost of Digging

Let $a_n$ be the cost of digging for the $n$-th metre.

Cost for the 1st metre ($a_1$) = $\textsf{₹} 150$.

Cost for the 2nd metre ($a_2$) = Cost for 1st metre + Rise = $150 + 50 = \textsf{₹} 200$.

Cost for the 3rd metre ($a_3$) = Cost for 2nd metre + Rise = $200 + 50 = \textsf{₹} 250$.

Cost for the 4th metre ($a_4$) = Cost for 3rd metre + Rise = $250 + 50 = \textsf{₹} 300$.

The list of costs for digging each subsequent metre is: 150, 200, 250, 300, ...

Check the difference between consecutive terms:

$a_2 - a_1 = 200 - 150 = 50$

$a_3 - a_2 = 250 - 200 = 50$

$a_4 - a_3 = 300 - 250 = 50$

Since the difference between consecutive terms is constant ($d = 50$), the list of costs for each metre of digging makes an arithmetic progression.

Why: The cost for digging each subsequent metre increases by a fixed amount (₹ 50).

(Note: If the question meant the total cost after digging 'n' metres, the sequence would be 150, 350, 600,... which is not an AP. However, the phrasing is more consistent with the cost per metre forming the AP, similar to how the cost per additional km was constant in part (i)).

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Solution (iv): Compound Interest

Let $A_n$ be the amount of money in the account after $n$ years.

Principal ($P$) = $\textsf{₹} 10000$.

Rate ($r$) = 8% per annum.

Amount after 1 year ($A_1$) = $P(1 + r/100)^1 = 10000(1 + 8/100)^1 = 10000(1.08)^1 = \textsf{₹} 10800$.

Amount after 2 years ($A_2$) = $P(1 + r/100)^2 = 10000(1.08)^2 = 10000(1.1664) = \textsf{₹} 11664$.

Amount after 3 years ($A_3$) = $P(1 + r/100)^3 = 10000(1.08)^3 = 10000(1.259712) = \textsf{₹} 12597.12$.

The list of amounts after each year is: 10800, 11664, 12597.12, ...

Check the difference between consecutive terms:

$A_2 - A_1 = 11664 - 10800 = 864$

$A_3 - A_2 = 12597.12 - 11664 = 933.12$

Since the difference between consecutive terms (864 and 933.12) is not constant, the list of numbers involved does not make an arithmetic progression.

Why: In compound interest, the interest earned each year is added to the principal, and the next year's interest is calculated on the new, larger principal. Thus, the increase in the amount is not constant.

The formula for the $n$-th term of an Arithmetic Progression (AP) is given by $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.

We need to find the first four terms, which are $a_1$, $a_2$, $a_3$, and $a_4$.

$a_1 = a + (1-1)d = a$

$a_2 = a + (2-1)d = a + d$

$a_3 = a + (3-1)d = a + 2d$

$a_4 = a + (4-1)d = a + 3d$

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(i) Given $a = 10$, $d = 10$.

First term, $a_1 = a = 10$.

Second term, $a_2 = a + d = 10 + 10 = 20$.

Third term, $a_3 = a + 2d = 10 + 2(10) = 10 + 20 = 30$.

Fourth term, $a_4 = a + 3d = 10 + 3(10) = 10 + 30 = 40$.

The first four terms of the AP are: $10, 20, 30, 40$.

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(ii) Given $a = -2$, $d = 0$.

First term, $a_1 = a = -2$.

Second term, $a_2 = a + d = -2 + 0 = -2$.

Third term, $a_3 = a + 2d = -2 + 2(0) = -2 + 0 = -2$.

Fourth term, $a_4 = a + 3d = -2 + 3(0) = -2 + 0 = -2$.

The first four terms of the AP are: $-2, -2, -2, -2$.

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(iii) Given $a = 4$, $d = -3$.

First term, $a_1 = a = 4$.

Second term, $a_2 = a + d = 4 + (-3) = 4 - 3 = 1$.

Third term, $a_3 = a + 2d = 4 + 2(-3) = 4 - 6 = -2$.

Fourth term, $a_4 = a + 3d = 4 + 3(-3) = 4 - 9 = -5$.

The first four terms of the AP are: $4, 1, -2, -5$.

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(iv) Given $a = -1$, $d = \frac{1}{2}$.

First term, $a_1 = a = -1$.

Second term, $a_2 = a + d = -1 + \frac{1}{2} = -\frac{2}{2} + \frac{1}{2} = -\frac{1}{2}$.

Third term, $a_3 = a + 2d = -1 + 2(\frac{1}{2}) = -1 + 1 = 0$.

Fourth term, $a_4 = a + 3d = -1 + 3(\frac{1}{2}) = -1 + \frac{3}{2} = -\frac{2}{2} + \frac{3}{2} = \frac{1}{2}$.

The first four terms of the AP are: $-1, -\frac{1}{2}, 0, \frac{1}{2}$.

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(v) Given $a = -1.25$, $d = -0.25$.

First term, $a_1 = a = -1.25$.

Second term, $a_2 = a + d = -1.25 + (-0.25) = -1.25 - 0.25 = -1.50$.

Third term, $a_3 = a + 2d = -1.25 + 2(-0.25) = -1.25 - 0.50 = -1.75$.

Fourth term, $a_4 = a + 3d = -1.25 + 3(-0.25) = -1.25 - 0.75 = -2.00$.

The first four terms of the AP are: $-1.25, -1.50, -1.75, -2.00$.

In an Arithmetic Progression (AP), the first term is the first number in the sequence, and the common difference is the constant difference between any term and its preceding term. The common difference $d$ can be found by subtracting any term from its subsequent term, i.e., $d = a_{n+1} - a_n$.

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(i) Given AP: $3, 1, -1, -3, \dots$

The first term $a$ is the first number in the sequence.

First term, $a = 3$.

The common difference $d$ is the difference between consecutive terms.

$d = 1 - 3 = -2$.

Alternatively, $d = -1 - 1 = -2$, or $d = -3 - (-1) = -3 + 1 = -2$.

The first term is $3$ and the common difference is $-2$.

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(ii) Given AP: $-5, -1, 3, 7, \dots$

The first term $a$ is the first number in the sequence.

First term, $a = -5$.

The common difference $d$ is the difference between consecutive terms.

$d = -1 - (-5) = -1 + 5 = 4$.

Alternatively, $d = 3 - (-1) = 3 + 1 = 4$, or $d = 7 - 3 = 4$.

The first term is $-5$ and the common difference is $4$.

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(iii) Given AP: $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots$

The first term $a$ is the first number in the sequence.

First term, $a = \frac{1}{3}$.

The common difference $d$ is the difference between consecutive terms.

$d = \frac{5}{3} - \frac{1}{3} = \frac{5-1}{3} = \frac{4}{3}$.

Alternatively, $d = \frac{9}{3} - \frac{5}{3} = \frac{9-5}{3} = \frac{4}{3}$, or $d = \frac{13}{3} - \frac{9}{3} = \frac{13-9}{3} = \frac{4}{3}$.

The first term is $\frac{1}{3}$ and the common difference is $\frac{4}{3}$.

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(iv) Given AP: $0.6, 1.7, 2.8, 3.9, \dots$

The first term $a$ is the first number in the sequence.

First term, $a = 0.6$.

The common difference $d$ is the difference between consecutive terms.

$d = 1.7 - 0.6 = 1.1$.

Alternatively, $d = 2.8 - 1.7 = 1.1$, or $d = 3.9 - 2.8 = 1.1$.

The first term is $0.6$ and the common difference is $1.1$.

A sequence is an Arithmetic Progression (AP) if the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by $d$. To check if a sequence is an AP, we calculate the difference between successive terms ($a_{n+1} - a_n$) for consecutive pairs.

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(i) Given sequence: $2, 4, 8, 16, \dots$

Difference between 2nd and 1st term: $4 - 2 = 2$

Difference between 3rd and 2nd term: $8 - 4 = 4$

Since the differences are not equal ($2 \neq 4$), the sequence is not an AP.

Conclusion: Not an AP.

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(ii) Given sequence: $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$

Difference between 2nd and 1st term: $\frac{5}{2} - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2}$

Difference between 3rd and 2nd term: $3 - \frac{5}{2} = \frac{6}{2} - \frac{5}{2} = \frac{1}{2}$

Difference between 4th and 3rd term: $\frac{7}{2} - 3 = \frac{7}{2} - \frac{6}{2} = \frac{1}{2}$

Since the difference is constant, the sequence is an AP with common difference $d = \frac{1}{2}$.

The first four terms are $a_1 = 2, a_2 = \frac{5}{2}, a_3 = 3, a_4 = \frac{7}{2}$.

The next three terms are:

$a_5 = a_4 + d = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4$

$a_6 = a_5 + d = 4 + \frac{1}{2} = \frac{8}{2} + \frac{1}{2} = \frac{9}{2}$

$a_7 = a_6 + d = \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5$

Conclusion: It is an AP. Common difference $d = \frac{1}{2}$. Next three terms: $4, \frac{9}{2}, 5$.

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(iii) Given sequence: $-1.2, -3.2, -5.2, -7.2, \dots$

Difference between 2nd and 1st term: $-3.2 - (-1.2) = -3.2 + 1.2 = -2.0$

Difference between 3rd and 2nd term: $-5.2 - (-3.2) = -5.2 + 3.2 = -2.0$

Difference between 4th and 3rd term: $-7.2 - (-5.2) = -7.2 + 5.2 = -2.0$

Since the difference is constant, the sequence is an AP with common difference $d = -2.0$.

The first four terms are $a_1 = -1.2, a_2 = -3.2, a_3 = -5.2, a_4 = -7.2$.

The next three terms are:

$a_5 = a_4 + d = -7.2 + (-2.0) = -7.2 - 2.0 = -9.2$

$a_6 = a_5 + d = -9.2 + (-2.0) = -9.2 - 2.0 = -11.2$

$a_7 = a_6 + d = -11.2 + (-2.0) = -11.2 - 2.0 = -13.2$

Conclusion: It is an AP. Common difference $d = -2.0$. Next three terms: $-9.2, -11.2, -13.2$.

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(iv) Given sequence: $-10, -6, -2, 2, \dots$

Difference between 2nd and 1st term: $-6 - (-10) = -6 + 10 = 4$

Difference between 3rd and 2nd term: $-2 - (-6) = -2 + 6 = 4$

Difference between 4th and 3rd term: $2 - (-2) = 2 + 2 = 4$

Since the difference is constant, the sequence is an AP with common difference $d = 4$.

The first four terms are $a_1 = -10, a_2 = -6, a_3 = -2, a_4 = 2$.

The next three terms are:

$a_5 = a_4 + d = 2 + 4 = 6$

$a_6 = a_5 + d = 6 + 4 = 10$

$a_7 = a_6 + d = 10 + 4 = 14$

Conclusion: It is an AP. Common difference $d = 4$. Next three terms: $6, 10, 14$.

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(v) Given sequence: $3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \dots$

Difference between 2nd and 1st term: $(3 + \sqrt{2}) - 3 = \sqrt{2}$

Difference between 3rd and 2nd term: $(3 + 2\sqrt{2}) - (3 + \sqrt{2}) = 3 + 2\sqrt{2} - 3 - \sqrt{2} = \sqrt{2}$

Difference between 4th and 3rd term: $(3 + 3\sqrt{2}) - (3 + 2\sqrt{2}) = 3 + 3\sqrt{2} - 3 - 2\sqrt{2} = \sqrt{2}$

Since the difference is constant, the sequence is an AP with common difference $d = \sqrt{2}$.

The first four terms are $a_1 = 3, a_2 = 3+\sqrt{2}, a_3 = 3+2\sqrt{2}, a_4 = 3+3\sqrt{2}$.

The next three terms are:

$a_5 = a_4 + d = (3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2}$

$a_6 = a_5 + d = (3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2}$

$a_7 = a_6 + d = (3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2}$

Conclusion: It is an AP. Common difference $d = \sqrt{2}$. Next three terms: $3 + 4\sqrt{2}, 3 + 5\sqrt{2}, 3 + 6\sqrt{2}$.

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(vi) Given sequence: $0.2, 0.22, 0.222, 0.2222, \dots$

Difference between 2nd and 1st term: $0.22 - 0.2 = 0.02$

Difference between 3rd and 2nd term: $0.222 - 0.22 = 0.002$

Since the differences are not equal ($0.02 \neq 0.002$), the sequence is not an AP.

Conclusion: Not an AP.

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(vii) Given sequence: $0, -4, -8, -12, \dots$

Difference between 2nd and 1st term: $-4 - 0 = -4$

Difference between 3rd and 2nd term: $-8 - (-4) = -8 + 4 = -4$

Difference between 4th and 3rd term: $-12 - (-8) = -12 + 8 = -4$

Since the difference is constant, the sequence is an AP with common difference $d = -4$.

The first four terms are $a_1 = 0, a_2 = -4, a_3 = -8, a_4 = -12$.

The next three terms are:

$a_5 = a_4 + d = -12 + (-4) = -16$

$a_6 = a_5 + d = -16 + (-4) = -20$

$a_7 = a_6 + d = -20 + (-4) = -24$

Conclusion: It is an AP. Common difference $d = -4$. Next three terms: $-16, -20, -24$.

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(viii) Given sequence: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots$

Difference between 2nd and 1st term: $-\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$

Difference between 3rd and 2nd term: $-\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$

Difference between 4th and 3rd term: $-\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$

Since the difference is constant, the sequence is an AP with common difference $d = 0$.

The first four terms are $a_1 = -\frac{1}{2}, a_2 = -\frac{1}{2}, a_3 = -\frac{1}{2}, a_4 = -\frac{1}{2}$.

The next three terms are:

$a_5 = a_4 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$

$a_6 = a_5 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$

$a_7 = a_6 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$

Conclusion: It is an AP. Common difference $d = 0$. Next three terms: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}$.

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(ix) Given sequence: $1, 3, 9, 27, \dots$

Difference between 2nd and 1st term: $3 - 1 = 2$

Difference between 3rd and 2nd term: $9 - 3 = 6$

Since the differences are not equal ($2 \neq 6$), the sequence is not an AP.

Conclusion: Not an AP.

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(x) Given sequence: $a, 2a, 3a, 4a, \dots$

Difference between 2nd and 1st term: $2a - a = a$

Difference between 3rd and 2nd term: $3a - 2a = a$

Difference between 4th and 3rd term: $4a - 3a = a$

Since the difference is constant (assuming $a$ is a constant value), the sequence is an AP with common difference $d = a$.

The first four terms are $a_1 = a, a_2 = 2a, a_3 = 3a, a_4 = 4a$.

The next three terms are:

$a_5 = a_4 + d = 4a + a = 5a$

$a_6 = a_5 + d = 5a + a = 6a$

$a_7 = a_6 + d = 6a + a = 7a$

Conclusion: It is an AP. Common difference $d = a$. Next three terms: $5a, 6a, 7a$.

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(xi) Given sequence: $a, a^2, a^3, a^4, \dots$

Difference between 2nd and 1st term: $a^2 - a = a(a-1)$

Difference between 3rd and 2nd term: $a^3 - a^2 = a^2(a-1)$

For this to be an AP, $a(a-1)$ must be equal to $a^2(a-1)$.

$a(a-1) = a^2(a-1)$

$a^2(a-1) - a(a-1) = 0$

$a(a-1)(a - 1) = 0$

$a(a-1)^2 = 0$

This is only true if $a=0$ or $a=1$. For any other value of $a$, the difference is not constant.

Conclusion: Not an AP (unless $a=0$ or $a=1$, in which case it is an AP with $d=0$; however, as a general sequence based on $a$, it is not). Let's assume it's not an AP for arbitrary $a$.

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(xii) Given sequence: $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$

Rewrite the terms in simplest radical form:

$\sqrt{2}$

$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$

$\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$

$\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2}$

The sequence is $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$

Difference between 2nd and 1st term: $2\sqrt{2} - \sqrt{2} = \sqrt{2}$

Difference between 3rd and 2nd term: $3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$

Difference between 4th and 3rd term: $4\sqrt{2} - 3\sqrt{2} = \sqrt{2}$

Since the difference is constant, the sequence is an AP with common difference $d = \sqrt{2}$.

The first four terms are $a_1 = \sqrt{2}, a_2 = 2\sqrt{2}, a_3 = 3\sqrt{2}, a_4 = 4\sqrt{2}$.

The next three terms are:

$a_5 = a_4 + d = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50}$

$a_6 = a_5 + d = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{36 \times 2} = \sqrt{72}$

$a_7 = a_6 + d = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98}$

Conclusion: It is an AP. Common difference $d = \sqrt{2}$. Next three terms: $\sqrt{50}, \sqrt{72}, \sqrt{98}$.

---

(xiii) Given sequence: $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$

Rewrite the terms: $\sqrt{3}, \sqrt{6}, 3, \sqrt{12} = 2\sqrt{3}$

Difference between 2nd and 1st term: $\sqrt{6} - \sqrt{3}$

Difference between 3rd and 2nd term: $3 - \sqrt{6}$

To check if $\sqrt{6} - \sqrt{3} = 3 - \sqrt{6}$, we add $\sqrt{6}$ and $\sqrt{3}$ to both sides:

$2\sqrt{6} = 3 + \sqrt{3}$

Squaring both sides: $(2\sqrt{6})^2 = (3 + \sqrt{3})^2$

$4 \times 6 = 3^2 + 2(3)\sqrt{3} + (\sqrt{3})^2$

$24 = 9 + 6\sqrt{3} + 3$

$24 = 12 + 6\sqrt{3}$

$12 = 6\sqrt{3}$

$2 = \sqrt{3}$

This is false ($2^2 = 4$, $(\sqrt{3})^2 = 3$). Therefore, the differences are not equal.

Conclusion: Not an AP.

---

(xiv) Given sequence: $1^2, 3^2, 5^2, 7^2, \dots$

Rewrite the terms: $1, 9, 25, 49, \dots$

Difference between 2nd and 1st term: $9 - 1 = 8$

Difference between 3rd and 2nd term: $25 - 9 = 16$

Since the differences are not equal ($8 \neq 16$), the sequence is not an AP.

Conclusion: Not an AP.

---

(xv) Given sequence: $1^2, 5^2, 7^2, 7^3, \dots$

Rewrite the terms: $1, 25, 49, 343, \dots$

Difference between 2nd and 1st term: $25 - 1 = 24$

Difference between 3rd and 2nd term: $49 - 25 = 24$

Difference between 4th and 3rd term: $343 - 49 = 294$

Since the differences are not equal ($24 \neq 294$), the sequence is not an AP.

Conclusion: Not an AP.

The given Arithmetic Progression (AP) is: $2, 7, 12, \dots$

---

Here, the first term is $a = 2$.

---

The common difference $d$ is the difference between consecutive terms.

$d = 7 - 2 = 5$.

We can verify this with the next pair of terms: $12 - 7 = 5$.

---

The formula for the $n$-th term of an AP is given by:

---

We need to find the 10th term of this AP, so we need to find $a_{10}$. Here, $n = 10$.

Substitute $a=2$, $d=5$, and $n=10$ into the formula (i):

$a_{10} = 2 + (10-1) \times 5$

$a_{10} = 2 + (9) \times 5$

$a_{10} = 2 + 45$

$a_{10} = 47$

---

Thus, the 10th term of the given AP is $47$.

The given Arithmetic Progression (AP) is: $21, 18, 15, \dots$

---

Here, the first term is $a = 21$.

---

The common difference $d$ is the difference between consecutive terms.

$d = 18 - 21 = -3$.

We can verify this with the next pair of terms: $15 - 18 = -3$. So, $d = -3$.

---

The formula for the $n$-th term of an AP is given by:

---

Part 1: Which term is -81?

We want to find the value of $n$ for which $a_n = -81$.

Substitute $a = 21$, $d = -3$, and $a_n = -81$ into formula (i):

$-81 = 21 + (n-1)(-3)$

Subtract 21 from both sides:

$-81 - 21 = (n-1)(-3)$

$-102 = -3(n-1)$

Divide both sides by -3:

$\frac{-102}{-3} = n-1$

$34 = n-1$

Add 1 to both sides:

$n = 34 + 1$

$n = 35$

Since $n$ is a positive integer, $-81$ is a term of the AP. Specifically, it is the 35th term.

---

Part 2: Is any term 0?

We want to find if there is a positive integer $m$ such that $a_m = 0$.

Substitute $a = 21$, $d = -3$, and $a_m = 0$ into formula (i) (using $m$ instead of $n$):

$0 = 21 + (m-1)(-3)$

Subtract 21 from both sides:

$0 - 21 = (m-1)(-3)$

$-21 = -3(m-1)$

Divide both sides by -3:

$\frac{-21}{-3} = m-1$

$7 = m-1$

Add 1 to both sides:

$m = 7 + 1$

$m = 8$

Since $m=8$ is a positive integer, $0$ is a term of the AP. It is the 8th term.

Reason: A number is a term in an AP if and only if the index $n$ calculated using the formula $a_n = a + (n-1)d$ is a positive integer. In both cases ($a_n = -81$ and $a_n = 0$), we obtained positive integer values for $n$ (35 and 8 respectively), which means both $-81$ and $0$ are terms in the given AP.

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

---

The formula for the $n$-th term of an AP is given by:

---

We are given that the 3rd term of the AP is 5. Using formula (1) with $n=3$:

$a_3 = a + (3-1)d$

$a_3 = a + 2d$

Since $a_3 = 5$, we have the equation:

---

We are also given that the 7th term of the AP is 9. Using formula (1) with $n=7$:

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

$a_7 = a + 6d$

Since $a_7 = 9$, we have the equation:

---

Now we have a system of two linear equations with two variables, $a$ and $d$:

$a + 2d = 5$

$a + 6d = 9$

To solve for $d$, we can subtract equation (i) from equation (ii):

$(a + 6d) - (a + 2d) = 9 - 5$

$a + 6d - a - 2d = 4$

$4d = 4$

$d = \frac{4}{4}$

---

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

$a + 2(1) = 5$

$a + 2 = 5$

$a = 5 - 2$

---

The Arithmetic Progression is determined by its first term ($a$) and its common difference ($d$).

The first term is $a = 3$ and the common difference is $d = 1$.

The terms of the AP are $a, a+d, a+2d, a+3d, \dots$

Substituting the values of $a$ and $d$, the AP is:

$3, 3+1, 3+2(1), 3+3(1), \dots$

which is:

$3, 4, 5, 6, \dots$

---

The AP is $\mathbf{3, 4, 5, 6, \dots}$.

The given list of numbers is: $5, 11, 17, 23, \dots$

---

First, we check if this list forms an Arithmetic Progression (AP) by finding the difference between consecutive terms.

Difference between 2nd and 1st term: $11 - 5 = 6$

Difference between 3rd and 2nd term: $17 - 11 = 6$

Difference between 4th and 3rd term: $23 - 17 = 6$

Since the difference is constant, the list of numbers is an AP.

---

The first term of this AP is $a = 5$.

The common difference is $d = 6$.

---

To check if 301 is a term of this AP, we need to determine if there exists a positive integer $n$ such that the $n$-th term, $a_n$, is equal to 301.

The formula for the $n$-th term of an AP is:

---

We set $a_n = 301$ and substitute the values of $a$ and $d$ into the formula (1):

$301 = 5 + (n-1)6$

Now, we solve this equation for $n$:

$301 - 5 = (n-1)6$

$296 = (n-1)6$

Divide both sides by 6:

Simplify the fraction on the left side:

$\frac{\cancel{296}^{148}}{\cancel{6}_{3}} = \frac{148}{3}$

So, equation (i) becomes:

$\frac{148}{3} = n-1$

Add 1 to both sides to find $n$:

$n = \frac{148}{3} + 1$

$n = \frac{148}{3} + \frac{3}{3}$

$n = \frac{148+3}{3}$

$n = \frac{151}{3}$

---

For 301 to be a term of the AP, the value of $n$ must be a positive integer.

The value we obtained for $n$ is $\frac{151}{3}$.

Since $\frac{151}{3}$ is not an integer (151 divided by 3 leaves a remainder), 301 is not a term of the given AP.

---

Conclusion: 301 is not a term of the list of numbers $5, 11, 17, 23, \dots$.

We are looking for the number of two-digit numbers that are divisible by 3.

The two-digit numbers start from 10 and end at 99.

---

The first two-digit number divisible by 3 is 12.

The next two-digit number divisible by 3 is 15.

The numbers divisible by 3 form an Arithmetic Progression (AP).

The common difference $d$ is 3.

The last two-digit number divisible by 3 is 99.

---

So, the list of two-digit numbers divisible by 3 is:

$12, 15, 18, \dots, 99$.

This is an AP with:

First term, $a = 12$.

