Chapter 5 Arithmetic Progressions

Given:

The given Arithmetic Progression (AP) is 5, 8, 11, 14, ...

To Find:

The 10th term of the given AP.

Solution:

In the given AP: 5, 8, 11, 14, ...

The first term is $a = 5$.

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

$d = 8 - 5 = 3$

$d = 11 - 8 = 3$

$d = 14 - 11 = 3$

So, the common difference is $d = 3$.

We need to find the 10th term of the AP. This means $n = 10$.

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

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

Substitute the values $a=5$, $d=3$, and $n=10$ into the formula:

$a_{10} = 5 + (10-1) \times 3$

$a_{10} = 5 + (9) \times 3$

$a_{10} = 5 + 27$

$a_{10} = 32$

Thus, the 10th term of the given AP is 32.

Comparing this result with the given options, we find that 32 corresponds to option (A).

Final Answer:

The 10th term of the AP is 32.

The correct option is (A) 32.

Given:

In an Arithmetic Progression (AP), we are given:

First term, $a = -7.2$

Common difference, $d = 3.6$

The $n$-th term, $a_n = 7.2$

To Find:

The value of $n$, which represents the position of the term $a_n = 7.2$ in the sequence.

Solution:

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

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

Substitute the given values into this formula:

$7.2 = -7.2 + (n-1)(3.6)$

Now, we need to solve this equation for $n$.

Add $7.2$ to both sides of the equation:

$7.2 + 7.2 = (n-1)(3.6)$

$14.4 = (n-1)(3.6)$

Divide both sides by $3.6$:

$\frac{14.4}{3.6} = n-1$

Performing the division:

$4 = n-1$

Add $1$ to both sides to isolate $n$:

$4 + 1 = n$

$n = 5$

So, the value of $n$ is 5.

This means that the 5th term of the AP is 7.2.

Comparing this result with the given options, we find that 5 corresponds to option (D).

Final Answer:

The value of $n$ is 5.

The correct option is (D) 5.

Given:

In an Arithmetic Progression (AP), we are given:

Common difference, $d = -4$

Number of terms, $n = 7$

The $n$-th term (7th term), $a_n = a_7 = 4$

To Find:

The first term, $a$ of the AP.

Solution:

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

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

Substitute the given values into this formula:

$4 = a + (7-1)(-4)$

Simplify the expression:

$4 = a + (6)(-4)$

$4 = a - 24$

To find $a$, add 24 to both sides of the equation:

$4 + 24 = a$

$28 = a$

So, the first term of the AP is 28.

Comparing this result with the given options, we find that 28 corresponds to option (D).

Final Answer:

The value of $a$ is 28.

The correct option is (D) 28.

Given:

In an Arithmetic Progression (AP), we are given:

First term, $a = 3.5$

Common difference, $d = 0$

Number of terms, $n = 101$

To Find:

The $n$-th term ($a_n$), which is the 101st term ($a_{101}$) of the AP.

Solution:

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

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

Substitute the given values $a = 3.5$, $d = 0$, and $n = 101$ into this formula:

$a_{101} = 3.5 + (101-1) \times 0$

Simplify the expression:

$a_{101} = 3.5 + (100) \times 0$

$a_{101} = 3.5 + 0$

$a_{101} = 3.5$

So, the 101st term of the AP is 3.5.

When the common difference $d$ is 0, every term in the AP is equal to the first term $a$. This is because no value is added to the preceding term to get the next term.

Comparing this result with the given options, we find that 3.5 corresponds to option (B).

Final Answer:

The value of $a_n$ (or $a_{101}$) is 3.5.

The correct option is (B) 3.5.

Given:

The given list of numbers is –10, –6, –2, 2,...

To Determine:

Whether the given list of numbers forms an Arithmetic Progression (AP) and if so, what the common difference is.

Solution:

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

Let the terms of the sequence be $a_1, a_2, a_3, a_4, ...$

Here, $a_1 = -10$, $a_2 = -6$, $a_3 = -2$, $a_4 = 2$, ...

We calculate the difference between consecutive terms:

Difference between the second term and the first term ($a_2 - a_1$):

$d_1 = a_2 - a_1 = -6 - (-10) = -6 + 10 = 4$

Difference between the third term and the second term ($a_3 - a_2$):

$d_2 = a_3 - a_2 = -2 - (-6) = -2 + 6 = 4$

Difference between the fourth term and the third term ($a_4 - a_3$):

$d_3 = a_4 - a_3 = 2 - (-2) = 2 + 2 = 4$

Since the difference between consecutive terms is constant ($d_1 = d_2 = d_3 = 4$), the given list of numbers is an AP.