Common difference, $d = 3$.

Last term (or the $n$-th term), $a_n = 99$.

---

We need to find the number of terms in this AP, which is $n$.

The formula for the $n$-th term of an AP is given by:

---

Substitute the values of $a$, $d$, and $a_n$ into formula (1):

$99 = 12 + (n-1)3$

Subtract 12 from both sides:

$99 - 12 = (n-1)3$

$87 = 3(n-1)$

Divide both sides by 3:

Perform the division:

$29 = n-1$

Add 1 to both sides:

$n = 29 + 1$

$n = 30$

---

Since $n$ represents the number of terms in the AP, there are 30 two-digit numbers that are divisible by 3.

---

Conclusion: There are 30 two-digit numbers divisible by 3.

The given Arithmetic Progression (AP) is: $10, 7, 4, \dots, -62$.

---

The first term of this AP is $a = 10$.

The common difference $d$ is the difference between consecutive terms:

$d = 7 - 10 = -3$.

The last term of the AP is $l = -62$.

---

We need to find the 11th term from the last term of this AP.

One way to find the $k$-th term from the last of an AP is to use the formula $l - (k-1)d$, where $l$ is the last term and $d$ is the common difference. Here, $k=11$.

Using the formula for the $k$-th term from the end:

---

Substitute $l = -62$, $d = -3$, and $k = 11$ into formula (1):

$11$-th term from the end $= -62 - (11-1)(-3)$

$= -62 - (10)(-3)$

$= -62 - (-30)$

$= -62 + 30$

$= -32$

---

Thus, the 11th term from the last term of the AP is -32.

---

Alternate Solution:

Another way to solve this is to consider the AP in reverse order.

If we reverse the AP, the new first term ($a'$) is the last term of the original AP, and the new common difference ($d'$) is the negative of the original common difference.

Reversed AP: $-62, \dots, 4, 7, 10$.

First term of the reversed AP, $a' = -62$.

Common difference of the reversed AP, $d' = -d = -(-3) = 3$.

---

The 11th term from the last of the original AP is the same as the 11th term from the beginning of the reversed AP.

We use the formula for the $k$-th term of an AP, $a'_k = a' + (k-1)d'$, where $k$ is the term number we want (11th).

---

Substitute $a' = -62$, $d' = 3$, and $k = 11$ into formula (2):

$a'_{11} = -62 + (11-1) \times 3$

$a'_{11} = -62 + (10) \times 3$

$a'_{11} = -62 + 30$

$a'_{11} = -32$

---

Using both methods, we find that the 11th term from the last term of the AP is -32.

Given: Principal amount $P = \textsf{₹}1000$.

Rate of simple interest $R = 8\%$ per year.

We need to calculate the simple interest at the end of each year.

The formula for simple interest for $T$ years is $I = \frac{P \times R \times T}{100}$.

---

Interest at the end of 1 year ($T=1$):

$I_1 = \frac{1000 \times 8 \times 1}{100} = \frac{8000}{100} = 80$.

Interest at the end of 2 years ($T=2$):

$I_2 = \frac{1000 \times 8 \times 2}{100} = \frac{16000}{100} = 160$.

Interest at the end of 3 years ($T=3$):

$I_3 = \frac{1000 \times 8 \times 3}{100} = \frac{24000}{100} = 240$.

And so on.

---

The interests at the end of 1, 2, 3, $\dots$ years are $80, 160, 240, \dots$.

Let's check if this list of numbers forms an AP.

Difference between 2nd and 1st term: $160 - 80 = 80$.

Difference between 3rd and 2nd term: $240 - 160 = 80$.

Since the difference between consecutive terms is constant (which is 80), the interests form an Arithmetic Progression (AP).

---

This AP is $80, 160, 240, \dots$.

The first term of this AP is $a = 80$.

The common difference is $d = 80$.

---

We need to find the interest at the end of 30 years. In this AP, the interest at the end of $n$ years is the $n$-th term ($a_n$). So, we need to find the 30th term ($a_{30}$) of this AP.

The formula for the $n$-th term of an AP is given by:

---

Substitute $a = 80$, $d = 80$, and $n = 30$ into formula (1):

$a_{30} = 80 + (30-1) \times 80$

$a_{30} = 80 + (29) \times 80$

$a_{30} = 80 + 2320$

$a_{30} = 2400$

---

The 30th term of the AP is 2400.

Therefore, the interest at the end of 30 years is $\textsf{₹}2400$.

---

Conclusion: The annual interests form an AP. The interest at the end of 30 years is $\textsf{₹}2400$.

The number of rose plants in the rows are given as: $23, 21, 19, \dots, 5$.

---

Let's check if this sequence is an Arithmetic Progression (AP).

Difference between the second and first term: $21 - 23 = -2$.

Difference between the third and second term: $19 - 21 = -2$.

Since the difference between consecutive terms is constant, the sequence is an AP.

---

In this AP:

The first term is $a = 23$.

The common difference is $d = -2$.

The last term (let's call it the $n$-th term) is $a_n = 5$.

---

We need to find the number of rows in the flower bed, which is the number of terms ($n$) in this AP.

The formula for the $n$-th term of an AP is given by:

---

Substitute the values of $a$, $d$, and $a_n$ into formula (1):

$5 = 23 + (n-1)(-2)$

Subtract 23 from both sides:

$5 - 23 = (n-1)(-2)$

$-18 = -2(n-1)$

Divide both sides by -2:

$\frac{-18}{-2} = n-1$

$9 = n-1$

Add 1 to both sides:

$n = 9 + 1$

$n = 10$

---

Since $n$ represents the number of terms (rows), there are 10 rows in the flower bed.

---

Conclusion: There are 10 rows in the flower bed.

The formula for the $n$-th term of an Arithmetic Progression (AP) is given by $a_n = a + (n-1)d$, where $a$ is the first term, $d$ is the common difference, and $n$ is the term number.

---

(i)

Given: $a = 7$, $d = 3$, $n = 8$.

To find: $a_n$ (specifically $a_8$).

Using the formula $a_n = a + (n-1)d$:

$a_8 = 7 + (8-1) \times 3$

$a_8 = 7 + (7) \times 3$

$a_8 = 7 + 21$

$a_8 = 28$

The value of $a_n$ is $\mathbf{28}$.

---

(ii)

Given: $a = -18$, $n = 10$, $a_n = 0$.

To find: $d$.

Using the formula $a_n = a + (n-1)d$:

$0 = -18 + (10-1)d$

$0 = -18 + 9d$

Add 18 to both sides:

$18 = 9d$

Divide both sides by 9:

$d = \frac{18}{9}$

$d = 2$

The value of $d$ is $\mathbf{2}$.

---

(iii)

Given: $d = -3$, $n = 18$, $a_n = -5$.

To find: $a$.

Using the formula $a_n = a + (n-1)d$:

$-5 = a + (18-1)(-3)$

$-5 = a + (17)(-3)$

$-5 = a - 51$

Add 51 to both sides:

$a = -5 + 51$

$a = 46$

The value of $a$ is $\mathbf{46}$.

---

(iv)

Given: $a = -18.9$, $d = 2.5$, $a_n = 3.6$.

To find: $n$.

Using the formula $a_n = a + (n-1)d$:

$3.6 = -18.9 + (n-1)2.5$

Add 18.9 to both sides:

$3.6 + 18.9 = (n-1)2.5$

$22.5 = (n-1)2.5$

Divide both sides by 2.5:

$n-1 = \frac{22.5}{2.5}$

To simplify the division, multiply numerator and denominator by 10:

$n-1 = \frac{22.5 \times 10}{2.5 \times 10} = \frac{225}{25}$

$n-1 = 9$

Add 1 to both sides:

$n = 9 + 1$

$n = 10$

The value of $n$ is $\mathbf{10}$.

---

(v)

Given: $a = 3.5$, $d = 0$, $n = 105$.

To find: $a_n$ (specifically $a_{105}$).

Using the formula $a_n = a + (n-1)d$:

$a_{105} = 3.5 + (105-1) \times 0$

$a_{105} = 3.5 + (104) \times 0$

$a_{105} = 3.5 + 0$

$a_{105} = 3.5$

The value of $a_n$ is $\mathbf{3.5}$.

To find the $n$-th term of an Arithmetic Progression (AP), we use the formula $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.

---

(i) Given AP: $10, 7, 4, \dots$

The first term is $a = 10$.

The common difference $d$ is $7 - 10 = -3$.

We need to find the 30th term, so $n = 30$.

Using the formula $a_n = a + (n-1)d$:

$a_{30} = 10 + (30-1)(-3)$

$a_{30} = 10 + (29)(-3)$

$a_{30} = 10 - 87$

$a_{30} = -77$

The 30th term of the AP is $-77$.

Comparing with the given options, the correct choice is (C) –77.

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(ii) Given AP: $-3, -\frac{1}{2}, 2, \dots$

The first term is $a = -3$.

The common difference $d$ is the difference between consecutive terms:

$d = -\frac{1}{2} - (-3) = -\frac{1}{2} + 3 = -\frac{1}{2} + \frac{6}{2} = \frac{5}{2}$.

We need to find the 11th term, so $n = 11$.

Using the formula $a_n = a + (n-1)d$:

$a_{11} = -3 + (11-1)(\frac{5}{2})$

$a_{11} = -3 + (10)(\frac{5}{2})$

$a_{11} = -3 + \frac{10 \times 5}{2}$

$a_{11} = -3 + \frac{50}{2}$

$a_{11} = -3 + 25$

$a_{11} = 22$

The 11th term of the AP is $22$.

Comparing with the given options, the correct choice is (B) 22.

In an Arithmetic Progression (AP), the $n$-th term is given by the formula $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.

---

(i) Given AP: $2, \square, 26$

This AP has 3 terms. The first term is $a_1 = 2$ and the third term is $a_3 = 26$. The missing term is the second term, $a_2$.

Using the formula for the $n$-th term:

$a_3 = a_1 + (3-1)d$

$26 = 2 + 2d$

Subtract 2 from both sides:

$26 - 2 = 2d$

$24 = 2d$

Divide by 2:

$d = \frac{24}{2} = 12$.

The common difference is $d = 12$.

The missing term is $a_2$, which is $a_1 + d$:

$a_2 = 2 + 12 = 14$.

The complete AP is $2, 14, 26$.

The missing term is $\mathbf{14}$.

---

(ii) Given AP: $\square, 13, \square, 3$

This AP has 4 terms. The second term is $a_2 = 13$ and the fourth term is $a_4 = 3$. The missing terms are the first term ($a_1$ or $a$) and the third term ($a_3$).

Using the formula for the $n$-th term:

$a_2 = a + (2-1)d \implies 13 = a + d$ ... (1)

$a_4 = a + (4-1)d \implies 3 = a + 3d$ ... (2)

Subtract equation (1) from equation (2):

$(a + 3d) - (a + d) = 3 - 13$

$2d = -10$

Divide by 2:

$d = \frac{-10}{2} = -5$.

The common difference is $d = -5$.

Substitute $d=-5$ into equation (1) to find $a$:

$13 = a + (-5)$

$13 = a - 5$

$a = 13 + 5 = 18$.

The first term is $a_1 = 18$.

The missing third term is $a_3 = a_2 + d$ (or $a + 2d$):

$a_3 = 13 + (-5) = 13 - 5 = 8$.

The complete AP is $18, 13, 8, 3$.

The missing terms are $\mathbf{18}$ and $\mathbf{8}$.

---

(iii) Given AP: $5, \square, \square, 9\frac{1}{2}$

This AP has 4 terms. The first term is $a_1 = 5$ and the fourth term is $a_4 = 9\frac{1}{2} = \frac{19}{2}$. The missing terms are the second ($a_2$) and third ($a_3$) terms.

Using the formula for the $n$-th term:

$a_4 = a_1 + (4-1)d$

$\frac{19}{2} = 5 + 3d$

Subtract 5 from both sides:

$\frac{19}{2} - 5 = 3d$

$\frac{19}{2} - \frac{10}{2} = 3d$

$\frac{9}{2} = 3d$

Divide by 3:

$d = \frac{9/2}{3} = \frac{9}{2 \times 3} = \frac{9}{6} = \frac{3}{2}$.

The common difference is $d = \frac{3}{2}$.

The missing second term is $a_2 = a_1 + d$:

$a_2 = 5 + \frac{3}{2} = \frac{10}{2} + \frac{3}{2} = \frac{13}{2}$.

The missing third term is $a_3 = a_2 + d$:

$a_3 = \frac{13}{2} + \frac{3}{2} = \frac{16}{2} = 8$.

The complete AP is $5, \frac{13}{2}, 8, \frac{19}{2}$.

The missing terms are $\mathbf{\frac{13}{2}}$ (or 6.5) and $\mathbf{8}$.

---

(iv) Given AP: – 4 , $\square$, $\square$, $\square$, $\square$, 6

This AP has 6 terms. The first term is $a_1 = -4$ and the sixth term is $a_6 = 6$. The missing terms are the second ($a_2$), third ($a_3$), fourth ($a_4$), and fifth ($a_5$) terms.

Using the formula for the $n$-th term:

$a_6 = a_1 + (6-1)d$

$6 = -4 + 5d$

Add 4 to both sides:

$6 + 4 = 5d$

$10 = 5d$

Divide by 5:

$d = \frac{10}{5} = 2$.

The common difference is $d = 2$.

Now find the missing terms:

$a_2 = a_1 + d = -4 + 2 = -2$.

$a_3 = a_2 + d = -2 + 2 = 0$.

$a_4 = a_3 + d = 0 + 2 = 2$.

$a_5 = a_4 + d = 2 + 2 = 4$.

The complete AP is $-4, -2, 0, 2, 4, 6$.

The missing terms are $\mathbf{-2, 0, 2, 4}$.

---

(v) Given AP: $\square, 38, \square, \square, \square, – 22$

This AP has 6 terms. The second term is $a_2 = 38$ and the sixth term is $a_6 = -22$. The missing terms are the first ($a_1$ or $a$), third ($a_3$), fourth ($a_4$), and fifth ($a_5$) terms.

Using the formula for the $n$-th term:

$a_2 = a + (2-1)d \implies 38 = a + d$ ... (1)

$a_6 = a + (6-1)d \implies -22 = a + 5d$ ... (2)

Subtract equation (1) from equation (2):

$(a + 5d) - (a + d) = -22 - 38$

$4d = -60$

Divide by 4:

$d = \frac{-60}{4} = -15$.

The common difference is $d = -15$.

Substitute $d=-15$ into equation (1) to find $a$:

$38 = a + (-15)$

$38 = a - 15$

$a = 38 + 15 = 53$.

The first term is $a_1 = 53$.

Now find the missing terms:

$a_3 = a_2 + d = 38 + (-15) = 38 - 15 = 23$.

$a_4 = a_3 + d = 23 + (-15) = 23 - 15 = 8$.

$a_5 = a_4 + d = 8 + (-15) = 8 - 15 = -7$.

The complete AP is $53, 38, 23, 8, -7, -22$.

The missing terms are $\mathbf{53, 23, 8, -7}$.

The given Arithmetic Progression (AP) is: $3, 8, 13, 18, \dots$

---

The first term is $a = 3$.

---

The common difference $d$ is the difference between consecutive terms:

$d = 8 - 3 = 5$.

We can check this with the next pair: $13 - 8 = 5$. So, $d = 5$.

---

We need to find which term of this AP is 78. Let the $n$-th term be 78, so $a_n = 78$.

---

The formula for the $n$-th term of an AP is given by:

---

Substitute $a = 3$, $d = 5$, and $a_n = 78$ into formula (1):

$78 = 3 + (n-1)5$

Subtract 3 from both sides:

$78 - 3 = (n-1)5$

$75 = 5(n-1)$

Divide both sides by 5:

$\frac{75}{5} = n-1$

$15 = n-1$

Add 1 to both sides:

$n = 15 + 1$

$n = 16$

---

Since $n=16$ is a positive integer, 78 is a term in the AP. It is the 16th term.

---

Conclusion: The 16th term of the AP is 78.

To find the number of terms ($n$) in an Arithmetic Progression (AP), we use the formula for the $n$-th term: $a_n = a + (n-1)d$, where $a$ is the first term, $d$ is the common difference, and $a_n$ is the last term.

---

(i) Given AP: $7, 13, 19, \dots, 205$

Here, the first term $a = 7$.

The common difference $d = 13 - 7 = 6$. (Also, $19 - 13 = 6$)

The last term $a_n = 205$.

We need to find $n$.

Using the formula $a_n = a + (n-1)d$:

$205 = 7 + (n-1)6$

Subtract 7 from both sides:

$205 - 7 = (n-1)6$

$198 = 6(n-1)$

Divide both sides by 6:

$\frac{198}{6} = n-1$

$33 = n-1$

Add 1 to both sides:

$n = 33 + 1$

$n = 34$

The number of terms in the AP is 34.

---

(ii) Given AP: $18, 15\frac{1}{2}, 13, \dots, -47$

Here, the first term $a = 18$.

The common difference $d = 15\frac{1}{2} - 18$.

$15\frac{1}{2} = \frac{31}{2}$.

$d = \frac{31}{2} - 18 = \frac{31}{2} - \frac{36}{2} = \frac{31 - 36}{2} = -\frac{5}{2}$.

The last term $a_n = -47$.

We need to find $n$.

Using the formula $a_n = a + (n-1)d$:

$-47 = 18 + (n-1)(-\frac{5}{2})$

Subtract 18 from both sides:

$-47 - 18 = (n-1)(-\frac{5}{2})$

$-65 = (n-1)(-\frac{5}{2})$

Multiply both sides by $-\frac{2}{5}$:

$-65 \times (-\frac{2}{5}) = n-1$

$\frac{65 \times 2}{5} = n-1$

$\frac{\cancel{65}^{13} \times 2}{\cancel{5}_{1}} = n-1$

$13 \times 2 = n-1$

$26 = n-1$

Add 1 to both sides:

$n = 26 + 1$

$n = 27$

The number of terms in the AP is 27.

The given Arithmetic Progression (AP) is: $11, 8, 5, 2, \dots$

---

The first term of this AP is $a = 11$.

---

The common difference $d$ is the difference between consecutive terms:

$d = 8 - 11 = -3$.

We can check this with the next pair: $5 - 8 = -3$. So, $d = -3$.

---

To check if $-150$ is a term of this AP, we need to determine if there exists a positive integer $n$ such that the $n$-th term, $a_n$, is equal to $-150$.

---

The formula for the $n$-th term of an AP is given by:

---

Substitute $a = 11$, $d = -3$, and $a_n = -150$ into formula (1):

$-150 = 11 + (n-1)(-3)$

Subtract 11 from both sides:

$-150 - 11 = (n-1)(-3)$

$-161 = -3(n-1)$

Divide both sides by -3:

Simplify the expression:

$\frac{161}{3} = n-1$

Add 1 to both sides to find $n$:

$n = \frac{161}{3} + 1$

$n = \frac{161}{3} + \frac{3}{3}$

$n = \frac{161+3}{3}$

$n = \frac{164}{3}$

---

For $-150$ to be a term of the AP, the value of $n$ must be a positive integer.

The value we obtained for $n$ is $\frac{164}{3}$.

Since $\frac{164}{3}$ is not an integer (it is approximately $54.67$), $-150$ is not a term of the given AP.

---

Conclusion: – 150 is not a term of the AP : 11, 8, 5, 2 . . .

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

---

The formula for the $n$-th term of an AP is given by:

---

We are given that the 11th term ($a_{11}$) is 38.

Using formula (1) with $n=11$:

$a_{11} = a + (11-1)d$

$38 = a + 10d$

---

We are also given that the 16th term ($a_{16}$) is 73.

Using formula (1) with $n=16$:

$a_{16} = a + (16-1)d$

$73 = a + 15d$

---

Now we have a system of two linear equations with two variables, $a$ and $d$.

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

$(a + 15d) - (a + 10d) = 73 - 38$

$a + 15d - a - 10d = 35$

$5d = 35$

Divide by 5:

$d = \frac{35}{5}$

---

Substitute the value of $d=7$ into equation (i) to find $a$:

$a + 10(7) = 38$

$a + 70 = 38$

$a = 38 - 70$

---

Now that we have the first term ($a = -32$) and the common difference ($d = 7$), we can find the 31st term ($a_{31}$).

Using the formula $a_n = a + (n-1)d$ with $n=31$:

$a_{31} = -32 + (31-1) \times 7$

$a_{31} = -32 + (30) \times 7$

$a_{31} = -32 + 210$

$a_{31} = 178$

---

Thus, the 31st term of the AP is $\mathbf{178}$.

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

---

The AP consists of 50 terms, so the total number of terms is $N = 50$.

The last term is the 50th term, so $a_{50} = 106$.

We are given that the 3rd term is 12, so $a_3 = 12$.

---

The formula for the $n$-th term of an AP is given by:

---

Using formula (1) for the 3rd term ($n=3$):

$a_3 = a + (3-1)d$

$12 = a + 2d$

---

Using formula (1) for the last term (50th term, $n=50$):

$a_{50} = a + (50-1)d$

$106 = a + 49d$

---

Now we have a system of two linear equations:

$a + 2d = 12$ (i)

$a + 49d = 106$ (ii)

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

$(a + 49d) - (a + 2d) = 106 - 12$

$a + 49d - a - 2d = 94$

$47d = 94$

Divide by 47:

$d = \frac{94}{47}$

---

Substitute the value of $d=2$ into equation (i) to find $a$:

$a + 2(2) = 12$

$a + 4 = 12$

$a = 12 - 4$

---

Now that we have the first term ($a = 8$) and the common difference ($d = 2$), we can find the 29th term ($a_{29}$).

Using the formula $a_n = a + (n-1)d$ with $n=29$:

$a_{29} = 8 + (29-1) \times 2$

$a_{29} = 8 + (28) \times 2$

$a_{29} = 8 + 56$

$a_{29} = 64$

---

Thus, the 29th term of the AP is $\mathbf{64}$.

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

---

The formula for the $n$-th term of an AP is given by:

---

We are given that the 3rd term ($a_3$) is 4.

Using formula (1) with $n=3$:

$a_3 = a + (3-1)d$

$4 = a + 2d$

---

We are also given that the 9th term ($a_9$) is -8.

Using formula (1) with $n=9$:

$a_9 = a + (9-1)d$

$-8 = a + 8d$

---

Now we have a system of two linear equations:

$a + 2d = 4$ (i)

$a + 8d = -8$ (ii)

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

$(a + 8d) - (a + 2d) = -8 - 4$

$a + 8d - a - 2d = -12$

$6d = -12$

Divide by 6:

$d = \frac{-12}{6}$

---

Substitute the value of $d=-2$ into equation (i) to find $a$:

$a + 2(-2) = 4$

$a - 4 = 4$

$a = 4 + 4$

---

We need to find which term of this AP is zero. Let the $n$-th term be 0, i.e., $a_n = 0$.

Using the formula $a_n = a + (n-1)d$ with $a=8$, $d=-2$, and $a_n=0$:

$0 = 8 + (n-1)(-2)$

Subtract 8 from both sides:

$-8 = -2(n-1)$

Divide both sides by -2:

$\frac{-8}{-2} = n-1$

$4 = n-1$

Add 1 to both sides:

$n = 4 + 1$

$n = 5$

---

Since $n=5$ is a positive integer, 0 is a term in the AP. It is the 5th term.

---

Conclusion: The 5th term of the AP is zero.

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

---

The formula for the $n$-th term of an AP is given by:

---

The 17th term of the AP ($a_{17}$) is given by using formula (1) with $n=17$:

$a_{17} = a + (17-1)d$

$a_{17} = a + 16d$

---

The 10th term of the AP ($a_{10}$) is given by using formula (1) with $n=10$:

$a_{10} = a + (10-1)d$

$a_{10} = a + 9d$

---

According to the question, the 17th term exceeds its 10th term by 7.

---

Substitute the expressions for $a_{17}$ and $a_{10}$ from equations (i) and (ii) into the given relationship:

$(a + 16d) = (a + 9d) + 7$

Simplify the equation:

$a + 16d = a + 9d + 7$

Subtract $a$ from both sides:

$16d = 9d + 7$

Subtract $9d$ from both sides:

$16d - 9d = 7$

$7d = 7$

Divide both sides by 7:

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

$d = 1$

---

Thus, the common difference of the AP is $\mathbf{1}$.

The given Arithmetic Progression (AP) is: $3, 15, 27, 39, \dots$

---

The first term is $a = 3$.

---

The common difference $d$ is the difference between consecutive terms:

$d = 15 - 3 = 12$.

We can verify this with the next pair: $27 - 15 = 12$. So, $d = 12$.

---

The formula for the $n$-th term of an AP is given by:

---

First, we find the 54th term of the AP. Using formula (1) with $n=54$:

$a_{54} = a + (54-1)d$

$a_{54} = 3 + (53)(12)$

$a_{54} = 3 + 636$

$a_{54} = 639$

---

Let the term which is 132 more than its 54th term be the $n$-th term, $a_n$.

According to the question:

$a_n = 639 + 132$

$a_n = 771$

---

Now, we need to find the value of $n$ for which $a_n = 771$.

Using the formula $a_n = a + (n-1)d$ again, with $a=3$, $d=12$, and $a_n=771$:

$771 = 3 + (n-1)12$

Subtract 3 from both sides:

$771 - 3 = (n-1)12$

$768 = 12(n-1)$

Divide both sides by 12:

$\frac{768}{12} = n-1$

Performing the division:

$64 = n-1$

Add 1 to both sides:

$n = 64 + 1$

$n = 65$

---

Since $n=65$ is a positive integer, 771 is a term in the AP. It is the 65th term.

---

Conclusion: The 65th term of the AP will be 132 more than its 54th term.

Let the first AP be denoted by $AP_1$ and the second AP by $AP_2$.

---

Let the first term of $AP_1$ be $a_1$ and its common difference be $d$.

Let the first term of $AP_2$ be $a_2$ and its common difference be $d'$.

According to the question, the two APs have the same common difference, so $d = d'$. Let's just use $d$ for the common difference for both APs.

---

The $n$-th term of $AP_1$ is given by $a_{1,n} = a_1 + (n-1)d$.

The $n$-th term of $AP_2$ is given by $a_{2,n} = a_2 + (n-1)d$.

---

We are given that the difference between their 100th terms is 100.

The 100th term of $AP_1$ is $a_{1,100} = a_1 + (100-1)d = a_1 + 99d$.

The 100th term of $AP_2$ is $a_{2,100} = a_2 + (100-1)d = a_2 + 99d$.

The difference between their 100th terms is:

Substitute the expressions for the terms:

$(a_1 + 99d) - (a_2 + 99d) = 100$

$a_1 + 99d - a_2 - 99d = 100$

$a_1 - a_2 = 100$

---

We need to find the difference between their 1000th terms.

The 1000th term of $AP_1$ is $a_{1,1000} = a_1 + (1000-1)d = a_1 + 999d$.

The 1000th term of $AP_2$ is $a_{2,1000} = a_2 + (1000-1)d = a_2 + 999d$.

The difference between their 1000th terms is:

$a_{1,1000} - a_{2,1000} = (a_1 + 999d) - (a_2 + 999d)$

$a_{1,1000} - a_{2,1000} = a_1 + 999d - a_2 - 999d$

$a_{1,1000} - a_{2,1000} = a_1 - a_2$

---

From equation (i), we know that $a_1 - a_2 = 100$.

Therefore, the difference between their 1000th terms is also 100.

$a_{1,1000} - a_{2,1000} = 100$

---

Conclusion: The difference between their 1000th terms is $\mathbf{100}$.

This is because the common difference cancels out when calculating the difference between the $n$-th terms of two APs with the same common difference, i.e., $(a_1 + (n-1)d) - (a_2 + (n-1)d) = a_1 - a_2$. The difference between any corresponding terms is equal to the difference between their first terms.

We are looking for the number of three-digit numbers that are divisible by 7.

The three-digit numbers range from 100 to 999.

---

To find the first three-digit number divisible by 7, we find the smallest number greater than or equal to 100 that is a multiple of 7.

Divide 100 by 7: $100 \div 7 \approx 14.28$.

The smallest integer multiple of 7 greater than 100 is $7 \times 15 = 105$.

So, the first three-digit number divisible by 7 is 105.

---

To find the last three-digit number divisible by 7, we find the largest number less than or equal to 999 that is a multiple of 7.

Divide 999 by 7: $999 \div 7 \approx 142.71$.

The largest integer multiple of 7 less than 999 is $7 \times 142 = 994$.

So, the last three-digit number divisible by 7 is 994.

---

The list of three-digit numbers divisible by 7 is:

$105, 112, 119, \dots, 994$.

This sequence forms an Arithmetic Progression (AP) because the difference between consecutive terms is the common difference, which is 7 (since the numbers are multiples of 7).

---

In this AP:

The first term is $a = 105$.

The common difference is $d = 7$.

The last term (or the $n$-th term) is $a_n = 994$.

---

We need to find the number of terms in this AP, which is $n$.

The formula for the $n$-th term of an AP is given by:

---

Substitute the values of $a$, $d$, and $a_n$ into formula (1):

$994 = 105 + (n-1)7$

Subtract 105 from both sides:

$994 - 105 = (n-1)7$

$889 = 7(n-1)$

Divide both sides by 7:

$\frac{889}{7} = n-1$

Performing the division:

$127 = n-1$

Add 1 to both sides:

$n = 127 + 1$

$n = 128$

---

Since $n$ represents the number of terms in the AP, there are 128 three-digit numbers that are divisible by 7.

---

Conclusion: There are 128 three-digit numbers divisible by 7.

We are looking for the number of multiples of 4 that are strictly between 10 and 250. This means numbers greater than 10 and less than 250 that are divisible by 4.

---

To find the first multiple of 4 greater than 10, we check numbers starting from 11.

$11 \div 4 = 2$ with a remainder. $12 \div 4 = 3$.

The first multiple of 4 greater than 10 is 12.

---

To find the last multiple of 4 less than 250, we check numbers downwards from 249.

Divide 250 by 4: $250 \div 4 = 62$ with a remainder of 2.

So, $250 = 4 \times 62 + 2$.

To get a multiple of 4 less than 250, we subtract the remainder: $250 - 2 = 248$.

$248 \div 4 = 62$.

The last multiple of 4 less than 250 is 248.

---

The list of multiples of 4 between 10 and 250 is:

$12, 16, 20, \dots, 248$.

This sequence forms an Arithmetic Progression (AP) because the difference between consecutive terms is the common difference, which is 4 (since the numbers are multiples of 4).

---

In this AP:

The first term is $a = 12$.

The common difference is $d = 4$.

The last term (or the $n$-th term) is $a_n = 248$.

---

We need to find the number of terms in this AP, which is $n$.

The formula for the $n$-th term of an AP is given by:

---

Substitute the values of $a$, $d$, and $a_n$ into formula (1):

$248 = 12 + (n-1)4$

Subtract 12 from both sides:

$248 - 12 = (n-1)4$

$236 = 4(n-1)$

Divide both sides by 4:

$\frac{236}{4} = n-1$

Performing the division:

$59 = n-1$

Add 1 to both sides:

$n = 59 + 1$

$n = 60$

---

Since $n$ represents the number of terms in the AP, there are 60 multiples of 4 between 10 and 250.

---

Conclusion: There are 60 multiples of 4 between 10 and 250.

Let the first Arithmetic Progression (AP) be $AP_1$ and the second AP be $AP_2$.

---

For $AP_1$: $63, 65, 67, \dots$

The first term is $a_1 = 63$.

The common difference is $d_1 = 65 - 63 = 2$.

The $n$-th term of $AP_1$ is given by the formula $a_{1,n} = a_1 + (n-1)d_1$.

---

For $AP_2$: $3, 10, 17, \dots$

The first term is $a_2 = 3$.

The common difference is $d_2 = 10 - 3 = 7$.

The $n$-th term of $AP_2$ is given by the formula $a_{2,n} = a_2 + (n-1)d_2$.

---

We need to find the value of $n$ for which the $n$-th terms of the two APs are equal, i.e., $a_{1,n} = a_{2,n}$.

Setting the expressions from equations (1) and (2) equal:

$63 + (n-1)2 = 3 + (n-1)7$

Expand both sides of the equation:

$63 + 2n - 2 = 3 + 7n - 7$

$61 + 2n = 7n - 4$

Rearrange the terms to solve for $n$. Subtract $2n$ from both sides and add 4 to both sides:

$61 + 4 = 7n - 2n$

$65 = 5n$

Divide both sides by 5:

$n = \frac{65}{5}$

$n = 13$

---

Since $n=13$ is a positive integer, the 13th terms of the two APs are equal.

Let's verify the 13th term for both APs:

For $AP_1$: $a_{1,13} = 63 + (13-1)2 = 63 + 12 \times 2 = 63 + 24 = 87$.

For $AP_2$: $a_{2,13} = 3 + (13-1)7 = 3 + 12 \times 7 = 3 + 84 = 87$.

The 13th terms are indeed equal (87).

---

Conclusion: The $n$-th terms of the two APs are equal when $n = \mathbf{13}$.

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

---

The formula for the $n$-th term of an AP is given by:

---

We are given that the 3rd term ($a_3$) is 16.

Using formula (1) with $n=3$:

$a_3 = a + (3-1)d$

$16 = a + 2d$

---

We are also given that the 7th term exceeds the 5th term by 12.

---

Using formula (1) for the 7th term ($n=7$):

$a_7 = a + (7-1)d = a + 6d$

---

Using formula (1) for the 5th term ($n=5$):

$a_5 = a + (5-1)d = a + 4d$

---

Substitute the expressions for $a_7$ and $a_5$ into the given condition $a_7 = a_5 + 12$:

$(a + 6d) = (a + 4d) + 12$

Simplify the equation:

$a + 6d = a + 4d + 12$

Subtract $a$ from both sides:

$6d = 4d + 12$

Subtract $4d$ from both sides:

$6d - 4d = 12$

$2d = 12$

Divide both sides by 2:

$d = \frac{12}{2}$

---

Now substitute the value of $d=6$ into equation (i) ($a + 2d = 16$) to find $a$:

$a + 2(6) = 16$

$a + 12 = 16$

Subtract 12 from both sides:

$a = 16 - 12$

---

The Arithmetic Progression is determined by its first term ($a$) and its common difference ($d$).

The first term is $a = 4$ and the common difference is $d = 6$.

The terms of the AP are $a, a+d, a+2d, a+3d, \dots$

Substituting the values of $a$ and $d$, the AP is:

$4, 4+6, 4+2(6), 4+3(6), \dots$

which is:

$4, 10, 16, 22, \dots$

---

The AP is $\mathbf{4, 10, 16, 22, \dots}$.

The given Arithmetic Progression (AP) is: $3, 8, 13, \dots, 253$.

---

The first term is $a = 3$.

---

The common difference $d$ is the difference between consecutive terms:

$d = 8 - 3 = 5$.

We can verify this with the next pair: $13 - 8 = 5$. So, $d = 5$.

---

The last term of the AP is $l = 253$.

---

We need to find the 20th term from the last term of this AP. Let $k=20$.

The formula for the $k$-th term from the last of an AP is given by:

---

Substitute the values $l = 253$, $d = 5$, and $k = 20$ into formula (1):

$20$-th term from the end $= 253 - (20-1) \times 5$

$= 253 - (19) \times 5$

$= 253 - 95$

To calculate $253 - 95$:

$\begin{array}{cc} & 2 & 5 & 3 \\ - & & 9 & 5 \\ \hline & 1 & 5 & 8 \\ \hline \end{array}$

$= 158$

---

Thus, the 20th term from the last term of the AP is $\mathbf{158}$.

---

Alternate Solution:

We can also find the total number of terms in the AP and then determine which term from the beginning corresponds to the 20th term from the last.

Let the number of terms in the AP be $n$. The last term is $a_n = 253$.

Using the formula $a_n = a + (n-1)d$ with $a=3$, $d=5$, and $a_n=253$:

$253 = 3 + (n-1)5$

$253 - 3 = (n-1)5$

$250 = 5(n-1)$

$\frac{250}{5} = n-1$

$50 = n-1$

$n = 50 + 1 = 51$.

There are 51 terms in the AP.

---

The $k$-th term from the last of an AP with $n$ terms is the $(n - k + 1)$-th term from the beginning.

Here, $n=51$ and $k=20$.

The term number from the beginning is $51 - 20 + 1 = 31 + 1 = 32$.

So, the 20th term from the last is the 32nd term from the beginning ($a_{32}$).

---

Now, we find the 32nd term using the formula $a_n = a + (n-1)d$ with $a=3$, $d=5$, and $n=32$:

$a_{32} = 3 + (32-1) \times 5$

$a_{32} = 3 + (31) \times 5$

$a_{32} = 3 + 155$

$a_{32} = 158$

---

Using both methods, the 20th term from the last term of the AP is 158.

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

---

The formula for the $n$-th term of an AP is given by $a_n = a + (n-1)d$.

---

The 4th term of the AP is $a_4 = a + (4-1)d = a + 3d$.

The 8th term of the AP is $a_8 = a + (8-1)d = a + 7d$.

The sum of the 4th and 8th terms is 24:

$(a + 3d) + (a + 7d) = 24$

$2a + 10d = 24$

Divide the equation by 2:

---

The 6th term of the AP is $a_6 = a + (6-1)d = a + 5d$.

The 10th term of the AP is $a_{10} = a + (10-1)d = a + 9d$.

The sum of the 6th and 10th terms is 44:

$(a + 5d) + (a + 9d) = 44$

$2a + 14d = 44$

Divide the equation by 2:

---

Now we have a system of two linear equations with two variables, $a$ and $d$:

$a + 5d = 12$ (i)

$a + 7d = 22$ (ii)

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

$(a + 7d) - (a + 5d) = 22 - 12$

$a + 7d - a - 5d = 10$

$2d = 10$

Divide by 2:

$d = \frac{10}{2}$

---

Substitute the value of $d=5$ into equation (i) to find $a$:

$a + 5(5) = 12$

$a + 25 = 12$

Subtract 25 from both sides:

$a = 12 - 25$

---

Now that we have the first term ($a = -13$) and the common difference ($d = 5$), we can find the first three terms of the AP.

First term: $a_1 = a = -13$.

Second term: $a_2 = a + d = -13 + 5 = -8$.

Third term: $a_3 = a + 2d = -13 + 2(5) = -13 + 10 = -3$.

---

The first three terms of the AP are $\mathbf{-13, -8, -3}$.

Let the annual salary in the first year (1995) be the first term of an Arithmetic Progression (AP).

---

The initial salary in 1995 is $a_1 = \textsf{₹}5000$.

The annual increment is constant, which represents the common difference, $d = \textsf{₹}200$.

The annual salaries form an AP:

Year 1 (1995): $\textsf{₹}5000$ ($a_1$)

Year 2 (1996): $\textsf{₹}5000 + \textsf{₹}200 = \textsf{₹}5200$ ($a_2$)

Year 3 (1997): $\textsf{₹}5200 + \textsf{₹}200 = \textsf{₹}5400$ ($a_3$)

and so on.

The AP of salaries is: $5000, 5200, 5400, \dots$

---

We want to find the year in which his income reached $\textsf{₹}7000$. Let this salary be the $n$-th term of the AP, so $a_n = \textsf{₹}7000$.

---

The formula for the $n$-th term of an AP is given by:

---

Substitute the values $a = 5000$, $d = 200$, and $a_n = 7000$ into formula (1):

$7000 = 5000 + (n-1)200$

Subtract 5000 from both sides:

$7000 - 5000 = (n-1)200$

$2000 = 200(n-1)$

Divide both sides by 200:

$\frac{2000}{200} = n-1$

$\frac{\cancel{2000}^{10}}{\cancel{200}_{1}} = n-1$

$10 = n-1$

Add 1 to both sides:

$n = 10 + 1$

$n = 11$

---

The $n$-th term being $\textsf{₹}7000$ means it is the 11th term of the AP.

The first term corresponds to the year 1995 ($n=1$).

The 11th term corresponds to the year $1995 + (11-1)$ years.

Year $= 1995 + 10 = 2005$.

---

Conclusion: Subba Rao's income reached $\textsf{₹}7000$ in the year 2005.

Let the weekly savings in the first week be the first term of an Arithmetic Progression (AP).

---

Savings in the first week is $a_1 = \textsf{₹}5$.

The weekly increase in savings is constant, which represents the common difference, $d = \textsf{₹}1.75$.

The weekly savings form an AP:

Week 1: $\textsf{₹}5$ ($a_1$)

Week 2: $\textsf{₹}5 + \textsf{₹}1.75 = \textsf{₹}6.75$ ($a_2$)

Week 3: $\textsf{₹}6.75 + \textsf{₹}1.75 = \textsf{₹}8.50$ ($a_3$)

and so on.

The AP of weekly savings is: $5, 6.75, 8.50, \dots$

---

We are given that in the $n$-th week, her weekly savings become $\textsf{₹}20.75$. So, the $n$-th term is $a_n = \textsf{₹}20.75$.

---

The formula for the $n$-th term of an AP is given by:

---

Substitute the values $a = 5$, $d = 1.75$, and $a_n = 20.75$ into formula (1):

$20.75 = 5 + (n-1)1.75$

Subtract 5 from both sides:

$20.75 - 5 = (n-1)1.75$

$15.75 = 1.75(n-1)$

Divide both sides by 1.75:

$n-1 = \frac{15.75}{1.75}$

To simplify the division, multiply the numerator and denominator by 100 to remove decimals:

$n-1 = \frac{15.75 \times 100}{1.75 \times 100} = \frac{1575}{175}$

We can perform the division:

$\frac{1575}{175} = \frac{315}{35} = \frac{63}{7} = 9$.

$n-1 = 9$

Add 1 to both sides:

$n = 9 + 1$

$n = 10$

---

Since $n=10$ is a positive integer, in the 10th week, Ramkali's weekly savings become $\textsf{₹}20.75$.

---

Conclusion: The value of $n$ is $\mathbf{10}$.

The given Arithmetic Progression (AP) is: $8, 3, -2, \dots$

---

Here, the first term is $a = 8$.

The common difference $d$ is the difference between consecutive terms:

$d = 3 - 8 = -5$.

We need to find the sum of the first 22 terms, so $n = 22$.

---

The formula for the sum of the first $n$ terms of an AP is given by:

---

Substitute the values $a = 8$, $d = -5$, and $n = 22$ into formula (1) to find the sum of the first 22 terms ($S_{22}$):

$S_{22} = \frac{22}{2}[2(8) + (22-1)(-5)]$

$S_{22} = 11[16 + (21)(-5)]$

$S_{22} = 11[16 - 105]$

$S_{22} = 11[-89]$

To calculate $11 \times -89$:

$\begin{array}{cc} & & 8 & 9 \\ \times & & 1 & 1 \\ \hline & & 8 & 9 \\ & 8 & 9 & \times \\ \hline & 9 & 7 & 9 \\ \hline \end{array}$

So, $11 \times -89 = -979$.

$S_{22} = -979$

---

Thus, the sum of the first 22 terms of the AP is $\mathbf{-979}$.

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

---

We are given that the sum of the first 14 terms ($S_{14}$) is 1050, so $n=14$ and $S_{14} = 1050$.

The first term is $a = 10$.

---

The formula for the sum of the first $n$ terms of an AP is given by:

---

Substitute $S_n = 1050$, $n = 14$, and $a = 10$ into formula (1):

$1050 = \frac{14}{2}[2(10) + (14-1)d]$

$1050 = 7[20 + 13d]$

Divide both sides by 7:

$\frac{1050}{7} = 20 + 13d$

$150 = 20 + 13d$

Subtract 20 from both sides:

$150 - 20 = 13d$

$130 = 13d$

Divide both sides by 13:

$d = \frac{130}{13}$

---

Now that we have the first term ($a = 10$) and the common difference ($d = 10$), we can find the 20th term ($a_{20}$).

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

Using the formula with $n=20$:

$a_{20} = 10 + (20-1) \times 10$

$a_{20} = 10 + (19) \times 10$

$a_{20} = 10 + 190$

$a_{20} = 200$

---

Thus, the 20th term of the AP is $\mathbf{200}$.

The given Arithmetic Progression (AP) is: $24, 21, 18, \dots$

---

Here, the first term is $a = 24$.

The common difference $d$ is the difference between consecutive terms:

$d = 21 - 24 = -3$.

We want to find the number of terms ($n$) that must be taken so that their sum ($S_n$) is 78.

$S_n = 78$.

---

The formula for the sum of the first $n$ terms of an AP is given by:

---

Substitute $S_n = 78$, $a = 24$, and $d = -3$ into formula (1):

$78 = \frac{n}{2}[2(24) + (n-1)(-3)]$

$78 = \frac{n}{2}[48 - 3(n-1)]$

$78 = \frac{n}{2}[48 - 3n + 3]$

$78 = \frac{n}{2}[51 - 3n]$

Multiply both sides by 2:

$2 \times 78 = n(51 - 3n)$

$156 = 51n - 3n^2$

Rearrange the terms to form a quadratic equation:

$3n^2 - 51n + 156 = 0$

Divide the entire equation by 3 to simplify:

$\frac{3n^2}{3} - \frac{51n}{3} + \frac{156}{3} = \frac{0}{3}$

$n^2 - 17n + 52 = 0$

---

Now we solve this quadratic equation for $n$. We can use factorization or the quadratic formula.

Using factorization, we look for two numbers that multiply to 52 and add up to -17. These numbers are -4 and -13.

$n^2 - 4n - 13n + 52 = 0$

$n(n - 4) - 13(n - 4) = 0$

$(n - 4)(n - 13) = 0$

---

This gives two possible values for $n$:

$n - 4 = 0 \implies n = 4$

$n - 13 = 0 \implies n = 13$

---

Both $n=4$ and $n=13$ are positive integers, so both are valid solutions for the number of terms.

This happens because the common difference is negative, causing the terms to decrease. The terms eventually become negative. The sum reaches 78 at the 4th term and then decreases, but later, the sum of subsequent negative terms brings the total sum back to 78 at the 13th term.

The terms of the AP are $24, 21, 18, 15, 12, 9, 6, 3, 0, -3, -6, -9, -12, \dots$

Sum of first 4 terms: $24 + 21 + 18 + 15 = 78$.

Sum of first 13 terms: $S_{13} = \frac{13}{2}[2(24) + (13-1)(-3)] = \frac{13}{2}[48 + 12(-3)] = \frac{13}{2}[48 - 36] = \frac{13}{2}[12] = 13 \times 6 = 78$.

---

Conclusion: The number of terms that must be taken so that their sum is 78 is $\mathbf{4}$ or $\mathbf{13}$.

The sequence of positive integers is $1, 2, 3, 4, \dots$. This sequence forms an Arithmetic Progression (AP) with the first term $a = 1$ and the common difference $d = 1$.

---

The sum of the first $n$ terms of an AP is given by the formula:

Alternatively, if the last term ($a_n$) is known, the sum can be calculated using:

---

(i) The first 1000 positive integers:

The sequence is $1, 2, 3, \dots, 1000$.

Here, the first term $a = 1$, the number of terms $n = 1000$, and the last term $a_n = 1000$.

Using formula (2) for the sum of the first 1000 terms ($S_{1000}$):

$S_{1000} = \frac{1000}{2}(1 + 1000)$

$S_{1000} = \cancel{\frac{1000}{2}}^{500}(1001)$

$S_{1000} = 500 \times 1001$

$S_{1000} = 500500$

The sum of the first 1000 positive integers is $\mathbf{500500}$.

---

(ii) The first n positive integers:

The sequence is $1, 2, 3, \dots, n$.

Here, the first term $a = 1$, the number of terms is $n$, and the last term $a_n = n$.

Using formula (2) for the sum of the first $n$ terms ($S_n$):

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

This is the well-known formula for the sum of the first $n$ positive integers.

The sum of the first $n$ positive integers is $\mathbf{\frac{n(n+1)}{2}}$.

The $n$-th term of the list of numbers is given by the formula:

---

We can find the first few terms of the sequence by substituting $n=1, 2, 3, \dots$ into the formula (i):

For $n=1$, the first term $a_1 = 3 + 2(1) = 3 + 2 = 5$.

For $n=2$, the second term $a_2 = 3 + 2(2) = 3 + 4 = 7$.

For $n=3$, the third term $a_3 = 3 + 2(3) = 3 + 6 = 9$.

The list of numbers is $5, 7, 9, \dots$.

---

Let's check if this sequence is an Arithmetic Progression (AP).

Difference between the second and first term: $a_2 - a_1 = 7 - 5 = 2$.

Difference between the third and second term: $a_3 - a_2 = 9 - 7 = 2$.

Since the difference between consecutive terms is constant (which is 2), the list of numbers forms an AP.

---

In this AP:

The first term is $a = a_1 = 5$.

The common difference is $d = 2$.

We need to find the sum of the first 24 terms, so $n = 24$.