The common difference is $d = 4$.

Comparing this result with the given options, we find that the list is an AP with a common difference of 4, which corresponds to option (B).

Final Answer:

The given list of numbers is an AP with a common difference of 4.

The correct option is (B) an AP with d = 4.

Given:

The given Arithmetic Progression (AP) is –5, $\frac{-5}{2}$, 0, $\frac{5}{2}$, ...

To Find:

The 11th term of the given AP.

Solution:

In the given AP: –5, $\frac{-5}{2}$, 0, $\frac{5}{2}$, ...

The first term is $a = -5$.

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

$d = a_2 - a_1 = \frac{-5}{2} - (-5) = \frac{-5}{2} + 5 = \frac{-5 + 10}{2} = \frac{5}{2}$

Check with the next pair of terms:

$d = a_3 - a_2 = 0 - (\frac{-5}{2}) = 0 + \frac{5}{2} = \frac{5}{2}$

The common difference is $d = \frac{5}{2}$.

We need to find the 11th term of the AP. This means $n = 11$.

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

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

Substitute the values $a = -5$, $d = \frac{5}{2}$, and $n = 11$ into the formula:

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

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

$a_{11} = -5 + \cancel{10}^{5} \times \frac{5}{\cancel{2}_{1}}$

$a_{11} = -5 + 5 \times 5$

$a_{11} = -5 + 25$

$a_{11} = 20$

Thus, the 11th term of the given AP is 20.

Comparing this result with the given options, we find that 20 corresponds to option (B).

Final Answer:

The 11th term of the AP is 20.

The correct option is (B) 20.

Given:

The first term of an AP, $a = -2$.

The common difference, $d = -2$.

To Find:

The first four terms of the AP.

Solution:

The terms of an Arithmetic Progression can be found using the formula $a_n = a + (n-1)d$, where $a$ is the first term, $d$ is the common difference, and $n$ is the term number.

Alternatively, each term after the first is obtained by adding the common difference to the preceding term.

The first term ($n=1$) is $a_1 = a$.

$a_1 = -2$

The second term ($n=2$) is $a_2 = a_1 + d$.

$a_2 = -2 + (-2)$

$a_2 = -2 - 2$

$a_2 = -4$

The third term ($n=3$) is $a_3 = a_2 + d$.

$a_3 = -4 + (-2)$

$a_3 = -4 - 2$

$a_3 = -6$

The fourth term ($n=4$) is $a_4 = a_3 + d$.

$a_4 = -6 + (-2)$

$a_4 = -6 - 2$

$a_4 = -8$

So, the first four terms of the AP are –2, –4, –6, and –8.

Comparing the calculated terms with the given options, we see that the sequence –2, –4, –6, –8 matches option (C).

Final Answer:

The first four terms of the AP are –2, –4, –6, –8.

The correct option is (C) – 2, – 4, – 6, – 8.

Given:

The first term of the AP, $a_1 = -3$.

The second term of the AP, $a_2 = 4$.

To Find:

The 21st term of the AP ($a_{21}$).

Solution:

In an Arithmetic Progression (AP), the common difference $d$ is the difference between any term and its preceding term.

Using the first two terms, the common difference $d$ is:

$d = a_2 - a_1$

$d = 4 - (-3)$

$d = 4 + 3$

$d = 7$

The first term of the AP is $a = a_1 = -3$.

The common difference is $d = 7$.

We need to find the 21st term of the AP. This means $n = 21$.

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

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

Substitute the values $a = -3$, $d = 7$, and $n = 21$ into the formula:

$a_{21} = -3 + (21-1) \times 7$

$a_{21} = -3 + (20) \times 7$

$a_{21} = -3 + 140$

$a_{21} = 137$

Thus, the 21st term of the AP is 137.

Comparing this result with the given options, we find that 137 corresponds to option (B).

Final Answer:

The 21st term of the AP is 137.

The correct option is (B) 137.

Given:

The 2nd term of an AP, $a_2 = 13$.

The 5th term of an AP, $a_5 = 25$.

To Find:

The 7th term of the AP, $a_7$.

Solution:

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

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

where $a$ is the first term and $d$ is the common difference.