---

The formula for the sum of the first $n$ terms of an AP is given by:

---

Substitute the values $a = 5$, $d = 2$, and $n = 24$ into formula (2) to find the sum of the first 24 terms ($S_{24}$):

$S_{24} = \frac{24}{2}[2(5) + (24-1)2]$

$S_{24} = 12[10 + (23)2]$

$S_{24} = 12[10 + 46]$

$S_{24} = 12[56]$

Calculate $12 \times 56$:

$\begin{array}{cc} & & 5 & 6 \\ \times & & 1 & 2 \\ \hline & 1 & 1 & 2 \\ & 5 & 6 & \times \\ \hline & 6 & 7 & 2 \\ \hline \end{array}$

$S_{24} = 672$

---

Thus, the sum of the first 24 terms of the list of numbers is $\mathbf{672}$.

Let the production of TV sets in the first year be $a$.

Let the uniform increase in production each year (the common difference) be $d$.

The production in subsequent years forms an Arithmetic Progression (AP).

---

The production in the $n$-th year is given by the $n$-th term of the AP, $a_n = a + (n-1)d$.

---

We are given that the production in the third year is 600 sets. So, $a_3 = 600$.

Using the formula $a_n = a + (n-1)d$ with $n=3$:

$a_3 = a + (3-1)d$

$600 = a + 2d$

---

We are also given that the production in the seventh year is 700 sets. So, $a_7 = 700$.

Using the formula $a_n = a + (n-1)d$ with $n=7$:

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

$700 = a + 6d$

---

Now we have a system of two linear equations:

$a + 2d = 600$ (i)

$a + 6d = 700$ (ii)

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

$(a + 6d) - (a + 2d) = 700 - 600$

$a + 6d - a - 2d = 100$

$4d = 100$

Divide by 4:

$d = \frac{100}{4}$

---

Substitute the value of $d=25$ into equation (i) ($a + 2d = 600$) to find $a$:

$a + 2(25) = 600$

$a + 50 = 600$

Subtract 50 from both sides:

$a = 600 - 50$

---

(i) The production in the 1st year:

The production in the 1st year is the first term, $a$.

Production in 1st year = 550 sets.

---

(ii) The production in the 10th year:

We need to find the 10th term of the AP, $a_{10}$.

Using the formula $a_n = a + (n-1)d$ with $a=550$, $d=25$, and $n=10$:

$a_{10} = 550 + (10-1) \times 25$

$a_{10} = 550 + (9) \times 25$

$a_{10} = 550 + 225$

$a_{10} = 775$

The production in the 10th year was 775 sets.

---

(iii) The total production in first 7 years:

We need to find the sum of the first 7 terms of the AP, $S_7$.

We can use the formula for the sum of the first $n$ terms, $S_n = \frac{n}{2}(a + a_n)$, since we know the first term ($a=550$) and the 7th term ($a_7=700$). Here $n=7$.

$S_7 = \frac{7}{2}(a + a_7)$

$S_7 = \frac{7}{2}(550 + 700)$

$S_7 = \frac{7}{2}(1250)$

$S_7 = 7 \times \frac{1250}{2}$

$S_7 = 7 \times 625$

Calculate $7 \times 625$:

$\begin{array}{cc} & & 6 & 2 & 5 \\ \times & & & & 7 \\ \hline & 4 & 3 & 7 & 5 \\ \hline \end{array}$

$S_7 = 4375$

The total production in the first 7 years was 4375 sets.

The formula for the sum of the first $n$ terms of an Arithmetic Progression (AP) is given by:

where $a$ is the first term, $d$ is the common difference, and $n$ is the number of terms.

---

(i) Given AP: $2, 7, 12, \dots$, to 10 terms.

First term, $a = 2$.

Common difference, $d = 7 - 2 = 5$.

Number of terms, $n = 10$.

Using formula (1) to find the sum of the first 10 terms ($S_{10}$):

$S_{10} = \frac{10}{2}[2(2) + (10-1)5]$

$S_{10} = 5[4 + (9)5]$

$S_{10} = 5[4 + 45]$

$S_{10} = 5[49]$

$S_{10} = 245$

The sum of the first 10 terms is $\mathbf{245}$.

---

(ii) Given AP: $-37, -33, -29, \dots$, to 12 terms.

First term, $a = -37$.

Common difference, $d = -33 - (-37) = -33 + 37 = 4$.

Number of terms, $n = 12$.

Using formula (1) to find the sum of the first 12 terms ($S_{12}$):

$S_{12} = \frac{12}{2}[2(-37) + (12-1)4]$

$S_{12} = 6[-74 + (11)4]$

$S_{12} = 6[-74 + 44]$

$S_{12} = 6[-30]$

$S_{12} = -180$

The sum of the first 12 terms is $\mathbf{-180}$.

---

(iii) Given AP: $0.6, 1.7, 2.8, \dots$, to 100 terms.

First term, $a = 0.6$.

Common difference, $d = 1.7 - 0.6 = 1.1$.

Number of terms, $n = 100$.

Using formula (1) to find the sum of the first 100 terms ($S_{100}$):

$S_{100} = \frac{100}{2}[2(0.6) + (100-1)1.1]$

$S_{100} = 50[1.2 + (99)1.1]$

Calculate $99 \times 1.1$: $99 \times (1 + 0.1) = 99 + 9.9 = 108.9$.

$S_{100} = 50[1.2 + 108.9]$

$S_{100} = 50[110.1]$

Calculate $50 \times 110.1$: $50 \times 110.1 = 5 \times 1101 = 5505$.

$S_{100} = 5505$

The sum of the first 100 terms is $\mathbf{5505}$.

---

(iv) Given AP: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$, to 11 terms.

First term, $a = \frac{1}{15}$.

Common difference, $d = \frac{1}{12} - \frac{1}{15}$. The least common multiple of 12 and 15 is 60.

$d = \frac{1 \times 5}{12 \times 5} - \frac{1 \times 4}{15 \times 4} = \frac{5}{60} - \frac{4}{60} = \frac{1}{60}$.

Number of terms, $n = 11$.

Using formula (1) to find the sum of the first 11 terms ($S_{11}$):

$S_{11} = \frac{11}{2}[2(\frac{1}{15}) + (11-1)\frac{1}{60}]$

$S_{11} = \frac{11}{2}[\frac{2}{15} + (10)\frac{1}{60}]$

$S_{11} = \frac{11}{2}[\frac{2}{15} + \frac{10}{60}]$

Simplify $\frac{10}{60} = \frac{1}{6}$.

$S_{11} = \frac{11}{2}[\frac{2}{15} + \frac{1}{6}]$

The least common multiple of 15 and 6 is 30.

$S_{11} = \frac{11}{2}[\frac{2 \times 2}{15 \times 2} + \frac{1 \times 5}{6 \times 5}]$

$S_{11} = \frac{11}{2}[\frac{4}{30} + \frac{5}{30}]$

$S_{11} = \frac{11}{2}[\frac{4+5}{30}]$

$S_{11} = \frac{11}{2}[\frac{9}{30}]$

Simplify $\frac{9}{30} = \frac{\cancel{9}^3}{\cancel{30}_{10}} = \frac{3}{10}$.

$S_{11} = \frac{11}{2} \times \frac{3}{10}$

$S_{11} = \frac{11 \times 3}{2 \times 10}$

$S_{11} = \frac{33}{20}$

The sum of the first 11 terms is $\mathbf{\frac{33}{20}}$.

To find the sum of an Arithmetic Progression (AP), we first need to identify the first term ($a$), the common difference ($d$), and the number of terms ($n$). If the last term ($a_n$) is known, we can use the formula $S_n = \frac{n}{2}(a + a_n)$. Otherwise, we use $S_n = \frac{n}{2}[2a + (n-1)d]$. To find $n$, we use the formula $a_n = a + (n-1)d$.

---

(i) Given series: $7 + 10\frac{1}{2} + 14 + \dots + 84$

This is an AP with:

First term, $a = 7$.

The common difference, $d = 10\frac{1}{2} - 7 = \frac{21}{2} - \frac{14}{2} = \frac{7}{2}$.

The last term is $a_n = 84$.

We need to find the number of terms, $n$. Using $a_n = a + (n-1)d$:

$84 = 7 + (n-1)\frac{7}{2}$

$84 - 7 = (n-1)\frac{7}{2}$

$77 = (n-1)\frac{7}{2}$

Multiply both sides by $\frac{2}{7}$:

$77 \times \frac{2}{7} = n-1$

$\cancel{77}^{11} \times \frac{2}{\cancel{7}_1} = n-1$

$11 \times 2 = n-1$

$22 = n-1$

$n = 22 + 1 = 23$.

There are 23 terms in the AP.

Now find the sum of the first 23 terms ($S_{23}$) using $S_n = \frac{n}{2}(a + a_n)$:

$S_{23} = \frac{23}{2}(7 + 84)$

$S_{23} = \frac{23}{2}(91)$

$S_{23} = \frac{23 \times 91}{2}$

$S_{23} = \frac{2093}{2}$

$S_{23} = 1046.5$ or $1046\frac{1}{2}$.

The sum of the given series is $\mathbf{1046.5}$ or $\mathbf{1046\frac{1}{2}}$.

---

(ii) Given series: $34 + 32 + 30 + \dots + 10$

This is an AP with:

First term, $a = 34$.

Common difference, $d = 32 - 34 = -2$.

The last term is $a_n = 10$.

We need to find the number of terms, $n$. Using $a_n = a + (n-1)d$:

$10 = 34 + (n-1)(-2)$

$10 - 34 = (n-1)(-2)$

$-24 = -2(n-1)$

Divide both sides by -2:

$\frac{-24}{-2} = n-1$

$12 = n-1$

$n = 12 + 1 = 13$.

There are 13 terms in the AP.

Now find the sum of the first 13 terms ($S_{13}$) using $S_n = \frac{n}{2}(a + a_n)$:

$S_{13} = \frac{13}{2}(34 + 10)$

$S_{13} = \frac{13}{2}(44)$

$S_{13} = 13 \times \frac{44}{2}$

$S_{13} = 13 \times 22$

$S_{13} = 286$

The sum of the given series is $\mathbf{286}$.

---

(iii) Given series: $-5 + (-8) + (-11) + \dots + (-230)$

This is an AP with:

First term, $a = -5$.

Common difference, $d = -8 - (-5) = -8 + 5 = -3$.

The last term is $a_n = -230$.

We need to find the number of terms, $n$. Using $a_n = a + (n-1)d$:

$-230 = -5 + (n-1)(-3)$

$-230 - (-5) = (n-1)(-3)$

$-230 + 5 = (n-1)(-3)$

$-225 = -3(n-1)$

Divide both sides by -3:

$\frac{-225}{-3} = n-1$

$75 = n-1$

$n = 75 + 1 = 76$.

There are 76 terms in the AP.

Now find the sum of the first 76 terms ($S_{76}$) using $S_n = \frac{n}{2}(a + a_n)$:

$S_{76} = \frac{76}{2}(-5 + (-230))$

$S_{76} = 38(-5 - 230)$

$S_{76} = 38(-235)$

Calculate $38 \times (-235)$: $38 \times 235 = 8930$. So $38 \times (-235) = -8930$.

The sum of the given series is $\mathbf{-8930}$.

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

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

---

(i) Given: $a = 5$, $d = 3$, $a_n = 50$. Find $n$ and $S_n$.

Using $a_n = a + (n-1)d$:

$50 = 5 + (n-1)3$

$50 - 5 = 3(n-1)$

$45 = 3(n-1)$

$\frac{45}{3} = n-1$

$15 = n-1$

$n = 15 + 1 = 16$.

Now find $S_n$ (which is $S_{16}$) using $S_n = \frac{n}{2}(a + a_n)$:

$S_{16} = \frac{16}{2}(5 + 50)$

$S_{16} = 8(55)$

$S_{16} = 440$.

So, $n = \mathbf{16}$ and $S_n = \mathbf{440}$.

---

(ii) Given: $a = 7$, $a_{13} = 35$. Find $d$ and $S_{13}$.

Using $a_n = a + (n-1)d$ with $n=13$ and $a_{13}=35$:

$35 = 7 + (13-1)d$

$35 = 7 + 12d$

$35 - 7 = 12d$

$28 = 12d$

$d = \frac{28}{12} = \frac{7}{3}$.

Now find $S_{13}$ using $S_n = \frac{n}{2}(a + a_n)$ with $n=13$ and $a_{13}=35$:

$S_{13} = \frac{13}{2}(7 + 35)$

$S_{13} = \frac{13}{2}(42)$

$S_{13} = 13 \times 21$

$S_{13} = 273$.

So, $d = \mathbf{\frac{7}{3}}$ and $S_{13} = \mathbf{273}$.

---

(iii) Given: $a_{12} = 37$, $d = 3$. Find $a$ and $S_{12}$.

Using $a_n = a + (n-1)d$ with $n=12$ and $a_{12}=37$:

$37 = a + (12-1)3$

$37 = a + 11 \times 3$

$37 = a + 33$

$a = 37 - 33 = 4$.

Now find $S_{12}$ using $S_n = \frac{n}{2}(a + a_n)$ with $n=12$ and $a_{12}=37$:

$S_{12} = \frac{12}{2}(4 + 37)$

$S_{12} = 6(41)$

$S_{12} = 246$.

So, $a = \mathbf{4}$ and $S_{12} = \mathbf{246}$.

---

(iv) Given: $a_3 = 15$, $S_{10} = 125$. Find $d$ and $a_{10}$.

Using $a_n = a + (n-1)d$ for $a_3$ ($n=3$):

Using $S_n = \frac{n}{2}[2a + (n-1)d]$ for $S_{10}$ ($n=10$):

$125 = \frac{10}{2}[2a + (10-1)d]$

$125 = 5[2a + 9d]$

Divide by 5:

$\frac{125}{5} = 2a + 9d$

From (i), $a = 15 - 2d$. Substitute this into (ii):

$25 = 2(15 - 2d) + 9d$

$25 = 30 - 4d + 9d$

$25 = 30 + 5d$

$25 - 30 = 5d$

$-5 = 5d$

$d = \frac{-5}{5} = -1$.

Now find $a$ using $a = 15 - 2d$:

$a = 15 - 2(-1) = 15 + 2 = 17$.

Finally, find $a_{10}$ using $a_n = a + (n-1)d$ with $a=17$, $d=-1$, and $n=10$:

$a_{10} = 17 + (10-1)(-1)$

$a_{10} = 17 + (9)(-1)$

$a_{10} = 17 - 9 = 8$.

So, $d = \mathbf{-1}$ and $a_{10} = \mathbf{8}$.

---

(v) Given: $d = 5$, $S_9 = 75$. Find $a$ and $a_9$.

Using $S_n = \frac{n}{2}[2a + (n-1)d]$ for $S_9$ ($n=9$):

$75 = \frac{9}{2}[2a + (9-1)5]$

$75 = \frac{9}{2}[2a + 8 \times 5]$

$75 = \frac{9}{2}[2a + 40]$

Multiply by 2 and divide by 9:

$\frac{75 \times 2}{9} = 2a + 40$

$\frac{150}{9} = 2a + 40$

$\frac{50}{3} = 2a + 40$

Subtract 40 from both sides:

$\frac{50}{3} - 40 = 2a$

$\frac{50}{3} - \frac{120}{3} = 2a$

$\frac{50 - 120}{3} = 2a$

$\frac{-70}{3} = 2a$

$a = \frac{-70}{3 \times 2} = \frac{-70}{6} = \frac{-35}{3}$.

Now find $a_9$ using $a_n = a + (n-1)d$ with $a = -\frac{35}{3}$, $d=5$, and $n=9$:

$a_9 = -\frac{35}{3} + (9-1)5$

$a_9 = -\frac{35}{3} + 8 \times 5$

$a_9 = -\frac{35}{3} + 40$

$a_9 = -\frac{35}{3} + \frac{120}{3}$

$a_9 = \frac{-35 + 120}{3} = \frac{85}{3}$.

So, $a = \mathbf{-\frac{35}{3}}$ and $a_9 = \mathbf{\frac{85}{3}}$.

---

(vi) Given: $a = 2$, $d = 8$, $S_n = 90$. Find $n$ and $a_n$.

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

$90 = \frac{n}{2}[2(2) + (n-1)8]$

$90 = \frac{n}{2}[4 + 8n - 8]$

$90 = \frac{n}{2}[8n - 4]$

$90 = n(4n - 2)$

$90 = 4n^2 - 2n$

Rearrange into a quadratic equation:

$4n^2 - 2n - 90 = 0$

Divide by 2:

$2n^2 - n - 45 = 0$.

Use the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=2$, $b=-1$, $c=-45$.

$n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-45)}}{2(2)}$

$n = \frac{1 \pm \sqrt{1 + 360}}{4}$

$n = \frac{1 \pm \sqrt{361}}{4}$

$n = \frac{1 \pm 19}{4}$.

Two possible values for $n$:

$n_1 = \frac{1 + 19}{4} = \frac{20}{4} = 5$

$n_2 = \frac{1 - 19}{4} = \frac{-18}{4} = -\frac{9}{2}$.

Since $n$ must be a positive integer, $n=5$ is the only valid solution.

Now find $a_n$ (which is $a_5$) using $a_n = a + (n-1)d$ with $a=2$, $d=8$, and $n=5$:

$a_5 = 2 + (5-1)8$

$a_5 = 2 + 4 \times 8$

$a_5 = 2 + 32 = 34$.

So, $n = \mathbf{5}$ and $a_n = \mathbf{34}$.

---

(vii) Given: $a = 8$, $a_n = 62$, $S_n = 210$. Find $n$ and $d$.

Using $S_n = \frac{n}{2}(a + a_n)$:

$210 = \frac{n}{2}(8 + 62)$

$210 = \frac{n}{2}(70)$

$210 = 35n$

$n = \frac{210}{35} = 6$.

Now use $a_n = a + (n-1)d$ with $a=8$, $a_n=62$, and $n=6$ to find $d$:

$62 = 8 + (6-1)d$

$62 = 8 + 5d$

$62 - 8 = 5d$

$54 = 5d$

$d = \frac{54}{5}$.

So, $n = \mathbf{6}$ and $d = \mathbf{\frac{54}{5}}$ (or 10.8).

---

(viii) Given: $a_n = 4$, $d = 2$, $S_n = –14$. Find $n$ and $a$.

Using $a_n = a + (n-1)d$:

Using $S_n = \frac{n}{2}(a + a_n)$:

$-14 = \frac{n}{2}(a + 4)$

Multiply by 2:

From (i), $4 = a + 2n - 2 \implies 6 = a + 2n \implies a = 6 - 2n$.

Substitute this expression for $a$ into equation (ii):

$-28 = n((6 - 2n) + 4)$

$-28 = n(10 - 2n)$

$-28 = 10n - 2n^2$

Rearrange into a quadratic equation:

$2n^2 - 10n - 28 = 0$

Divide by 2:

$n^2 - 5n - 14 = 0$.

Factorize the quadratic equation. We look for two numbers that multiply to -14 and add up to -5. These are -7 and 2.

$(n - 7)(n + 2) = 0$.

Two possible values for $n$:

$n - 7 = 0 \implies n = 7$

$n + 2 = 0 \implies n = -2$.

Since $n$ must be a positive integer, $n=7$ is the only valid solution.

Now find $a$ using $a = 6 - 2n$ with $n=7$:

$a = 6 - 2(7)$

$a = 6 - 14 = -8$.

So, $n = \mathbf{7}$ and $a = \mathbf{-8}$.

---

(ix) Given: $a = 3$, $n = 8$, $S = 192$. Find $d$.

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

$192 = \frac{8}{2}[2(3) + (8-1)d]$

$192 = 4[6 + 7d]$

Divide by 4:

$\frac{192}{4} = 6 + 7d$

$48 = 6 + 7d$

$48 - 6 = 7d$

$42 = 7d$

$d = \frac{42}{7} = 6$.

So, $d = \mathbf{6}$.

---

(x) Given: $l = 28$, $S = 144$, and there are total 9 terms. Find $a$.

The last term is given as $l$, which is the $n$-th term, $a_n$. So, $a_n = 28$.

The total number of terms is 9, so $n = 9$.

The sum is given as $S = 144$, so $S_9 = 144$.

We need to find the first term, $a$.

Using the formula $S_n = \frac{n}{2}(a + a_n)$, substitute $S_9=144$, $n=9$, and $a_n=28$:

$144 = \frac{9}{2}(a + 28)$

Multiply by 2 and divide by 9:

$\frac{144 \times 2}{9} = a + 28$

$\frac{288}{9} = a + 28$

$32 = a + 28$

Subtract 28 from both sides:

$a = 32 - 28 = 4$.

So, $a = \mathbf{4}$.

The given Arithmetic Progression (AP) is: $9, 17, 25, \dots$

---

Here, the first term is $a = 9$.

The common difference $d$ is the difference between consecutive terms:

$d = 17 - 9 = 8$.

We want to find the number of terms ($n$) that must be taken so that their sum ($S_n$) is 636.

$S_n = 636$.

---

The formula for the sum of the first $n$ terms of an AP is given by:

---

Substitute $S_n = 636$, $a = 9$, and $d = 8$ into formula (1):

$636 = \frac{n}{2}[2(9) + (n-1)8]$

$636 = \frac{n}{2}[18 + 8n - 8]$

$636 = \frac{n}{2}[10 + 8n]$

$636 = n(5 + 4n)$

$636 = 5n + 4n^2$

Rearrange the terms to form a quadratic equation:

$4n^2 + 5n - 636 = 0$

---

Now we solve this quadratic equation for $n$. We can use factorization or the quadratic formula.

Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=4$, $b=5$, $c=-636$.

First, calculate the discriminant $\Delta = b^2 - 4ac$:

$\Delta = (5)^2 - 4(4)(-636)$

$\Delta = 25 + 16(636)$

$\Delta = 25 + 10176$

$\Delta = 10201$

Now, find the square root of the discriminant:

$\sqrt{10201}$. We know $100^2 = 10000$, so the number is close to 100 and ends in 1. It could be 101.

$101 \times 101 = 10201$.

So, $\sqrt{10201} = 101$.

---

Now substitute the values into the quadratic formula for $n$:

$n = \frac{-5 \pm 101}{2(4)}$

$n = \frac{-5 \pm 101}{8}$.

Two possible values for $n$:

$n_1 = \frac{-5 + 101}{8} = \frac{96}{8} = 12$

$n_2 = \frac{-5 - 101}{8} = \frac{-106}{8} = -\frac{53}{4}$.

---

Since $n$ represents the number of terms, it must be a positive integer. Therefore, $n=12$ is the only valid solution.

---

Conclusion: $\mathbf{12}$ terms of the AP must be taken to give a sum of 636.

Given: First term $a = 5$.

Given: Last term $a_n = 45$.

Given: Sum of the terms $S_n = 400$.

---

We need to find the number of terms ($n$) and the common difference ($d$).

---

We can use the formula for the sum of the first $n$ terms when the first and last terms are known:

---

Substitute the given values into formula (1):

$400 = \frac{n}{2}(5 + 45)$

$400 = \frac{n}{2}(50)$

$400 = 25n$

Divide both sides by 25:

$n = \frac{400}{25}$

$n = 16$.

---

So, the number of terms in the AP is $\mathbf{16}$.

---

Now that we know $n=16$, the last term $a_n=45$ is actually the 16th term, $a_{16}=45$.

We can use the formula for the $n$-th term of an AP to find the common difference $d$:

---

Substitute $a_{16} = 45$, $a = 5$, and $n = 16$ into formula (2):

$45 = 5 + (16-1)d$

$45 = 5 + 15d$

Subtract 5 from both sides:

$45 - 5 = 15d$

$40 = 15d$

Divide both sides by 15:

$d = \frac{40}{15}$

Simplify the fraction by dividing numerator and denominator by their greatest common divisor, 5:

$d = \frac{40 \div 5}{15 \div 5} = \frac{8}{3}$.

---

So, the common difference is $\mathbf{\frac{8}{3}}$.

Given: First term $a = 17$.

Given: Last term $a_n = 350$.

Given: Common difference $d = 9$.

---

We need to find the number of terms ($n$) and the sum of the terms ($S_n$).

---

First, find the number of terms ($n$) using the formula for the $n$-th term of an AP:

---

Substitute the given values into formula (1):

$350 = 17 + (n-1)9$

Subtract 17 from both sides:

$350 - 17 = (n-1)9$

$333 = 9(n-1)$

Divide both sides by 9:

$\frac{333}{9} = n-1$

Performing the division:

$37 = n-1$

Add 1 to both sides:

$n = 37 + 1 = 38$.

---

So, the number of terms is $\mathbf{38}$.

---

Now, find the sum of these 38 terms ($S_{38}$). We can use the formula for the sum when the first term ($a$), the last term ($a_n$), and the number of terms ($n$) are known:

---

Substitute the values $n=38$, $a=17$, and $a_{38}=350$ into formula (2):

$S_{38} = \frac{38}{2}(17 + 350)$

$S_{38} = 19(367)$

Calculate $19 \times 367$:

$\begin{array}{cc}& & 3 & 6 & 7 \\ \times & & & 1 & 9 \\ \hline & 3 & 3 & 0 & 3 \\ & 3 & 6 & 7 & \times \\ \hline & 6 & 9 & 7 & 3 \\ \hline \end{array}$

$S_{38} = 6973$.

---

Thus, the sum of the 38 terms is $\mathbf{6973}$.

Given: Common difference $d = 7$.

Given: The 22nd term $a_{22} = 149$.

We need to find the sum of the first 22 terms, $S_{22}$. This means the number of terms is $n=22$.

---

To find the sum of the first 22 terms, we can use the formula $S_n = \frac{n}{2}(a + a_n)$, since we know $n$, $d$, and $a_n$ (where $n=22$). However, we first need to find the first term ($a$).

---

Use the formula for the $n$-th term of an AP to find the first term ($a$):

---

Substitute $a_{22} = 149$, $n = 22$, and $d = 7$ into formula (1):

$149 = a + (22-1)7$

$149 = a + (21)7$

$149 = a + 147$

Subtract 147 from both sides:

$a = 149 - 147$

$a = 2$.

---

So, the first term is $a = 2$.

---

Now, find the sum of the first 22 terms ($S_{22}$) using the formula $S_n = \frac{n}{2}(a + a_n)$ with $n=22$, $a=2$, and $a_{22}=149$:

---

$S_{22} = \frac{22}{2}(2 + 149)$

$S_{22} = 11(151)$

Calculate $11 \times 151$:

$\begin{array}{cc}& & 1 & 5 & 1 \\ \times & & & 1 & 1 \\ \hline & & 1 & 5 & 1 \\ & 1 & 5 & 1 & \times \\ \hline & 1 & 6 & 6 & 1 \\ \hline \end{array}$

$S_{22} = 1661$.

---

Thus, the sum of the first 22 terms of the AP is $\mathbf{1661}$.

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

---

We are given that the second term ($a_2$) is 14, and the third term ($a_3$) is 18.

---

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

$d = a_3 - a_2$

$d = 18 - 14$

$d = 4$.

---

Now we need to find the first term ($a$). The second term is $a_2 = a + d$.

$14 = a + 4$

Subtract 4 from both sides:

$a = 14 - 4$

$a = 10$.

---

So, the first term is $a = 10$ and the common difference is $d = 4$.

We need to find the sum of the first 51 terms, so $n = 51$.

---

The formula for the sum of the first $n$ terms of an AP is given by:

---

Substitute the values $a = 10$, $d = 4$, and $n = 51$ into formula (1) to find the sum of the first 51 terms ($S_{51}$):

$S_{51} = \frac{51}{2}[2(10) + (51-1)4]$

$S_{51} = \frac{51}{2}[20 + (50)4]$

$S_{51} = \frac{51}{2}[20 + 200]$

$S_{51} = \frac{51}{2}[220]$

$S_{51} = 51 \times \frac{220}{2}$

$S_{51} = 51 \times 110$

Calculate $51 \times 110$:

$51 \times 110 = 51 \times 11 \times 10$

$\begin{array}{cc}& & 5 & 1 \\ \times & & 1 & 1 \\ \hline & & 5 & 1 \\ & 5 & 1 & \times \\ \hline & 5 & 6 & 1 \\ \hline \end{array}$

$51 \times 11 = 561$.

So, $S_{51} = 561 \times 10 = 5610$.

---

Thus, the sum of the first 51 terms of the AP is $\mathbf{5610}$.

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

---

The formula for the sum of the first $n$ terms of an AP is given by:

---

We are given that the sum of the first 7 terms ($S_7$) is 49.

Using formula (1) with $n=7$ and $S_7=49$:

$49 = \frac{7}{2}[2a + (7-1)d]$

$49 = \frac{7}{2}[2a + 6d]$

$49 = \frac{7}{2} \times 2[a + 3d]$

$49 = 7[a + 3d]$

Divide both sides by 7:

$\frac{49}{7} = a + 3d$

---

We are also given that the sum of the first 17 terms ($S_{17}$) is 289.

Using formula (1) with $n=17$ and $S_{17}=289$:

$289 = \frac{17}{2}[2a + (17-1)d]$

$289 = \frac{17}{2}[2a + 16d]$

$289 = \frac{17}{2} \times 2[a + 8d]$

$289 = 17[a + 8d]$

Divide both sides by 17 ($289 = 17^2$, so $289 \div 17 = 17$):

$\frac{289}{17} = a + 8d$

---

Now we have a system of two linear equations:

$a + 3d = 7$ (i)

$a + 8d = 17$ (ii)

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

$(a + 8d) - (a + 3d) = 17 - 7$

$a + 8d - a - 3d = 10$

$5d = 10$

Divide by 5:

$d = \frac{10}{5}$

---

Substitute the value of $d=2$ into equation (i) ($a + 3d = 7$) to find $a$:

$a + 3(2) = 7$

$a + 6 = 7$

$a = 7 - 6$

---

Now that we have the first term ($a = 1$) and the common difference ($d = 2$), we can find the sum of the first $n$ terms ($S_n$).

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

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

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

$S_n = \frac{n}{2}[2n]$

$S_n = \frac{n}{\cancel{2}_1} \times \cancel{2}^1 n$

$S_n = n \times n$

$S_n = n^2$

---

Thus, the sum of the first $n$ terms of the AP is $\mathbf{n^2}$.

A sequence $a_1, a_2, a_3, \dots, a_n, \dots$ forms an Arithmetic Progression (AP) if the difference between consecutive terms is constant. That is, $a_{n+1} - a_n = d$ for all $n \geq 1$, where $d$ is the common difference.

---

(i) Given $a_n = 3 + 4n$.

To show it is an AP, we find the difference between the $(n+1)$-th term and the $n$-th term.

$a_{n+1} = 3 + 4(n+1) = 3 + 4n + 4 = 7 + 4n$.

$a_{n+1} - a_n = (7 + 4n) - (3 + 4n)$

$a_{n+1} - a_n = 7 + 4n - 3 - 4n$

$a_{n+1} - a_n = 4$.

Since the difference between consecutive terms is a constant (4), the sequence forms an AP with common difference $d = 4$.

---

Now find the sum of the first 15 terms ($S_{15}$).

The first term is $a_1 = 3 + 4(1) = 3 + 4 = 7$. So, $a=7$.

The common difference is $d=4$.

The number of terms is $n=15$.

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

$S_{15} = \frac{15}{2}[2(7) + (15-1)4]$

$S_{15} = \frac{15}{2}[14 + (14)4]$

$S_{15} = \frac{15}{2}[14 + 56]$

$S_{15} = \frac{15}{2}[70]$

$S_{15} = 15 \times \frac{70}{2}$

$S_{15} = 15 \times 35$

Calculate $15 \times 35$:

$\begin{array}{cc}& & 3 & 5 \\ \times & & 1 & 5 \\ \hline & 1 & 7 & 5 \\ & 3 & 5 & \times \\ \hline & 5 & 2 & 5 \\ \hline \end{array}$

$S_{15} = 525$.

The sum of the first 15 terms is $\mathbf{525}$.

---

(ii) Given $a_n = 9 – 5n$.

To show it is an AP, we find the difference between the $(n+1)$-th term and the $n$-th term.

$a_{n+1} = 9 - 5(n+1) = 9 - 5n - 5 = 4 - 5n$.

$a_{n+1} - a_n = (4 - 5n) - (9 - 5n)$

$a_{n+1} - a_n = 4 - 5n - 9 + 5n$

$a_{n+1} - a_n = -5$.

Since the difference between consecutive terms is a constant (-5), the sequence forms an AP with common difference $d = -5$.

---

Now find the sum of the first 15 terms ($S_{15}$).

The first term is $a_1 = 9 - 5(1) = 9 - 5 = 4$. So, $a=4$.

The common difference is $d=-5$.

The number of terms is $n=15$.

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

$S_{15} = \frac{15}{2}[2(4) + (15-1)(-5)]$

$S_{15} = \frac{15}{2}[8 + (14)(-5)]$

$S_{15} = \frac{15}{2}[8 - 70]$

$S_{15} = \frac{15}{2}[-62]$

$S_{15} = 15 \times \frac{-62}{2}$

$S_{15} = 15 \times (-31)$

Calculate $15 \times (-31)$: $15 \times 31 = 465$. So $15 \times (-31) = -465$.

$S_{15} = -465$.

The sum of the first 15 terms is $\mathbf{-465}$.

The sum of the first $n$ terms of an AP is given by $S_n = 4n - n^2$.

---

The first term ($a_1$) is the sum of the first 1 term ($S_1$).

$S_1 = 4(1) - (1)^2 = 4 - 1 = 3$.

So, the first term is $a_1 = S_1 = \mathbf{3}$.

---

The sum of the first two terms ($S_2$) is found by substituting $n=2$ into the formula for $S_n$.

$S_2 = 4(2) - (2)^2 = 8 - 4 = 4$.

The sum of the first two terms is $\mathbf{4}$.

---

The second term ($a_2$) is the difference between the sum of the first two terms ($S_2$) and the sum of the first one term ($S_1$).

$a_2 = S_2 - S_1$

$a_2 = 4 - 3 = 1$.

The second term is $\mathbf{1}$.

---

The common difference ($d$) of the AP is the difference between the second term and the first term.

$d = a_2 - a_1 = 1 - 3 = -2$.

---

The third term ($a_3$) can be found using $a_3 = a_2 + d$ or using the relationship $a_n = S_n - S_{n-1}$ for $n > 1$.

Using $a_3 = a_2 + d$: $a_3 = 1 + (-2) = 1 - 2 = -1$.

Alternatively, using $a_n = S_n - S_{n-1}$: We find $S_3$.

$S_3 = 4(3) - (3)^2 = 12 - 9 = 3$.

$a_3 = S_3 - S_2 = 3 - 4 = -1$.

The third term is $\mathbf{-1}$.

---

The 10th term ($a_{10}$) can be found using the formula $a_n = a_1 + (n-1)d$, with $a_1 = 3$ and $d = -2$, and $n=10$.

$a_{10} = 3 + (10-1)(-2)$

$a_{10} = 3 + (9)(-2)$

$a_{10} = 3 - 18 = -15$.

The 10th term is $\mathbf{-15}$.

---

The $n$-th term ($a_n$) can be found using the formula $a_n = a_1 + (n-1)d$ with $a_1 = 3$ and $d = -2$.

$a_n = 3 + (n-1)(-2)$

$a_n = 3 - 2n + 2$

$a_n = 5 - 2n$.

Alternatively, using $a_n = S_n - S_{n-1}$ for $n > 1$:

$S_{n-1} = 4(n-1) - (n-1)^2$

$S_{n-1} = 4n - 4 - (n^2 - 2n + 1)$

$S_{n-1} = 4n - 4 - n^2 + 2n - 1$

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

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

$a_n = (4n - n^2) - (-n^2 + 6n - 5)$

$a_n = 4n - n^2 + n^2 - 6n + 5$

$a_n = -2n + 5 = 5 - 2n$.

This formula holds for $n > 1$. For $n=1$, $a_1 = 5 - 2(1) = 3$, which matches $S_1$. So the formula $a_n = 5-2n$ is valid for all positive integers $n$.

The $n$-th term is $\mathbf{5-2n}$.

The positive integers divisible by 6 are the multiples of 6.

The first few positive integers divisible by 6 are $6, 12, 18, 24, \dots$.

This sequence forms an Arithmetic Progression (AP).

---

In this AP:

The first term is $a = 6$.

The common difference is $d = 12 - 6 = 6$.

We need to find the sum of the first 40 terms, so $n = 40$.

---

The formula for the sum of the first $n$ terms of an AP is given by:

---

Substitute the values $a = 6$, $d = 6$, and $n = 40$ into formula (1) to find the sum of the first 40 terms ($S_{40}$):

$S_{40} = \frac{40}{2}[2(6) + (40-1)6]$

$S_{40} = 20[12 + (39)6]$

Calculate $39 \times 6$:

$39 \times 6 = (40 - 1) \times 6 = 40 \times 6 - 1 \times 6 = 240 - 6 = 234$.

$S_{40} = 20[12 + 234]$

$S_{40} = 20[246]$

Calculate $20 \times 246$:

$20 \times 246 = 2 \times 10 \times 246 = 2 \times 2460 = 4920$.

$S_{40} = 4920$.

---

Thus, the sum of the first 40 positive integers divisible by 6 is $\mathbf{4920}$.

---

Alternate Method:

The first 40 positive integers divisible by 6 are $6 \times 1, 6 \times 2, 6 \times 3, \dots, 6 \times 40$.

The sequence is $6, 12, 18, \dots, 240$.

This is an AP with $a = 6$, $n = 40$, and the last term $a_{40} = 240$.

Using the sum formula $S_n = \frac{n}{2}(a + a_n)$:

$S_{40} = \frac{40}{2}(6 + 240)$

$S_{40} = 20(246)$

$S_{40} = 4920$.

The sum is 4920.

The first 15 multiples of 8 are $8 \times 1, 8 \times 2, 8 \times 3, \dots, 8 \times 15$.

The sequence is $8, 16, 24, \dots, 120$.

This sequence forms an Arithmetic Progression (AP).

---

In this AP:

The first term is $a = 8$.

The common difference is $d = 16 - 8 = 8$.

The number of terms is $n = 15$.

The last term is $a_{15} = 8 \times 15 = 120$.

---

We need to find the sum of the first 15 terms, $S_{15}$.

Using the formula for the sum of the first $n$ terms when the first and last terms are known:

---

Substitute the values $n=15$, $a=8$, and $a_{15}=120$ into formula (1):

$S_{15} = \frac{15}{2}(8 + 120)$

$S_{15} = \frac{15}{2}(128)$

$S_{15} = 15 \times \frac{128}{2}$

$S_{15} = 15 \times 64$

Calculate $15 \times 64$:

$\begin{array}{cc}& & 6 & 4 \\ \times & & 1 & 5 \\ \hline & 3 & 2 & 0 \\ & 6 & 4 & \times \\ \hline & 9 & 6 & 0 \\ \hline \end{array}$

$S_{15} = 960$.

---

Thus, the sum of the first 15 multiples of 8 is $\mathbf{960}$.

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Alternate Method:

Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with $a=8$, $d=8$, and $n=15$:

$S_{15} = \frac{15}{2}[2(8) + (15-1)8]$

$S_{15} = \frac{15}{2}[16 + (14)8]$

$S_{15} = \frac{15}{2}[16 + 112]$

$S_{15} = \frac{15}{2}[128]$

$S_{15} = 15 \times 64 = 960$.

The sum is 960.

The odd numbers between 0 and 50 are $1, 3, 5, \dots, 49$.

This sequence forms an Arithmetic Progression (AP) because the difference between consecutive odd numbers is constant (2).

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In this AP:

The first term is $a = 1$.

The common difference is $d = 3 - 1 = 2$.

The last term is $a_n = 49$.

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First, we need to find the number of terms ($n$) in this AP.

Using the formula for the $n$-th term of an AP:

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Substitute the values $a=1$, $d=2$, and $a_n=49$ into formula (1):

$49 = 1 + (n-1)2$

$49 - 1 = (n-1)2$

$48 = 2(n-1)$

Divide both sides by 2:

$\frac{48}{2} = n-1$

$24 = n-1$

Add 1 to both sides:

$n = 24 + 1 = 25$.

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So, there are 25 odd numbers between 0 and 50.

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Now, find the sum of these 25 terms ($S_{25}$). We can use the formula for the sum when the first term ($a$), the last term ($a_n$), and the number of terms ($n$) are known:

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Substitute the values $n=25$, $a=1$, and $a_{25}=49$ into formula (2):

$S_{25} = \frac{25}{2}(1 + 49)$

$S_{25} = \frac{25}{2}(50)$

$S_{25} = 25 \times \frac{50}{2}$

$S_{25} = 25 \times 25$

$S_{25} = 625$.

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Thus, the sum of the odd numbers between 0 and 50 is $\mathbf{625}$.

The penalties for delay for each day form a sequence:

Day 1 penalty: $\textsf{₹}200$

Day 2 penalty: $\textsf{₹}250$

Day 3 penalty: $\textsf{₹}300$

and so on.

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The penalty for each succeeding day is $\textsf{₹}50$ more than for the preceding day. This means the sequence of penalties forms an Arithmetic Progression (AP).

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In this AP:

The first term is $a = 200$.

The common difference is $d = 250 - 200 = 50$. (Also, $300 - 250 = 50$).

The contractor delayed the work by 30 days, so we need to find the total penalty for 30 days. This means we need to find the sum of the first 30 terms of this AP, i.e., $n=30$ and we need to calculate $S_{30}$.

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The formula for the sum of the first $n$ terms of an AP is given by:

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Substitute the values $a = 200$, $d = 50$, and $n = 30$ into formula (1) to find the total penalty for 30 days ($S_{30}$):

$S_{30} = \frac{30}{2}[2(200) + (30-1)50]$

$S_{30} = 15[400 + (29)50]$

Calculate $29 \times 50$: $29 \times 50 = 29 \times 5 \times 10 = 145 \times 10 = 1450$.

$S_{30} = 15[400 + 1450]$

$S_{30} = 15[1850]$

Calculate $15 \times 1850$:

$\begin{array}{cc}& & 1 & 8 & 5 & 0 \\ \times & & & & 1 & 5 \\ \hline & & 9 & 2 & 5 & 0 \\ & 1 & 8 & 5 & 0 & \times \\ \hline & 2 & 7 & 7 & 5 & 0 \\ \hline \end{array}$

$S_{30} = 27750$.

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The total money the contractor has to pay as penalty for delaying the work by 30 days is $\textsf{₹}27750$.

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Conclusion: The contractor has to pay a penalty of $\mathbf{\textsf{₹}27750}$.

There are seven cash prizes, so the number of prizes is $n=7$.

The total sum of the prizes is $\textsf{₹}700$. This is the sum of the first 7 terms of the sequence of prize values, $S_7 = 700$.

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Each prize is $\textsf{₹}20$ less than its preceding prize. This means the sequence of prize values forms an Arithmetic Progression (AP).

Let the value of the first prize be $a$ (the first term). The common difference is $d = -20$ (since the value decreases). The prizes are $a, a-20, a-40, \dots$.

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We need to find the value of each of the seven prizes. This means finding the terms $a_1, a_2, \dots, a_7$. First, we need to find the first term $a$ and the common difference $d$. We already know $d=-20$.

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Use the formula for the sum of the first $n$ terms of an AP:

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Substitute $S_7 = 700$, $n = 7$, and $d = -20$ into formula (1):

$700 = \frac{7}{2}[2a + (7-1)(-20)]$

$700 = \frac{7}{2}[2a + (6)(-20)]$

$700 = \frac{7}{2}[2a - 120]$

$700 = \frac{7}{\cancel{2}} \times \cancel{2}[a - 60]$

$700 = 7(a - 60)$

Divide both sides by 7:

$\frac{700}{7} = a - 60$

$100 = a - 60$

Add 60 to both sides:

$a = 100 + 60$

$a = 160$.

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The first term (the value of the first prize) is $a = \textsf{₹}160$.

The common difference is $d = -20$.

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Now we can find the value of each of the seven prizes:

1st Prize ($a_1$): $a = 160$.

2nd Prize ($a_2$): $a + d = 160 + (-20) = 160 - 20 = 140$.

3rd Prize ($a_3$): $a + 2d = 160 + 2(-20) = 160 - 40 = 120$.

4th Prize ($a_4$): $a + 3d = 160 + 3(-20) = 160 - 60 = 100$.

5th Prize ($a_5$): $a + 4d = 160 + 4(-20) = 160 - 80 = 80$.

6th Prize ($a_6$): $a + 5d = 160 + 5(-20) = 160 - 100 = 60$.

7th Prize ($a_7$): $a + 6d = 160 + 6(-20) = 160 - 120 = 40$.

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The values of the seven prizes are $\textsf{₹}160, \textsf{₹}140, \textsf{₹}120, \textsf{₹}100, \textsf{₹}80, \textsf{₹}60$, and $\textsf{₹}40$.

Let's check the sum: $160 + 140 + 120 + 100 + 80 + 60 + 40 = 300 + 220 + 140 + 40 = 520 + 140 + 40 = 660 + 40 = 700$. The sum is correct.

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Conclusion: The values of the prizes are $\textsf{₹}160, \textsf{₹}140, \textsf{₹}120, \textsf{₹}100, \textsf{₹}80, \textsf{₹}60, \textsf{₹}40$.

The number of trees planted by each section is equal to the class number. There are 12 classes (Class I to Class XII).

There are 3 sections for each class.

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Number of trees planted by each section of Class I = 1 tree.

Total trees planted by Class I = Number of sections $\times$ Trees per section = $3 \times 1 = 3$ trees.

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Number of trees planted by each section of Class II = 2 trees.

Total trees planted by Class II = Number of sections $\times$ Trees per section = $3 \times 2 = 6$ trees.

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Number of trees planted by each section of Class III = 3 trees.

Total trees planted by Class III = Number of sections $\times$ Trees per section = $3 \times 3 = 9$ trees.

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... and so on, up to Class XII.

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Number of trees planted by each section of Class XII = 12 trees.

Total trees planted by Class XII = Number of sections $\times$ Trees per section = $3 \times 12 = 36$ trees.

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The total number of trees planted by the students is the sum of the trees planted by each class:

Total trees = (Trees by Class I) + (Trees by Class II) + ... + (Trees by Class XII)

Total trees = $3 + 6 + 9 + \dots + 36$.

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This sequence of numbers $3, 6, 9, \dots, 36$ forms an Arithmetic Progression (AP).

In this AP:

The first term is $a = 3$.

The common difference is $d = 6 - 3 = 3$. (Also, $9 - 6 = 3$).

The last term is $a_{12} = 36$.

The number of terms is $n = 12$ (since there are 12 classes).

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We need to find the sum of this AP ($S_{12}$). We can use the formula for the sum when the first term ($a$), the last term ($a_n$), and the number of terms ($n$) are known:

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Substitute the values $n=12$, $a=3$, and $a_{12}=36$ into formula (1):

$S_{12} = \frac{12}{2}(3 + 36)$

$S_{12} = 6(39)$

Calculate $6 \times 39$:

$6 \times 39 = 6 \times (40 - 1) = 240 - 6 = 234$.

$S_{12} = 234$.

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Thus, the total number of trees planted by the students is 234.

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Conclusion: The total number of trees planted will be $\mathbf{234}$.

The spiral is made up of successive semicircles with radii $r_1 = 0.5$ cm, $r_2 = 1.0$ cm, $r_3 = 1.5$ cm, $r_4 = 2.0$ cm, and so on.

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The length of a semicircle with radius $r$ is given by $\pi r$.

The lengths of the successive semicircles are:

Length of the first semicircle ($l_1$) with radius $r_1 = 0.5$ cm $= \pi r_1 = \pi(0.5)$ cm.

Length of the second semicircle ($l_2$) with radius $r_2 = 1.0$ cm $= \pi r_2 = \pi(1.0)$ cm.

Length of the third semicircle ($l_3$) with radius $r_3 = 1.5$ cm $= \pi r_3 = \pi(1.5)$ cm.

Length of the fourth semicircle ($l_4$) with radius $r_4 = 2.0$ cm $= \pi r_4 = \pi(2.0)$ cm.

And so on.

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The lengths of the successive semicircles form a sequence:

$0.5\pi, 1.0\pi, 1.5\pi, 2.0\pi, \dots$

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Let's check if this sequence is an Arithmetic Progression (AP) by finding the difference between consecutive terms:

$l_2 - l_1 = 1.0\pi - 0.5\pi = 0.5\pi$

$l_3 - l_2 = 1.5\pi - 1.0\pi = 0.5\pi$

$l_4 - l_3 = 2.0\pi - 1.5\pi = 0.5\pi$

Since the difference between consecutive terms is constant, the sequence of lengths forms an AP.

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In this AP of lengths:

The first term is $a = l_1 = 0.5\pi$.

The common difference is $d = 0.5\pi$.

The spiral is made up of thirteen consecutive semicircles, so the number of terms is $n=13$.

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We need to find the total length of the spiral, which is the sum of the lengths of the thirteen semicircles. This is the sum of the first 13 terms of this AP ($S_{13}$).

The formula for the sum of the first $n$ terms of an AP is given by:

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Substitute the values $a = 0.5\pi$, $d = 0.5\pi$, and $n = 13$ into formula (1) to find the total length ($S_{13}$):

$S_{13} = \frac{13}{2}[2(0.5\pi) + (13-1)(0.5\pi)]$

$S_{13} = \frac{13}{2}[\pi + (12)(0.5\pi)]$

$S_{13} = \frac{13}{2}[\pi + 6\pi]$

$S_{13} = \frac{13}{2}[7\pi]$

$S_{13} = \frac{13 \times 7\pi}{2}$

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Now substitute the value of $\pi = \frac{22}{7}$ into equation (i):

$S_{13} = \frac{91}{2} \times \frac{22}{7}$

$S_{13} = \frac{\cancel{91}^{13}}{2} \times \frac{22}{\cancel{7}_1}$

$S_{13} = \frac{13}{2} \times 22$

$S_{13} = 13 \times \frac{22}{2}$

$S_{13} = 13 \times 11$

Calculate $13 \times 11$:

$\begin{array}{cc}& & 1 & 3 \\ \times & & 1 & 1 \\ \hline & & 1 & 3 \\ & 1 & 3 & \times \\ \hline & 1 & 4 & 3 \\ \hline \end{array}$

$S_{13} = 143$.

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Thus, the total length of the spiral made up of thirteen consecutive semicircles is 143 cm.

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Conclusion: The total length of the spiral is $\mathbf{143}$ cm.

The number of logs in each row, starting from the bottom, forms a sequence:

Bottom row: 20 logs

Next row: 19 logs

Row next to that: 18 logs

and so on.

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The sequence of the number of logs in the rows is $20, 19, 18, \dots$.

This sequence forms an Arithmetic Progression (AP).

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In this AP:

The first term is $a = 20$.

The common difference is $d = 19 - 20 = -1$.

The total number of logs is 200. This is the sum of the logs in all the rows. Let the number of rows be $n$. The sum of the first $n$ terms ($S_n$) is 200.

$S_n = 200$.

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We need to find the number of rows ($n$) and the number of logs in the top row (which is the $n$-th term, $a_n$).

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Use the formula for the sum of the first $n$ terms of an AP:

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Substitute the values $S_n = 200$, $a = 20$, and $d = -1$ into formula (1):

$200 = \frac{n}{2}[2(20) + (n-1)(-1)]$

$200 = \frac{n}{2}[40 - (n-1)]$

$200 = \frac{n}{2}[40 - n + 1]$

$200 = \frac{n}{2}[41 - n]$

Multiply both sides by 2:

$2 \times 200 = n(41 - n)$

$400 = 41n - n^2$

Rearrange the terms to form a quadratic equation:

$n^2 - 41n + 400 = 0$

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Now we solve this quadratic equation for $n$. We can use factorization or the quadratic formula. We look for two numbers that multiply to 400 and add up to -41. These numbers are -16 and -25.

$n^2 - 16n - 25n + 400 = 0$

$n(n - 16) - 25(n - 16) = 0$

$(n - 16)(n - 25) = 0$

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This gives two possible values for $n$:

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

$n - 25 = 0 \implies n = 25$

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Both $n=16$ and $n=25$ are positive integers. We need to check if both are valid solutions in the context of the problem.

The number of logs in each successive row is decreasing. Let's find the number of logs in the top row ($a_n$) for each value of $n$ using the formula $a_n = a + (n-1)d$, with $a=20$ and $d=-1$.

If $n=16$:

$a_{16} = 20 + (16-1)(-1)$

$a_{16} = 20 + (15)(-1)$

$a_{16} = 20 - 15 = 5$.

A row having 5 logs is possible.

If $n=25$:

$a_{25} = 20 + (25-1)(-1)$

$a_{25} = 20 + (24)(-1)$

$a_{25} = 20 - 24 = -4$.

The number of logs in a row cannot be negative. Therefore, $n=25$ is not a valid solution in this context.

Thus, the number of rows is 16.

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The number of rows is $n=16$.

The number of logs in the top row is the 16th term, $a_{16}$, which we calculated as 5.

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Conclusion: The 200 logs are placed in 16 rows, and there are 5 logs in the top row.

There are 10 potatoes in the line.

Distance of the first potato from the bucket = 5 m.

Distance between consecutive potatoes = 3 m.

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To pick up the first potato:

Distance run to the first potato = 5 m.

Distance run back to the bucket = 5 m.

Total distance for the first potato = $5 + 5 = 2 \times 5 = 10$ m.

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To pick up the second potato:

The second potato is at a distance of $5 + 3 = 8$ m from the bucket.

Distance run to the second potato = 8 m.

Distance run back to the bucket = 8 m.

Total distance for the second potato = $8 + 8 = 2 \times 8 = 16$ m.

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To pick up the third potato:

The third potato is at a distance of $5 + 3 + 3 = 5 + 2 \times 3 = 11$ m from the bucket.

Distance run to the third potato = 11 m.

Distance run back to the bucket = 11 m.

Total distance for the third potato = $11 + 11 = 2 \times 11 = 22$ m.

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The distances run to pick up each successive potato and return to the bucket form a sequence:

Distance for 1st potato: $2 \times 5 = 10$ m

Distance for 2nd potato: $2 \times (5+3) = 2 \times 8 = 16$ m

Distance for 3rd potato: $2 \times (5+3+3) = 2 \times 11 = 22$ m

Distance for 4th potato: $2 \times (5+3+3+3) = 2 \times 14 = 28$ m

and so on.

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The sequence of distances run for each potato is $10, 16, 22, 28, \dots$.

Let's check if this sequence is an Arithmetic Progression (AP) by finding the difference between consecutive terms:

$16 - 10 = 6$

$22 - 16 = 6$

$28 - 22 = 6$

Since the difference is constant, the sequence forms an AP with common difference $d = 6$.

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In this AP:

The first term is $a = 10$.

The common difference is $d = 6$.

There are 10 potatoes, so we need to find the total distance run to pick up all 10 potatoes. This is the sum of the first 10 terms of this AP ($S_{10}$).

The number of terms is $n=10$.

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The formula for the sum of the first $n$ terms of an AP is given by:

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Substitute the values $a = 10$, $d = 6$, and $n = 10$ into formula (1) to find the total distance ($S_{10}$):

$S_{10} = \frac{10}{2}[2(10) + (10-1)6]$

$S_{10} = 5[20 + (9)6]$

$S_{10} = 5[20 + 54]$

$S_{10} = 5[74]$

Calculate $5 \times 74$:

$5 \times 74 = 5 \times (70 + 4) = 350 + 20 = 370$.

$S_{10} = 370$.

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Thus, the total distance the competitor has to run is 370 meters.

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Conclusion: The total distance the competitor has to run is $\mathbf{370}$ m.

The given Arithmetic Progression (AP) is: $121, 117, 113, \dots$

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Here, the first term is $a = 121$.

The common difference $d$ is the difference between consecutive terms:

$d = 117 - 121 = -4$.

We can check this with the next pair: $113 - 117 = -4$. So, $d = -4$.

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We are looking for the first negative term. Let the $n$-th term ($a_n$) be the first term that is negative. This means we need to find the smallest positive integer $n$ such that $a_n < 0$.

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The formula for the $n$-th term of an AP is given by:

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We set $a_n < 0$ and substitute the values of $a$ and $d$ into formula (1):

$a + (n-1)d < 0$

$121 + (n-1)(-4) < 0$

$121 - 4(n-1) < 0$

Subtract 121 from both sides:

$-4(n-1) < -121$

Divide both sides by -4. When dividing an inequality by a negative number, the inequality sign reverses.

$n-1 > \frac{121}{4}$

Convert the fraction to a decimal or mixed number:

$\frac{121}{4} = \frac{120+1}{4} = \frac{120}{4} + \frac{1}{4} = 30 + 0.25 = 30.25$.

So, the inequality is:

$n-1 > 30.25$

Add 1 to both sides:

$n > 30.25 + 1$

$n > 31.25$

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Since $n$ must be an integer (representing the term number), the smallest integer greater than 31.25 is 32.

So, the first value of $n$ for which $a_n$ is negative is $n = 32$.

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Let's check the 31st and 32nd terms to confirm:

$a_{31} = 121 + (31-1)(-4) = 121 + 30(-4) = 121 - 120 = 1$ (positive).

$a_{32} = 121 + (32-1)(-4) = 121 + 31(-4) = 121 - 124 = -3$ (negative).

This confirms that the 32nd term is the first negative term.

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Conclusion: The 32nd term of the AP is its first negative term.

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

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

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The third term is $a_3 = a + (3-1)d = a + 2d$.

The seventh term is $a_7 = a + (7-1)d = a + 6d$.

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We are given that the sum of the third and seventh terms is 6:

$(a + 2d) + (a + 6d) = 6$

$2a + 8d = 6$

Divide the equation by 2:

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We are also given that the product of the third and seventh terms is 8:

$(a + 2d)(a + 6d) = 8$

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From equation (i), we can express $a$ in terms of $d$: $a = 3 - 4d$.

Substitute this expression for $a$ into the product equation:

$((3 - 4d) + 2d)((3 - 4d) + 6d) = 8$

$(3 - 2d)(3 + 2d) = 8$

This is in the form $(x-y)(x+y) = x^2 - y^2$ where $x=3$ and $y=2d$.

$3^2 - (2d)^2 = 8$

$9 - 4d^2 = 8$

Subtract 9 from both sides:

$-4d^2 = 8 - 9$

$-4d^2 = -1$

Divide by -4:

$d^2 = \frac{-1}{-4} = \frac{1}{4}$.

Take the square root of both sides:

$d = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$.

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We have two possible values for the common difference $d$. We will find the corresponding first term $a$ for each case using equation (i) ($a = 3 - 4d$).

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Case 1: $d = \frac{1}{2}$

$a = 3 - 4(\frac{1}{2}) = 3 - 2 = 1$.

In this case, the AP starts with $a=1$ and has $d=\frac{1}{2}$. The terms are $1, 1.5, 2, 2.5, \dots$

Let's check the 3rd and 7th terms: $a_3 = 1 + 2(\frac{1}{2}) = 1+1=2$, $a_7 = 1 + 6(\frac{1}{2}) = 1+3=4$. Sum = $2+4=6$. Product = $2 \times 4 = 8$. This case is valid.

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Case 2: $d = -\frac{1}{2}$

$a = 3 - 4(-\frac{1}{2}) = 3 - (-2) = 3 + 2 = 5$.

In this case, the AP starts with $a=5$ and has $d=-\frac{1}{2}$. The terms are $5, 4.5, 4, 3.5, \dots$

Let's check the 3rd and 7th terms: $a_3 = 5 + 2(-\frac{1}{2}) = 5-1=4$, $a_7 = 5 + 6(-\frac{1}{2}) = 5-3=2$. Sum = $4+2=6$. Product = $4 \times 2 = 8$. This case is also valid.

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Both cases satisfy the given conditions. We need to find the sum of the first sixteen terms ($S_{16}$).

The formula for the sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$. We need $S_{16}$, so $n=16$.

$S_{16} = \frac{16}{2}[2a + (16-1)d] = 8[2a + 15d]$.

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Case 1: $a = 1$, $d = \frac{1}{2}$

$S_{16} = 8[2(1) + 15(\frac{1}{2})]$

$S_{16} = 8[2 + \frac{15}{2}]$

$S_{16} = 8[\frac{4}{2} + \frac{15}{2}]$

$S_{16} = 8[\frac{19}{2}]$

$S_{16} = \cancel{8}^4 \times \frac{19}{\cancel{2}_1}$

$S_{16} = 4 \times 19 = 76$.

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Case 2: $a = 5$, $d = -\frac{1}{2}$

$S_{16} = 8[2(5) + 15(-\frac{1}{2})]$

$S_{16} = 8[10 - \frac{15}{2}]$

$S_{16} = 8[\frac{20}{2} - \frac{15}{2}]$

$S_{16} = 8[\frac{5}{2}]$

$S_{16} = \cancel{8}^4 \times \frac{5}{\cancel{2}_1}$

$S_{16} = 4 \times 5 = 20$.

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Both sums are possible depending on the specific AP that satisfies the conditions.

However, sometimes in problems like this, there is an implicit assumption that $d$ should be positive or the AP is increasing/decreasing in a consistent way unless specified otherwise, or the question might be worded to imply a unique AP. Given the phrasing "an AP", it suggests there might be a unique solution intended or that the question accepts both cases.

Let's re-read the question. It asks for "the sum of first sixteen terms of the AP". This phrasing suggests there might be only one possible sum. Let's check the conditions again: $a+4d = 3$. If $d=1/2$, $a=1$. If $d=-1/2$, $a=5$. Both pairs $(a, d)$ are valid.

Let's reconsider the equation $a+4d=3$. This is the 5th term of the AP, $a_5=3$.

So, $a_3 = a_5 - 2d = 3 - 2d$ and $a_7 = a_5 + 2d = 3 + 2d$.

The product is $a_3 a_7 = (3 - 2d)(3 + 2d) = 3^2 - (2d)^2 = 9 - 4d^2 = 8$. This leads to $d^2 = 1/4$, so $d = \pm 1/2$. Then $a=3-4d = 3 - 4(\pm 1/2) = 3 \mp 2$. So $(a, d) = (1, 1/2)$ or $(5, -1/2)$.

The sum of the first 16 terms is $S_{16} = \frac{16}{2}(a + a_{16})$. $a_{16} = a + 15d$.

$S_{16} = 8(a + a + 15d) = 8(2a + 15d)$.

Alternatively, $S_{16} = \frac{16}{2}[2(3-4d) + 15d] = 8[6 - 8d + 15d] = 8[6 + 7d]$.

If $d = 1/2$, $S_{16} = 8[6 + 7(1/2)] = 8[6 + 3.5] = 8[9.5] = 76$.

If $d = -1/2$, $S_{16} = 8[6 + 7(-1/2)] = 8[6 - 3.5] = 8[2.5] = 20$.

There is no indication in the question to prefer one case over the other. However, it's possible the question expects only one answer. Let's re-read the prompt guidelines - "Provide any alternate solution as well (if needed)". This suggests that multiple valid solutions might exist and should be presented.

Therefore, there are two possible APs that satisfy the conditions, leading to two possible sums for the first 16 terms.

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Conclusion: The sum of the first sixteen terms of the AP is either $\mathbf{76}$ (when $a=1$ and $d=1/2$) or $\mathbf{20}$ (when $a=5$ and $d=-1/2$).

The distance between consecutive rungs is 25 cm.

The distance between the top and bottom rungs is $2\frac{1}{2}$ m.

Convert the distance between top and bottom rungs to centimeters: $2\frac{1}{2}$ m $= 2.5$ m $= 2.5 \times 100$ cm $= 250$ cm.

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Let the number of rungs be $n$. The distance between $n$ rungs is $(n-1)$ times the distance between consecutive rungs.

Distance between top and bottom rungs = $(n-1) \times$ Distance between consecutive rungs.

$250 = (n-1) \times 25$

Divide both sides by 25:

$\frac{250}{25} = n-1$

$10 = n-1$

$n = 10 + 1 = 11$.

So, there are 11 rungs on the ladder.

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The lengths of the rungs decrease uniformly from bottom to top. This means the lengths form an Arithmetic Progression (AP).

The length of the bottom rung is 45 cm. This is the first term, $a = 45$.

The length of the top rung is 25 cm. This is the last term, $a_n$. Since there are 11 rungs, the top rung is the 11th term, $a_{11} = 25$.

The number of terms is $n = 11$.

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We need to find the total length of the wood required for the rungs, which is the sum of the lengths of all 11 rungs. This is the sum of the first 11 terms of this AP ($S_{11}$).

We can use the formula for the sum of the first $n$ terms when the first term ($a$), the last term ($a_n$), and the number of terms ($n$) are known:

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Substitute the values $n=11$, $a=45$, and $a_{11}=25$ into formula (1):

$S_{11} = \frac{11}{2}(45 + 25)$

$S_{11} = \frac{11}{2}(70)$

$S_{11} = 11 \times \frac{70}{2}$

$S_{11} = 11 \times 35$

Calculate $11 \times 35$:

$\begin{array}{cc}& & 3 & 5 \\ \times & & 1 & 1 \\ \hline & & 3 & 5 \\ & 3 & 5 & \times \\ \hline & 3 & 8 & 5 \\ \hline \end{array}$

$S_{11} = 385$.

---

Thus, the total length of the wood required for the rungs is 385 cm.

---

Conclusion: The length of the wood required for the rungs is $\mathbf{385}$ cm.

The house numbers are $1, 2, 3, \dots, 49$. This is an Arithmetic Progression (AP) with the first term $a=1$ and the common difference $d=1$.

---

The sum of the numbers of the houses preceding the house numbered $x$ is the sum of the first $x-1$ house numbers. This is the sum of the first $x-1$ terms of the AP, $S_{x-1}$.

For $x=1$, there are no houses preceding it, so the sum is 0. $S_{1-1} = S_0 = 0$.

For $x > 1$, $S_{x-1} = 1 + 2 + \dots + (x-1)$.

Using the sum formula $S_n = \frac{n}{2}(a + a_n)$ with $n=x-1$, $a=1$, and $a_{x-1}=x-1$:

---

The sum of the numbers of the houses following the house numbered $x$ is the sum of the house numbers from $x+1$ to 49. This sum can be found by subtracting the sum of the first $x$ house numbers ($S_x$) from the total sum of all 49 house numbers ($S_{49}$).

The houses following house $x$ are $x+1, x+2, \dots, 49$. This is also an AP with first term $a' = x+1$, last term $l'=49$, and number of terms $n' = 49 - (x+1) + 1 = 49 - x$. The sum is $\frac{49-x}{2}((x+1)+49) = \frac{49-x}{2}(x+50)$.

Alternatively, using the hint, the sum of houses following $x$ is $S_{49} - S_x$.

The total sum of houses from 1 to 49 ($S_{49}$) is the sum of an AP with $a=1$, $n=49$, and $a_{49}=49$.

$S_{49} = \frac{49}{2}(1 + 49) = \frac{49}{2}(50) = 49 \times 25$.

$49 \times 25 = (50-1) \times 25 = 1250 - 25 = 1225$.

The sum of the first $x$ house numbers ($S_x$) is the sum of an AP with $a=1$, $n=x$, and $a_x=x$.

The sum of the numbers of the houses following house $x$ is $S_{49} - S_x$.

---

We are given that the sum of the numbers of the houses preceding house $x$ is equal to the sum of the numbers of the houses following it.

Substitute the expressions from (i) and (iv) into this equation:

$\frac{x(x-1)}{2} = 1225 - \frac{x(x+1)}{2}$

Add $\frac{x(x+1)}{2}$ to both sides:

$\frac{x(x-1)}{2} + \frac{x(x+1)}{2} = 1225$

Combine the terms on the left side:

$\frac{x(x-1) + x(x+1)}{2} = 1225$

$\frac{x^2 - x + x^2 + x}{2} = 1225$

$\frac{2x^2}{2} = 1225$

$x^2 = 1225$

Take the square root of both sides:

$x = \pm \sqrt{1225}$.

To find $\sqrt{1225}$:

$\begin{array}{c|cc} & 3\ . \ 5 & \\ \hline \phantom{()} 3 & \overline{12} \; \overline{25} & \\ + \; 3 & 9\phantom{........)} \\ \hline \phantom{()} 6 \; 5 & 3 \; 25 \phantom{)} \\ \phantom{()} +5 & 3 \; 25 \phantom{)} \\ \hline \phantom{()} 70 & 0\phantom{)} \end{array}$

So, $\sqrt{1225} = 35$.

$x = \pm 35$.

---

Since $x$ represents a house number, it must be a positive integer. Also, the house numbered $x$ must be within the range of 1 to 49. Thus, $x=35$ is the only valid solution.

The value of $x=35$ lies between 1 and 49 (exclusive of 1, since $S_{x-1}$ requires $x-1 \geq 0$). For $x=35$, $x-1=34$, and houses following 35 exist (up to 49).

---

Conclusion: There is a value of $x$ that satisfies the condition, and that value is $\mathbf{35}$.

The terrace has 15 steps. Each step is 50 m long.

Each step has a rise (height) of $\frac{1}{4}$ m and a tread (depth) of $\frac{1}{2}$ m.

---

Let's consider the volume of concrete required for each step, starting from the bottom step (Step 1).

The steps are built one above the other. The bottom step (Step 1) has a height equal to the rise of one step ($\frac{1}{4}$ m) and a depth equal to the tread of one step ($\frac{1}{2}$ m). Its length is 50 m.

Volume of Step 1 = Length $\times$ Tread $\times$ Rise $= 50 \times \frac{1}{2} \times \frac{1}{4}$ m$^3$.

---

Step 2 is built on top of Step 1. Step 2 adds another 'layer' of concrete with the same tread ($\frac{1}{2}$ m) and length (50 m), but its height is the rise of one step ($\frac{1}{4}$ m).

However, the problem implies that the steps are built as successive layers of concrete. The first layer forms the first step (bottom). The second layer forms the second step on top of the first. The height of the concrete added for the second step is its rise ($\frac{1}{4}$ m), but its tread is the same as the previous step ($\frac{1}{2}$ m).

Let's interpret the volumes based on the visual representation and the hint.

The first step (bottom) has dimensions: Length = 50 m, Tread = $\frac{1}{2}$ m, Height = $\frac{1}{4}$ m.

Volume of 1st step = $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$ m$^3$.

The second step sits on the first. Its dimensions relative to the ground are: Length = 50 m, Tread = $\frac{1}{2}$ m, Height = $2 \times \frac{1}{4} = \frac{2}{4}$ m (total height from ground).

The volume added to create the second step is the concrete block with Length = 50 m, Tread = $\frac{1}{2}$ m, and incremental height = $\frac{1}{4}$ m.

Volume of concrete for 2nd step = $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$ m$^3$.

This interpretation seems incorrect based on the picture. The steps are building up.

---

Let's re-read: "Each step has a rise of $\frac{1}{4}$ m and a tread of $\frac{1}{2}$ m". This describes the dimensions of each individual block of concrete added to form the step relative to the layer below it, EXCEPT for the bottom layer which forms the "first step".

Consider the cross-section area. The first step (bottom) has a rectangular cross-section of height $\frac{1}{4}$ and width $\frac{1}{2}$.

Area of cross-section of 1st step = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ m$^2$.

Volume of 1st step = Area $\times$ Length = $\frac{1}{8} \times 50 = \frac{50}{8} = \frac{25}{4}$ m$^3$. This matches the hint.

---

The second step is built on top. It has the same tread ($\frac{1}{2}$ m) and length (50 m), but its height starts from the top of the first step. The step structure is like a staircase. Each step effectively adds a block of concrete of size: Length = 50 m, Tread = $\frac{1}{2}$ m, Rise = $\frac{1}{4}$ m.

However, the figure shows the steps are getting wider at the bottom, not that the width of each block is $\frac{1}{2}$. The "tread" is the horizontal depth of the step.

Let's analyze the volumes of concrete needed for each step layer, starting from the bottom.

The first step (bottom) is a block of concrete with dimensions Length = 50 m, Tread = $\frac{1}{2}$ m, Rise = $\frac{1}{4}$ m.

Volume of 1st step = $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$ m$^3$.

---

The second step has a rise of $\frac{1}{4}$ and tread of $\frac{1}{2}$. It sits on the first step. The total height from the ground up to the top of the second step is $2 \times \frac{1}{4}$. The total depth from the front edge back is $2 \times \frac{1}{2}$.

This implies the steps are like nested rectangular prisms subtracted from a larger prism. Or, looking at the side profile, the concrete area is like a series of rectangles added.

Let's look at the cross-sectional area from the side.

The first step (bottom) has a rectangle of width $\frac{1}{2}$ and height $\frac{1}{4}$. Area = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$. Volume = $\frac{1}{8} \times 50 = \frac{25}{4}$.

The second step adds a rectangle of width $\frac{1}{2}$ on top of the first step's tread, with a height up to $2 \times \frac{1}{4} = \frac{1}{2}$. But the diagram suggests that the steps are built such that the front edge of each step is vertically above the front edge of the step below it.

The diagram suggests that the volume of concrete added for each successive step (starting from the bottom as step 1) has a constant length (50 m) and tread ($\frac{1}{2}$ m), but the height of the 'block' added increases. No, the height added is constant ($\frac{1}{4}$ m), but the number of 'tread' blocks increases.

Let's re-read again: "Each step has a rise of $\frac{1}{4}$ m and a tread of $\frac{1}{2}$ m". This describes the geometry of the step itself. The total structure is a set of stacked layers of concrete.

The bottom layer (Step 1) is a block of length 50 m, width (tread) $\frac{1}{2}$ m, and height (rise) $\frac{1}{4}$ m. Volume = $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$ m$^3$.

The second step adds a block on top. This block has length 50 m, tread $\frac{1}{2}$ m, and rise $\frac{1}{4}$ m. It sits on top of the first step's tread, but offset to the back to create the step shape. This doesn't seem right for calculating the total volume.

Consider the volume of the concrete forming the structure up to the top of each step.

Volume up to the top of Step 1 (bottom): Length=50, Width=$\frac{1}{2}$, Height=$\frac{1}{4}$. Volume = $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$.

Volume up to the top of Step 2: The total height is $2 \times \frac{1}{4} = \frac{1}{2}$. The base length along the slope is 50. The horizontal distance covered by the first two treads is $2 \times \frac{1}{2} = 1$. The structure up to step 2 looks like a block with dimensions $50 \times 1 \times \frac{1}{2}$. Volume = $50 \times 1 \times \frac{1}{2} = 25$. No, this is not correct either.

Let's look at the volume of each horizontal layer of concrete needed to build the steps, starting from the bottom.

The bottom-most layer corresponds to the total depth and total height. This is confusing.

Let's consider the volume of each individual block of concrete that constitutes each step layer, starting from the bottom. The first step is the lowest block. It has length 50 m. Its base is $\frac{1}{2}$ m wide (tread). Its height is $\frac{1}{4}$ m (rise).

Volume of concrete for Step 1 = $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$.

The second step sits on top of the first step's tread. It has length 50 m, tread $\frac{1}{2}$ m, and rise $\frac{1}{4}$ m. This seems incorrect as it would mean all steps have the same volume.

Let's look at the structure from the side. Each step adds a rectangle of concrete. The first step adds a rectangle of base $\frac{1}{2}$ and height $\frac{1}{4}$. The second step adds another rectangle of base $\frac{1}{2}$ but at a height of $\frac{1}{4}$ to $\frac{1}{2}$. This means the second layer of concrete (which forms the 2nd step's tread) has a base of $\frac{1}{2}$ and a height of $\frac{1}{4}$.

The diagram shows that the structure has a uniform length of 50m. The side profile is a staircase shape.

Consider the volume of each block added to form the staircase, starting from the bottom layer.

Bottom layer (height $\frac{1}{4}$): It covers the full horizontal extent of the staircase. The horizontal extent of the staircase is determined by the sum of the treads.

No, the diagram shows the bottom step has a tread of $\frac{1}{2}$ and a rise of $\frac{1}{4}$. The step above it adds another tread of $\frac{1}{2}$ and rise of $\frac{1}{4}$.

The volume of concrete in each step is the volume of the rectangular block that forms that step.

Step 1 (bottom): Length = 50m, Tread = $\frac{1}{2}$m, Rise = $\frac{1}{4}$m. Volume $V_1 = 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$ m$^3$.

Step 2: This step is built on top of the first step. It has the same dimensions relative to its position: Length = 50m, Tread = $\frac{1}{2}$m, Rise = $\frac{1}{4}$m. Volume $V_2 = 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$ m$^3$.

If all steps have the same volume, the total volume would be $15 \times \frac{25}{4}$. But the lengths decrease uniformly in length from 45cm at the bottom to 25cm at the top. This is the key! The 50m length mentioned earlier might be the width of the terrace, not the length of the rungs.

Let's assume the "length of each step" means the width of the concrete block from side to side. This width is 50 m. The rungs are 25 cm apart and their lengths decrease. The rungs are the wooden parts you step on. The concrete forms the steps themselves.

The statement "rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top" describes the wooden rungs, which are placed on top of the concrete steps.

Let's assume the question is about the concrete steps. Each step has a constant length (50 m), a constant tread ($\frac{1}{2}$ m), and a constant rise ($\frac{1}{4}$ m). The volume of each step is the volume of the concrete block that forms it.

Volume of one step (block) = Length $\times$ Tread $\times$ Rise $= 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$ m$^3$.

There are 15 steps.

Total volume of concrete = Sum of the volumes of the 15 steps.

If each step is a separate block of concrete with the same dimensions, then the total volume is $15 \times \frac{25}{4}$. But the figure shows a continuous concrete structure.

Let's interpret the volume of concrete required for each layer of the staircase structure, starting from the bottom.

Layer 1 (height $\frac{1}{4}$): This layer has a width determined by the total tread of 15 steps ($15 \times \frac{1}{2}$) and a length of 50. Volume = $50 \times (15 \times \frac{1}{2}) \times \frac{1}{4}$. No, this interpretation doesn't match the diagram or the hint.

Let's assume the steps are built such that the volume of each successive layer (starting from the bottom) forms an AP. The width of the terrace is 50m.

The bottom step is a rectangular block with height $\frac{1}{4}$ and tread $\frac{1}{2}$. Its volume is $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$.

The second step adds another block on top. This block also has height $\frac{1}{4}$ and tread $\frac{1}{2}$. The volume of concrete for the second step seems to be the volume of the block needed to create the step shape at the second level.

Consider the layers of concrete added. The bottom layer has height $\frac{1}{4}$ and goes back a certain distance. The second layer has height $\frac{1}{4}$ and goes back a smaller distance, and so on.

Let's assume the figure implies that the depth of the concrete structure at each level decreases. The total depth at the bottom is the sum of the treads of all steps. The height of the lowest layer is the rise of one step.

The volume of the bottom layer of concrete with height $\frac{1}{4}$ is its area in the side profile multiplied by the length 50m. The area of the bottom layer in the side profile is the area of the rectangle with width equal to the total horizontal extent and height $\frac{1}{4}$. The total horizontal extent is $15 \times \frac{1}{2}$.

No, the diagram shows that the horizontal extent decreases as you go up.

Let's look at the volume of concrete at each level, starting from the bottom step which is at height 0 to $\frac{1}{4}$. This step has a full 'tread' length. The next step is at height $\frac{1}{4}$ to $\frac{1}{2}$.

Let's consider the volume of concrete added at each "rise" level, starting from the bottom. The first layer of concrete is at height 0 to $\frac{1}{4}$. This layer has a uniform width of 50m. Its depth is the total horizontal distance covered by all 15 treads, which is $15 \times \frac{1}{2}$. No, that is not correct visually.

Let's assume the question describes the volume of concrete in each step layer, starting from the bottom.

The first step (bottom layer of concrete) has length 50m, thickness (rise) $\frac{1}{4}$m. Its depth is the total horizontal length of concrete at that height. The total horizontal extent of the steps at the base is the sum of all treads if they were laid out horizontally, but they are stacked. The diagram shows the steps forming a staircase shape.

Let's interpret the volume of concrete for each "block" that makes up the steps, starting from the bottom.

The bottom layer of concrete has length 50m, height $\frac{1}{4}$m, and its depth is the total horizontal extent, which is the sum of the treads of all 15 steps, $15 \times \frac{1}{2}$. No, that's not right.

Let's go back to the hint and the first interpretation that matches it.

Volume of concrete for the first step (bottom) = $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$.

The second step is above the first. It has the same length 50 and tread $\frac{1}{2}$. Its height is also $\frac{1}{4}$. But how does this create a decreasing volume?

Let's reconsider the structure as layers of concrete. The bottom-most layer has height $\frac{1}{4}$. What is its horizontal area? It extends the full width of the terrace (50m) and back by a certain amount.

Let's assume the volume of concrete for each successive horizontal layer of height $\frac{1}{4}$ forms an AP.

The bottom-most horizontal layer of concrete has height $\frac{1}{4}$. Its horizontal area is $50 \times (\text{total tread at this level})$. The total tread at the very bottom is the sum of all treads, which is $15 \times \frac{1}{2}$. No, this is not correct. The diagram does not show this.

Let's assume the hint means the volume of concrete for the first "unit" step (at the lowest level) is $50 \times \frac{1}{2} \times \frac{1}{4}$.

Consider the volumes of concrete layers from bottom to top. Each layer has height $\frac{1}{4}$ and length 50. The depth of each layer is the total horizontal length of concrete at that height. The bottom layer goes back the full extent ($15 \times \frac{1}{2}$ if they were stacked like that). But the diagram shows a decreasing horizontal extent as height increases.

Let's assume the volume of concrete for the 1st step (bottom) is $V_1$. For the 2nd step is $V_2$, and so on, up to the 15th step (top). The sequence $V_1, V_2, \dots, V_{15}$ forms an AP.

The shape of the concrete is a prism with a staircase-shaped cross-section. The length of the prism is 50m.

Let's find the area of the cross-section. The cross-section is made of 15 rectangles stacked and shifted. The bottom rectangle has width $\frac{1}{2}$ and height $\frac{1}{4}$. The second layer of concrete adds another rectangle of width $\frac{1}{2}$ and height $\frac{1}{4}$ next to the first, and on top of the previous height.

Let's look at the total horizontal distance covered by the treads at each level, starting from the bottom.

At the level of the 1st step (from height 0 to $\frac{1}{4}$), the horizontal extent is the tread of the 1st step + tread of 2nd step + ... + tread of 15th step? No, that's wrong.

Consider the volume of each horizontal slab of concrete of thickness $\frac{1}{4}$.

The bottom slab (height 0 to $\frac{1}{4}$): Its horizontal area is Length $\times$ Total horizontal distance at this height. The total horizontal distance is the sum of the treads. No.

Let's assume the problem implies that the volume of concrete for the steps, starting from the bottom step as the "first step" in terms of volume calculation, forms an AP.

The first step (bottom) has height $\frac{1}{4}$, tread $\frac{1}{2}$, length 50. Volume = $50 \times \frac{1}{2} \times \frac{1}{4}$.

The second step is built on top. It has height $\frac{1}{4}$ and tread $\frac{1}{2}$. The concrete structure for the second step has a different shape from the first step. The volume of concrete added to form the second step layer has height $\frac{1}{4}$ and tread $\frac{1}{2}$ and length 50. But the total volume up to step 2 includes the first step.

Let's assume the volumes of the *individual blocks of concrete* that are stacked form an AP. The bottom block has volume $V_1$. The next block $V_2$, etc. The diagram shows that the horizontal dimension of these blocks is decreasing.

Let's assume the volume of the concrete in each step (layer), starting from the bottom, forms an AP. The bottom step is a block of concrete with dimensions 50m $\times \frac{1}{2}$m $\times \frac{1}{4}$m. $V_1 = 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$.

The second step adds a layer of concrete on top of the first step's tread. This layer has height $\frac{1}{4}$, length 50. What is its tread/depth? The steps recede.

Let's reconsider the hint: "Volume of concrete required to build the first step = $\frac{1}{4}$ × $\frac{1}{2}$ × 50 m³". This is height $\times$ tread $\times$ length. This suggests the dimensions of the bottom step block are $\frac{1}{4}$ (height), $\frac{1}{2}$ (tread), and 50 (length).

The rungs' lengths decrease from 45 cm to 25 cm. This refers to the wooden pieces. The concrete steps are what the rungs are attached to.

Let's assume the volume of concrete required for each step, starting from the bottom, forms an AP. The volumes might decrease because the dimensions of the steps might change. The length of the terrace is 50m. The rise is $\frac{1}{4}$m, tread is $\frac{1}{2}$m.

Consider the side profile area. The first rectangle has base $\frac{1}{2}$ and height $\frac{1}{4}$. The second rectangle sits on top and next to it, with base $\frac{1}{2}$ and height $\frac{1}{4}$.

Let the total height be $H = 15 \times \frac{1}{4} = \frac{15}{4}$ m. Let the total horizontal extent be $T = 15 \times \frac{1}{2} = \frac{15}{2}$ m.

The structure is a series of stacked rectangular blocks of concrete, each 50m long.

The first step is a block of size 50m $\times \frac{1}{2}$m $\times \frac{1}{4}$m. Volume $V_1 = \frac{25}{4}$.

The second step is the concrete added above the first step's tread. It has dimensions 50m $\times \frac{1}{2}$m $\times \frac{1}{4}$m. Volume $V_2 = \frac{25}{4}$.

If the volume of each step is the same, $15 \times \frac{25}{4} = \frac{375}{4} = 93.75$ m$^3$. But the rungs' lengths changing suggests something about the steps might be changing uniformly.

Let's assume the volumes of the steps themselves form an AP. Why would they form an AP? The structure looks like a solid object. The volume should be the sum of the volumes of the layers.

Consider the layers of concrete added, starting from the bottom. Each layer has a height equal to the rise, $\frac{1}{4}$ m. Its length is 50 m. Its depth is the total horizontal distance of concrete at that height.

Layer 1 (height 0 to $\frac{1}{4}$): Depth = $15 \times \frac{1}{2}$ if it extended all the way? No.

Let's look at the volumes of the steps from top to bottom instead. The top step is a block of size 50 $\times \frac{1}{2} \times \frac{1}{4}$. Volume = $\frac{25}{4}$.

The step below it has the same rise and tread, but it extends further back. The structure up to the top of the second step (from the top) has a tread of $2 \times \frac{1}{2} = 1$ and a height of $2 \times \frac{1}{4} = \frac{1}{2}$. The volume up to the top of the second step (from top) seems to be $50 \times 1 \times \frac{1}{2} = 25$. This is not the volume of just the second step.

Let's assume the volumes of concrete for the layers, starting from the bottom, form an AP because the length of the rungs changes uniformly. This connection is not clear. The concrete is the steps, the rungs are on top.

Let's assume the volumes of the *steps* as blocks of concrete form an AP.

Volume of the first step (bottom): $V_1 = 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$.

Volume of the second step: It is built on top. It has length 50, tread $\frac{1}{2}$, rise $\frac{1}{4}$. Is its volume the same? No. The rungs lengths changing must be a clue about the steps.

Let's assume the question means the lengths of the concrete steps are decreasing uniformly. However, the problem states "Each step is 50 m long". This seems contradictory with "rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top". Let's assume the 50m is the width of the terrace, and the "length" of the rungs refers to the dimension perpendicular to the 50m width, i.e., the length of the step itself in the side view.

If the "length" refers to the dimension of the step parallel to the run, i.e., the length perpendicular to the 50m width. Then the steps have a varying dimension. The bottom step's length is related to the 45cm rung. The top step's length is related to the 25cm rung.

Let's assume the length of the concrete steps varies, and the given lengths (45 cm and 25 cm) are related to the lengths of the steps, not the rungs directly.

Let the length of the bottom concrete step be $L_1$ and the length of the top concrete step be $L_{15}$. The total number of steps is 15. The steps have a uniform decrease in length. This sequence of lengths $L_1, L_2, \dots, L_{15}$ forms an AP.

Let's assume the lengths of the steps correspond to the rung lengths mentioned: $L_1 = 45$ cm $= 0.45$ m, $L_{15} = 25$ cm $= 0.25$ m.

The tread is $\frac{1}{2}$ m and the rise is $\frac{1}{4}$ m. The width of the terrace is 50 m.

Let's assume the volume of a step is its length $\times$ tread $\times$ rise. But the diagram shows the steps are stacked.

Let's assume the volume of concrete in each horizontal layer of height $\frac{1}{4}$ forms an AP because the length of the rung at that level changes uniformly. The length of the rung is the horizontal dimension of the step at that level.

Let the length of the concrete step at the level of the $k$-th rung (from bottom) be $L_k$. The rungs are placed on the steps.

Let's assume the volume of the concrete in the $k$-th step (from the bottom) is proportional to the length of the rung at that level. This is not directly stated.

Let's assume the problem means the volume of concrete in each horizontal slice of thickness $\frac{1}{4}$m forms an AP. The area of the horizontal slice at a certain height will depend on the horizontal extent at that height (the length of the step). The length of the step (perpendicular to the 50m width) decreases uniformly.

Let the length of the step at the bottom be $L_{bottom} = 0.45$ m. Let the length of the step at the top be $L_{top} = 0.25$ m. There are 15 steps. The lengths of the steps form an AP.

The step at height $\frac{1}{4}$ from the bottom corresponds to the bottom rung length. The step at total height $15 \times \frac{1}{4}$ corresponds to the top rung length.

Let's assume the lengths of the steps form an AP with 15 terms, $a=0.45$, $a_{15}=0.25$. We need the total volume. Each step layer has a thickness of $\frac{1}{4}$ and a width of 50. The varying dimension is the "length" (tread direction).

Let's assume the volume of the $k$-th layer of concrete from the bottom has a depth that corresponds to the $k$-th term in an AP. The first layer has depth related to 45cm, the last related to 25cm. This is still confusing.

Let's re-evaluate the simplest interpretation that fits the hint: The volume of concrete for each step (layer) might form an AP.

Volume of the first step (bottom layer with height $\frac{1}{4}$) = $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$. This is the first term of the sequence of volumes.

Consider the steps are built layer by layer, each layer having a rise of $\frac{1}{4}$ and a tread of $\frac{1}{2}$ and length 50. The total volume is the sum of the volumes of these layers. The layers are stacked such that they form steps. The first layer (bottom) has height $\frac{1}{4}$. The second layer is placed on the tread of the first, adding another $\frac{1}{4}$ height. The volume of concrete in each layer might be decreasing as the "length" of the step decreases.

Let's assume the volume of concrete in the $k$-th step from the bottom is $V_k$. The total volume is $\sum_{k=1}^{15} V_k$. The hint suggests the first term of the volume series is $V_1 = \frac{25}{4}$.

The length of the rungs decreases uniformly. This suggests that the horizontal dimension of the concrete steps also decreases uniformly as we go up.

Let the length of the step at the level of the $k$-th rung from the bottom be $L_k$. The length of the bottom step is 45cm $= 0.45$m. The length of the top step is 25cm $= 0.25$m. There are 15 steps.

Let's assume the horizontal length of the concrete step at the level of the $k$-th rung from the bottom is given by an AP. The length of the lowest step is 0.45m, the length of the highest step is 0.25m.

Let's assume the question implies that the volume of concrete in the $k$-th horizontal layer of thickness $\frac{1}{4}$ (starting from the bottom) has a horizontal area $A_k$, and $A_k \times 50$ is the volume. The area $A_k$ corresponds to the cross-section in the side view.

The area of the cross-section of the $k$-th rectangle from the bottom (height $\frac{1}{4}$, width $\frac{1}{2}$) is $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$. This area seems constant for each layer.

Let's reconsider the problem statement and diagram carefully. The rungs are on the steps. The steps are concrete. The lengths decreasing uniformly from 45cm to 25cm refers to the rungs, which suggests the steps might have a varying width (the dimension a person walks on).

Let's assume the problem means the volumes of concrete used for each step are related to the rung lengths. Perhaps the "length" 50m is the width of the steps, and the varying dimension is the depth (tread).

Let the depth of the bottom step be $D_1$ related to 45cm, and the depth of the top step be $D_{15}$ related to 25cm. The rise is $\frac{1}{4}$, the width is 50. If the depth decreases uniformly, the sequence $D_1, D_2, \dots, D_{15}$ forms an AP. The volumes are $50 \times D_k \times \frac{1}{4}$. If $D_k$ forms an AP, then $V_k = \frac{50}{4} D_k = \frac{25}{2} D_k$ also forms an AP.

Let's assume the lengths given (45cm and 25cm) are the depths of the bottom and top steps respectively. Convert to meters: $D_1 = 0.45$ m, $D_{15} = 0.25$ m.

The volume of the first step (bottom) = Width $\times$ Depth $\times$ Rise $= 50 \times 0.45 \times \frac{1}{4}$ m$^3$.

The volume of the last step (top) = Width $\times$ Depth $\times$ Rise $= 50 \times 0.25 \times \frac{1}{4}$ m$^3$.

Volume of 1st step $= 50 \times \frac{45}{100} \times \frac{1}{4} = \frac{50 \times 45}{400} = \frac{1 \times 45}{8} = \frac{45}{8}$ m$^3$.

Volume of 15th step $= 50 \times \frac{25}{100} \times \frac{1}{4} = \frac{50 \times 25}{400} = \frac{1 \times 25}{8} = \frac{25}{8}$ m$^3$.

The depths $D_k$ form an AP with $D_1 = 0.45$ and $D_{15} = 0.25$. The volumes $V_k = 50 \times D_k \times \frac{1}{4} = \frac{25}{2} D_k$. Since $D_k$ forms an AP, $V_k$ also forms an AP with the first term $V_1 = \frac{25}{2} \times 0.45 = \frac{25}{2} \times \frac{45}{100} = \frac{1}{2} \times \frac{45}{4} = \frac{45}{8}$. The last term is $V_{15} = \frac{25}{2} \times 0.25 = \frac{25}{2} \times \frac{25}{100} = \frac{1}{2} \times \frac{25}{4} = \frac{25}{8}$.

The volumes of the 15 steps form an AP with first term $a = \frac{45}{8}$, last term $a_{15} = \frac{25}{8}$, and number of terms $n = 15$.

We need to find the total volume, which is the sum of this AP, $S_{15}$.

Using the formula $S_n = \frac{n}{2}(a + a_n)$:

$S_{15} = \frac{15}{2}(\frac{45}{8} + \frac{25}{8})$

$S_{15} = \frac{15}{2}(\frac{45 + 25}{8})$

$S_{15} = \frac{15}{2}(\frac{70}{8})$

Simplify $\frac{70}{8} = \frac{35}{4}$.

$S_{15} = \frac{15}{2} \times \frac{35}{4}$

$S_{15} = \frac{15 \times 35}{2 \times 4}$

$S_{15} = \frac{525}{8}$ m$^3$.

Convert to decimal: $525 \div 8 = 65.625$.

$S_{15} = 65.625$ m$^3$.

---

Let's check if the calculated number of rungs makes sense with the distance between top and bottom rungs. The distance between top and bottom rungs is $2\frac{1}{2}$ m $= 250$ cm. If rungs are 25 cm apart, the number of gaps is $250 \div 25 = 10$. The number of rungs is number of gaps + 1 = $10 + 1 = 11$. This contradicts the fact that there are 15 steps/rungs. The problem phrasing is slightly ambiguous ("comprises of 15 steps", "There are 15 steps", but the rung calculation gives 11). Let's assume there are 15 steps, and 15 rungs placed on them, and the dimensions relate to the steps.

Let's assume the problem implies 15 steps and the lengths 45cm and 25cm refer to the lengths of the bottom and top steps (the dimension perpendicular to the 50m width and parallel to the ground). The width is 50m. The rise is $\frac{1}{4}$m. The tread is $\frac{1}{2}$m.

This means the cross-sectional area of each horizontal layer of thickness $\frac{1}{4}$ is $L_k \times \frac{1}{2}$, where $L_k$ is the length of the step at that layer's level. The length $L_k$ forms an AP from 45cm to 25cm over 15 layers. This doesn't seem right either based on the diagram.

Let's go back to the interpretation where the steps are horizontal layers of concrete, each with thickness $\frac{1}{4}$ m and width 50 m. The length (depth) of these layers is decreasing. The bottom layer has a length equal to the total tread distance, and the top layer has a length equal to the tread of the top step.

Let's assume the volume of concrete in each horizontal slab of thickness $\frac{1}{4}$ m (starting from the bottom) forms an AP. The bottom slab's volume is $V_1$, the next is $V_2$, ..., the top slab's volume is $V_{15}$. The sum is $S_{15}$.

Each slab has height $\frac{1}{4}$ and width 50. The length of the slab decreases. The total horizontal distance covered by the treads is $15 \times \frac{1}{2}$.

Let's assume the horizontal length of the concrete layer at height $k \times \frac{1}{4}$ (measured from the level of the top of the first step, so at heights $\frac{1}{4}, \frac{2}{4}, \dots, \frac{15}{4}$) is related to the rung lengths. This is still confusing.

Let's look at the problem again. "Each step has a rise of $\frac{1}{4}$ m and a tread of $\frac{1}{2}$ m". This sounds like the dimensions of the rectangular block that forms a step in the staircase profile.

The total volume is the sum of the volumes of the 15 steps. Let's assume the lengths of the steps (perpendicular to the 50m width) decrease uniformly, forming an AP from 45cm to 25cm. Let these lengths be $l_1, l_2, \dots, l_{15}$, where $l_1 = 0.45$ m and $l_{15} = 0.25$ m.

The volume of the $k$-th step block is $50 \times \frac{1}{2} \times l_k$. No, this does not match the diagram.

Let's consider the volume of concrete for each step (layer) from the bottom. Each layer has a height of $\frac{1}{4}$ m and a width of 50 m. The depth of the layer is the total horizontal extent of concrete at that height.

The bottom layer (height 0 to $\frac{1}{4}$) has depth equal to the total tread of 15 steps, $15 \times \frac{1}{2}$. Volume $V_1 = 50 \times (15 \times \frac{1}{2}) \times \frac{1}{4} = 50 \times \frac{15}{2} \times \frac{1}{4} = \frac{750}{8} = \frac{375}{4}$.

The second layer (height $\frac{1}{4}$ to $\frac{1}{2}$) has depth equal to the total tread of 14 steps, $14 \times \frac{1}{2}$. Volume $V_2 = 50 \times (14 \times \frac{1}{2}) \times \frac{1}{4} = 50 \times 7 \times \frac{1}{4} = \frac{350}{4} = \frac{175}{2}$.

The third layer (height $\frac{1}{2}$ to $\frac{3}{4}$) has depth $13 \times \frac{1}{2}$. Volume $V_3 = 50 \times (13 \times \frac{1}{2}) \times \frac{1}{4} = 50 \times \frac{13}{2} \times \frac{1}{4} = \frac{650}{8} = \frac{325}{4}$.

The volumes are $\frac{375}{4}, \frac{350}{4}, \frac{325}{4}, \dots$. Let's check the common difference:

$V_2 - V_1 = \frac{350}{4} - \frac{375}{4} = -\frac{25}{4}$.

$V_3 - V_2 = \frac{325}{4} - \frac{350}{4} = -\frac{25}{4}$.

This sequence of volumes forms an AP with first term $a = \frac{375}{4}$ and common difference $d = -\frac{25}{4}$.

The number of terms is 15 (since there are 15 layers of height $\frac{1}{4}$).

We need to find the sum of these 15 volumes, $S_{15}$.

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

$S_{15} = \frac{15}{2}[2(\frac{375}{4}) + (15-1)(-\frac{25}{4})]$

$S_{15} = \frac{15}{2}[\frac{375}{2} + (14)(-\frac{25}{4})]$

$S_{15} = \frac{15}{2}[\frac{375}{2} - \frac{14 \times 25}{4}]$

$S_{15} = \frac{15}{2}[\frac{375}{2} - \frac{350}{4}]$

$S_{15} = \frac{15}{2}[\frac{375 \times 2}{2 \times 2} - \frac{350}{4}]$

$S_{15} = \frac{15}{2}[\frac{750}{4} - \frac{350}{4}]$

$S_{15} = \frac{15}{2}[\frac{750 - 350}{4}]$

$S_{15} = \frac{15}{2}[\frac{400}{4}]$

$S_{15} = \frac{15}{2}[100]$

$S_{15} = 15 \times \frac{100}{2}$

$S_{15} = 15 \times 50 = 750$.

The total volume of concrete required is 750 m$^3$.

---

Let's recheck the volume calculation for the last layer (top). The top layer has height $\frac{1}{4}$ and depth equal to the tread of the top step, which corresponds to 1 rung length. Let's assume the top rung length is the depth of the top concrete step, 25cm $= 0.25$m. Then the volume of the top step is $50 \times 0.25 \times \frac{1}{4} = 50 \times \frac{1}{4} \times \frac{1}{4} = \frac{50}{16} = \frac{25}{8}$. This is the 15th term $a_{15}$.

If $a_n = a + (n-1)d$, then $a_{15} = a + 14d$.

$\frac{25}{8} = \frac{375}{4} + 14(-\frac{25}{4})$

$\frac{25}{8} = \frac{750}{8} - \frac{14 \times 25}{4} = \frac{750}{8} - \frac{350}{4} = \frac{750}{8} - \frac{700}{8} = \frac{50}{8} = \frac{25}{4}$.

Wait, $\frac{25}{4}$ is not $\frac{25}{8}$. My assumption about the last term is wrong based on the calculated common difference.

Let's reconsider the volumes $V_k$ for $k=1, \dots, 15$. The lowest layer (Step 1) has height $\frac{1}{4}$. Its horizontal area is $50 \times$ (depth). The total horizontal depth at this level seems to be the sum of the treads of all 15 steps if viewed from above, but this doesn't make sense for volume calculation in layers.

Let's stick with the interpretation that the volume of the concrete in each horizontal layer of thickness $\frac{1}{4}$ forms an AP. The area of the horizontal cross-section at height $h = k \times \frac{1}{4}$ (from the bottom) is $50 \times L_k$, where $L_k$ is the horizontal extent at that height. The lengths of the steps decrease uniformly. The bottom step has length related to 45cm. The top step has length related to 25cm.

Let the horizontal length of the concrete at height $\frac{1}{4}$ be $x_1$. At height $\frac{2}{4}$ be $x_2$, ..., at height $\frac{15}{4}$ be $x_{15}$. These lengths form an AP. The volume of the $k$-th layer is $50 \times x_k \times \frac{1}{4}$. If $x_k$ forms an AP, the volumes form an AP.

Let's assume the horizontal lengths of the concrete layers (corresponding to the length of the rung at that level) form an AP: $l_1, l_2, \dots, l_{15}$ where $l_1 = 0.45$ and $l_{15} = 0.25$. The volume of the $k$-th layer is $50 \times l_k \times \frac{1}{4}$.

Volume of 1st layer $= 50 \times 0.45 \times \frac{1}{4} = \frac{45}{8}$.

Volume of 15th layer $= 50 \times 0.25 \times \frac{1}{4} = \frac{25}{8}$.

The volumes form an AP with $a = \frac{45}{8}$, $a_{15} = \frac{25}{8}$, $n = 15$.

The sum is $S_{15} = \frac{15}{2}(a + a_{15}) = \frac{15}{2}(\frac{45}{8} + \frac{25}{8}) = \frac{15}{2}(\frac{70}{8}) = \frac{15}{2}(\frac{35}{4}) = \frac{525}{8} = 65.625$.

This seems like a plausible interpretation, assuming the lengths 45cm and 25cm refer to the depths of the bottom and top steps, and the staircase structure is built with constant rise ($\frac{1}{4}$) and width (50), while the depth decreases uniformly.

---

Let's verify the hint again. "Volume of concrete required to build the first step = $\frac{1}{4}$ × $\frac{1}{2}$ × 50 m$^3$". This is height $\times$ tread $\times$ length. This matches the dimensions of the bottom step if its height is $\frac{1}{4}$, tread is $\frac{1}{2}$, and length is 50. But the lengths of rungs are 45cm and 25cm. This strongly suggests the lengths of the steps vary.

Let's assume the 50m is the width. The rise is $\frac{1}{4}$, the tread is $\frac{1}{2}$. The length (perpendicular to the 50m width) decreases. The bottom step length is 45cm $= 0.45$m, the top step length is 25cm $= 0.25$m. There are 15 steps.

Volume of the bottom step = Width $\times$ Tread $\times$ Length $= 50 \times \frac{1}{2} \times 0.45 = 25 \times 0.45 = 11.25$ m$^3$.

Volume of the top step = Width $\times$ Tread $\times$ Length $= 50 \times \frac{1}{2} \times 0.25 = 25 \times 0.25 = 6.25$ m$^3$.

The lengths $0.45, \dots, 0.25$ form an AP. The volumes $50 \times \frac{1}{2} \times (\text{length})$ form an AP. The volumes are $11.25, \dots, 6.25$. Let the volumes be $V_1, V_2, \dots, V_{15}$.

$V_k = 50 \times \frac{1}{2} \times L_k = 25 L_k$, where $L_k$ is the length of the $k$-th step (from bottom).

$L_1 = 0.45$, $L_{15} = 0.25$. This AP has $n=15$.

The sum of the lengths is $S_L = \frac{15}{2}(0.45 + 0.25) = \frac{15}{2}(0.70) = 15 \times 0.35 = 5.25$ m.

The total volume is the sum of $V_k = 25 L_k$. Total Volume = $\sum_{k=1}^{15} 25 L_k = 25 \sum_{k=1}^{15} L_k = 25 \times S_L$.

Total Volume $= 25 \times 5.25$ m$^3$.

$25 \times 5.25 = 25 \times 5 + 25 \times 0.25 = 125 + 6.25 = 131.25$ m$^3$.

This seems a more consistent interpretation with "lengths decrease uniformly". Let's check the hint again. "Volume of concrete required to build the first step = $\frac{1}{4}$ × $\frac{1}{2}$ × 50 m$^3$". This is $6.25$ m$^3$. This matches $50 \times \frac{1}{2} \times 0.25$, which is the volume of the top step (length 0.25m). This contradicts the bottom step having length 0.45m.

Let's assume the hint volume is the volume of the BOTTOM step. So $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4} = 6.25$. If this is the volume of the bottom step, and its length is 0.45m, its width is 50m, its rise is $\frac{1}{4}$m, then Volume = width $\times$ length $\times$ rise = $50 \times 0.45 \times \frac{1}{4} = 11.25$. This is not 6.25.

Let's go back to the first successful interpretation where horizontal layers of height $\frac{1}{4}$ form an AP of volumes: $\frac{375}{4}, \frac{350}{4}, \frac{325}{4}, \dots$. The sum was 750 m$^3$. This interpretation assumes the horizontal length of the concrete at height $k \times \frac{1}{4}$ corresponds to the total tread of the top $15-k+1$ steps. At height 0, it's 15 treads. At height $\frac{14}{4}$, it's 1 tread.

Let's check if the last term of this AP is correct. $a = \frac{375}{4}$, $d = -\frac{25}{4}$. $a_{15} = a + 14d = \frac{375}{4} + 14(-\frac{25}{4}) = \frac{375 - 350}{4} = \frac{25}{4}$. This is the volume of the top layer, which is $50 \times (\text{length}) \times \frac{1}{4}$. If the length is $\frac{1}{2}$ (tread), the volume is $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$.

So, the volume of concrete in the $k$-th horizontal layer of thickness $\frac{1}{4}$ m (from the bottom) is $50 \times \frac{1}{2} \times (16-k) \times \frac{1}{2}$? No.

The interpretation where the volume of concrete for each horizontal layer of height $\frac{1}{4}$ forms an AP seems to work. The bottom layer volume is $V_1 = 50 \times (\text{total tread}) \times \frac{1}{4}$? No.

Let's reconsider the horizontal layers of thickness $\frac{1}{4}$. The bottom layer has depth $15 \times \frac{1}{2}$, length 50, height $\frac{1}{4}$. Volume = $50 \times (15 \times \frac{1}{2}) \times \frac{1}{4} = 50 \times \frac{15}{2} \times \frac{1}{4} = \frac{750}{8}$.

The next layer has depth $14 \times \frac{1}{2}$. Volume = $50 \times (14 \times \frac{1}{2}) \times \frac{1}{4} = \frac{700}{8}$.

The top layer has depth $1 \times \frac{1}{2}$. Volume = $50 \times (1 \times \frac{1}{2}) \times \frac{1}{4} = \frac{50}{8}$.

The volumes are $\frac{750}{8}, \frac{700}{8}, \frac{650}{8}, \dots, \frac{50}{8}$. This forms an AP with $a = \frac{750}{8}$, $d = -\frac{50}{8}$, and $n=15$. The last term is indeed $\frac{50}{8}$.

The sum of this AP is $S_{15} = \frac{15}{2}(a + a_{15}) = \frac{15}{2}(\frac{750}{8} + \frac{50}{8}) = \frac{15}{2}(\frac{800}{8}) = \frac{15}{2}(100) = 15 \times 50 = 750$.

This interpretation aligns with the diagram where the concrete extends back further at lower levels. The total horizontal depth at each level is the sum of the treads from that level upwards. At the bottom, it's 15 treads. At the top, it's 1 tread.

Volume of 1st layer (bottom, height $\frac{1}{4}$): Width = 50, Height = $\frac{1}{4}$, Depth = $15 \times \frac{1}{2}$. $V_1 = 50 \times \frac{1}{4} \times 15 \times \frac{1}{2} = \frac{750}{8}$.

Volume of 2nd layer (height $\frac{1}{4}$ to $\frac{1}{2}$): Width = 50, Height = $\frac{1}{4}$, Depth = $14 \times \frac{1}{2}$. $V_2 = 50 \times \frac{1}{4} \times 14 \times \frac{1}{2} = \frac{700}{8}$.

... Volume of 15th layer (top, height $\frac{14}{4}$ to $\frac{15}{4}$): Width = 50, Height = $\frac{1}{4}$, Depth = $1 \times \frac{1}{2}$. $V_{15} = 50 \times \frac{1}{4} \times 1 \times \frac{1}{2} = \frac{50}{8}$.

This AP of volumes has $a=\frac{750}{8}$, $d=\frac{700}{8} - \frac{750}{8} = -\frac{50}{8}$, and $n=15$. The last term is $a_{15} = a + 14d = \frac{750}{8} + 14(-\frac{50}{8}) = \frac{750 - 700}{8} = \frac{50}{8}$. This confirms the AP.

The sum is $S_{15} = \frac{15}{2}(\frac{750}{8} + \frac{50}{8}) = \frac{15}{2}(\frac{800}{8}) = \frac{15}{2}(100) = 750$.

---

Let's try to connect this to the hint: "Volume of concrete required to build the first step = $\frac{1}{4}$ × $\frac{1}{2}$ × 50 m$^3$". This volume is $\frac{25}{4}$. In the AP of volumes calculated above ($\frac{750}{8}, \frac{700}{8}, \dots, \frac{50}{8}$), $\frac{25}{4} = \frac{50}{8}$ is the volume of the top layer (the 15th term). The hint refers to the "first step". If "first step" refers to the top step, the hint is correct, and the sequence of volumes is the reverse of the one calculated, i.e., starting from the top.

Let the volume of the step layers, starting from the top, be an AP. The top step volume is $V_1' = 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$. This is the first term $a' = \frac{25}{4}$.

The volumes increase as we go down. The difference between consecutive layers (from top to bottom) would be the same as the difference from bottom to top, but positive: $d' = \frac{50}{8} = \frac{25}{4}$.

The sequence of volumes from top to bottom is $\frac{25}{4}, \frac{25}{4} + \frac{25}{4}, \frac{25}{4} + 2\frac{25}{4}, \dots$.

This is $\frac{25}{4}, \frac{50}{4}, \frac{75}{4}, \dots$. The common difference is $\frac{25}{4}$.

The first term (top) is $a' = \frac{25}{4}$. The common difference is $d' = \frac{25}{4}$. The number of terms is $n=15$.

The sum of the 15 terms is $S_{15} = \frac{15}{2}[2a' + (15-1)d'] = \frac{15}{2}[2(\frac{25}{4}) + 14(\frac{25}{4})]$.

$S_{15} = \frac{15}{2}[\frac{50}{4} + \frac{350}{4}]$

$S_{15} = \frac{15}{2}[\frac{400}{4}]$

$S_{15} = \frac{15}{2}[100] = 15 \times 50 = 750$.

This interpretation is consistent with the hint if "first step" refers to the top step. The volume of each horizontal layer of thickness $\frac{1}{4}$ is $50 \times \frac{1}{2} \times k \times \frac{1}{4}$ where $k$ is the number of treads from the top. No, this is also wrong.

Let's go back to the AP of volumes $\frac{750}{8}, \frac{700}{8}, \dots, \frac{50}{8}$. This sequence has first term $\frac{750}{8}$ (bottom layer volume) and last term $\frac{50}{8}$ (top layer volume). The sum is 750.

The hint "Volume of concrete required to build the first step = $\frac{1}{4}$ × $\frac{1}{2}$ × 50 m$^3$" refers to the volume of the top step if interpreted as a block with dimensions rise $\times$ tread $\times$ length. Let's assume the hint is defining the volume unit or the volume of the simplest step block.

The total volume is the sum of volumes of 15 layers, each of height $\frac{1}{4}$ and width 50. The length of the $k$-th layer from the bottom is $15 \times \frac{1}{2} - (k-1) \times \frac{1}{2} = (15 - k + 1) \times \frac{1}{2} = (16-k) \times \frac{1}{2}$.

Volume of the $k$-th layer $V_k = 50 \times (16-k) \times \frac{1}{2} \times \frac{1}{4} = 50 \times (16-k) \times \frac{1}{8} = \frac{25}{4}(16-k)$.

$V_1 = \frac{25}{4}(16-1) = \frac{25}{4}(15) = \frac{375}{4}$.

$V_2 = \frac{25}{4}(16-2) = \frac{25}{4}(14) = \frac{350}{4}$.

$V_{15} = \frac{25}{4}(16-15) = \frac{25}{4}(1) = \frac{25}{4}$.

The volumes are $\frac{375}{4}, \frac{350}{4}, \dots, \frac{25}{4}$. This is an AP with $a = \frac{375}{4}$, $d = -\frac{25}{4}$, $n=15$. The last term is $a_{15} = a + 14d = \frac{375}{4} + 14(-\frac{25}{4}) = \frac{375-350}{4} = \frac{25}{4}$. This AP is correct.

The sum is $S_{15} = \frac{15}{2}(a + a_{15}) = \frac{15}{2}(\frac{375}{4} + \frac{25}{4}) = \frac{15}{2}(\frac{400}{4}) = \frac{15}{2}(100) = 750$.

This interpretation fits the decreasing volume and leads to a calculation involving an AP. The hint "Volume of concrete required to build the first step = $\frac{1}{4}$ × $\frac{1}{2}$ × 50 m$^3$" seems to refer to the volume of the top-most layer if interpreted as a block with dimensions rise $\times$ tread $\times$ length. This is $\frac{1}{4} \times \frac{1}{2} \times 50 = \frac{25}{4}$. This volume is the last term ($V_{15}$) in our sequence, which is the volume of the top layer with depth $\frac{1}{2}$. This fits perfectly if the depth of the top layer is $\frac{1}{2}$m. The problem states "a tread of $\frac{1}{2}$ m", which is the horizontal dimension of each step.

So, the volume of the $k$-th layer from the bottom (thickness $\frac{1}{4}$, width 50) has a length equal to the sum of the treads of the top $16-k$ steps. No.

The volume of the $k$-th layer (from bottom) of thickness $\frac{1}{4}$ is $50 \times (\text{horizontal extent}) \times \frac{1}{4}$. The horizontal extent at this level is the tread $\times$ number of steps from this level upwards. At the bottom, the horizontal extent is the sum of treads of 15 steps if they were stacked back to back. The diagram shows the steps recede.

Let's assume the volume of concrete for each horizontal layer of height $\frac{1}{4}$ is $V_k$. The lowest layer has volume $V_1$. The next layer $V_2$, ..., the top layer $V_{15}$. These volumes form an AP. The hint gives the volume of the top layer (the "first step" from the top) as $\frac{25}{4}$.

The volume of the top layer is $V_{15} = 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}$. This is the last term of the AP of volumes from bottom to top. So $a_{15} = \frac{25}{4}$.

What is the first term? The volume of the bottom layer. It has height $\frac{1}{4}$ and width 50. Its depth is the total horizontal extent. The total horizontal extent of the 15 steps is $15 \times \frac{1}{2}$. So the bottom layer might be $50 \times 15 \times \frac{1}{2} \times \frac{1}{4}$? No.

Let's use the volumes of the layers of height $\frac{1}{4}$, starting from the bottom. The number of treads at the bottom level is 15 times the tread $\frac{1}{2}$? No. The horizontal length of concrete at the level of the $k$-th rise from the bottom is the sum of treads from that level to the top? No.

The total volume is the sum of the areas of the cross-section slices (each of width 50). The area of the cross-section is the sum of 15 rectangles of width $\frac{1}{2}$ and heights $\frac{1}{4}, \frac{2}{4}, \dots, \frac{15}{4}$. No, this is wrong.

The area of the cross-section is the sum of the areas of 15 rectangles with base $\frac{1}{2}$ and height $\frac{1}{4}$, stacked. The first rectangle (bottom) has area $\frac{1}{2} \times \frac{1}{4}$. The second is on top, area $\frac{1}{2} \times \frac{1}{4}$. The third is on top, area $\frac{1}{2} \times \frac{1}{4}$.

The volume of concrete in the $k$-th step from the bottom is $50 \times (\text{area of } k \text{-th rectangle in cross-section})$. The area of the $k$-th rectangle (from bottom) is $\frac{1}{2} \times \frac{1}{4}$. This area is constant for each step layer.

Let's re-read the problem again. "Each step has a rise of $\frac{1}{4}$ m and a tread of $\frac{1}{2}$ m". This describes the shape of each step.

Consider the volume of concrete in each step. The bottom step is a rectangular prism with dimensions $50 \times \frac{1}{2} \times \frac{1}{4}$. Volume is $\frac{25}{4}$.

The step above it is also a rectangular prism with dimensions $50 \times \frac{1}{2} \times \frac{1}{4}$. Volume is $\frac{25}{4}$.

If each of the 15 steps is a separate block of concrete with these dimensions, the total volume is $15 \times \frac{25}{4} = \frac{375}{4} = 93.75$. But the rungs decreasing in length suggests the structure is not uniform like this.

Let's go back to the interpretation that the volumes of the horizontal layers of thickness $\frac{1}{4}$ form an AP. The $k$-th layer from the bottom has depth $(16-k) \times \frac{1}{2}$ and width 50 and height $\frac{1}{4}$.

Volume $V_k = 50 \times (16-k) \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4}(16-k)$.

Volumes are $\frac{25}{4}(15), \frac{25}{4}(14), \dots, \frac{25}{4}(1)$.

This is an AP with first term $a = \frac{25}{4} \times 15 = \frac{375}{4}$, common difference $d = \frac{25}{4} \times 14 - \frac{25}{4} \times 15 = \frac{25}{4} (14 - 15) = -\frac{25}{4}$. The number of terms is 15.

The sum is $S_{15} = \frac{15}{2}(a + a_{15}) = \frac{15}{2}(\frac{375}{4} + \frac{25}{4}) = \frac{15}{2}(\frac{400}{4}) = \frac{15}{2}(100) = 750$.

The hint "Volume of concrete required to build the first step = $\frac{1}{4}$ × $\frac{1}{2}$ × 50 m$^3$" is $\frac{25}{4}$. This is the volume of the layer with depth $\frac{1}{2}$, which is the top layer ($k=15$). So the hint gives the volume of the top layer.

So the sequence of volumes from bottom to top is $V_1, V_2, \dots, V_{15}$, where $V_{15} = \frac{25}{4}$. The common difference is $d = -\frac{25}{4}$. The first term $V_1$ must be such that $V_1 + 14(-\frac{25}{4}) = \frac{25}{4}$. $V_1 - \frac{350}{4} = \frac{25}{4}$. $V_1 = \frac{375}{4}$. This matches the AP found earlier.

The total volume is the sum of the terms of this AP.

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The volumes of the horizontal layers of concrete, starting from the bottom layer, form an Arithmetic Progression (AP).

Each layer has a height (rise) of $\frac{1}{4}$ m and a width (length of the step) of 50 m.

The depth (tread direction) of the $k$-th layer from the bottom ($V_k$) corresponds to the horizontal extent of the concrete at that level. Based on the structure, the depth of the $k$-th layer from the bottom is the sum of the treads of the top $(15 - k + 1)$ steps, each tread being $\frac{1}{2}$ m.

Depth of the $k$-th layer $= (16 - k) \times \frac{1}{2}$ m.

Volume of the $k$-th layer, $V_k = \text{Width} \times \text{Depth} \times \text{Height} = 50 \times (16 - k) \times \frac{1}{2} \times \frac{1}{4}$ m$^3$.

$V_k = \frac{50}{8}(16 - k) = \frac{25}{4}(16 - k)$.

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The volumes of the layers for $k=1, 2, \dots, 15$ are:

For $k=1$ (bottom layer): $V_1 = \frac{25}{4}(16-1) = \frac{25}{4} \times 15 = \frac{375}{4}$ m$^3$.

For $k=2$: $V_2 = \frac{25}{4}(16-2) = \frac{25}{4} \times 14 = \frac{350}{4}$ m$^3$.

For $k=15$ (top layer): $V_{15} = \frac{25}{4}(16-15) = \frac{25}{4} \times 1 = \frac{25}{4}$ m$^3$.

The sequence of volumes is $\frac{375}{4}, \frac{350}{4}, \dots, \frac{25}{4}$. This is an AP with first term $a = \frac{375}{4}$, common difference $d = \frac{350}{4} - \frac{375}{4} = -\frac{25}{4}$, and number of terms $n = 15$. The last term is $a_{15} = \frac{25}{4}$.

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The total volume of concrete is the sum of the volumes of these 15 layers ($S_{15}$).

Using the formula for the sum of the first $n$ terms of an AP when the first and last terms are known:

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Substitute the values $n=15$, $a=\frac{375}{4}$, and $a_{15}=\frac{25}{4}$ into the sum formula:

$S_{15} = \frac{15}{2}(\frac{375}{4} + \frac{25}{4})$

$S_{15} = \frac{15}{2}(\frac{375 + 25}{4})$

$S_{15} = \frac{15}{2}(\frac{400}{4})$

$S_{15} = \frac{15}{2}(100)$

$S_{15} = 15 \times \frac{100}{2}$

$S_{15} = 15 \times 50$

$S_{15} = 750$.

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The total volume of concrete required is 750 m$^3$. This matches the result obtained with the previous consistent interpretation.

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Conclusion: The total volume of concrete required to build the terrace is $\mathbf{750}$ m$^3$.