Using the given information, we can write two equations:

For the 2nd term ($n=2$):

Substituting the value of $a_2$:

$13 = a + d$

For the 5th term ($n=5$):

Substituting the value of $a_5$:

$25 = a + 4d$

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

Equation A: $a + d = 13$

Equation B: $a + 4d = 25$

Subtract Equation A from Equation B:

$(a + 4d) - (a + d) = 25 - 13$

$a + 4d - a - d = 12$

$3d = 12$

Divide by 3:

$d = \frac{12}{3}$

$d = 4$

Now substitute the value of $d = 4$ into Equation A:

$a + 4 = 13$

Subtract 4 from both sides:

$a = 13 - 4$

$a = 9$

So, the first term of the AP is $a = 9$ and the common difference is $d = 4$.

We need to find the 7th term of the AP ($n=7$).

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

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

$a_7 = a + 6d$

Substitute the values $a = 9$ and $d = 4$:

$a_7 = 9 + 6 \times 4$

$a_7 = 9 + 24$

$a_7 = 33$

Thus, the 7th term of the AP is 33.

Comparing this result with the given options, we find that 33 corresponds to option (B).

Final Answer:

The 7th term of the AP is 33.

The correct option is (B) 33.

Given:

The given Arithmetic Progression (AP) is 21, 42, 63, 84, ...

The $n$-th term of the AP is $a_n = 210$.

To Find:

The term number $n$ for which $a_n = 210$.

Solution:

In the given AP: 21, 42, 63, 84, ...

The first term is $a = 21$.

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

$d = a_2 - a_1 = 42 - 21 = 21$

$d = a_3 - a_2 = 63 - 42 = 21$

So, the common difference is $d = 21$.

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

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

We are given $a_n = 210$, $a = 21$, and $d = 21$. Substitute these values into the formula:

$210 = 21 + (n-1)(21)$

Subtract 21 from both sides of the equation:

$210 - 21 = (n-1)(21)$

$189 = (n-1)(21)$

Divide both sides by 21:

$\frac{189}{21} = n-1$

Performing the division:

$9 = n-1$

Add 1 to both sides to find $n$:

$9 + 1 = n$

$n = 10$

So, the 10th term of the AP is 210.

Comparing this result with the given options, we find that the 10th term corresponds to option (B).

Final Answer:

The term number is 10.

The correct option is (B) 10^th.

Given:

The common difference of an AP is $d = 5$.

To Find:

The value of the difference $a_{18} - a_{13}$.

Solution:

The formula for the $n$-th term of an Arithmetic Progression 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 expression for the 18th term ($a_{18}$) by setting $n=18$:

$a_{18} = a + (18-1)d$

$a_{18} = a + 17d$

Next, we find the expression for the 13th term ($a_{13}$) by setting $n=13$:

$a_{13} = a + (13-1)d$

$a_{13} = a + 12d$

Now, we calculate the difference $a_{18} - a_{13}$:

$a_{18} - a_{13} = (a + 17d) - (a + 12d)$

Remove the parentheses:

$a_{18} - a_{13} = a + 17d - a - 12d$

Combine like terms (the $a$ terms cancel out):

$a_{18} - a_{13} = (17d - 12d)$

$a_{18} - a_{13} = 5d$

Substitute the given value of the common difference, $d = 5$, into this expression:

$a_{18} - a_{13} = 5 \times 5$

$a_{18} - a_{13} = 25$

So, the difference between the 18th term and the 13th term is 25.

Comparing this result with the given options, we find that 25 corresponds to option (C).

Final Answer:

The value of $a_{18} - a_{13}$ is 25.

The correct option is (C) 25.

Given:

In an Arithmetic Progression (AP), the difference between the 18th term and the 14th term is given as $a_{18} - a_{14} = 32$.

To Find:

The common difference ($d$) of the AP.

Solution:

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

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

where $a$ is the first term and $d$ is the common difference.

We can write the expression for the 18th term ($a_{18}$) by setting $n=18$:

$a_{18} = a + (18-1)d$

$a_{18} = a + 17d$

We can also write the expression for the 14th term ($a_{14}$) by setting $n=14$:

$a_{14} = a + (14-1)d$

$a_{14} = a + 13d$

We are given the difference $a_{18} - a_{14} = 32$. Substitute the expressions for $a_{18}$ and $a_{14}$ into this equation:

$(a + 17d) - (a + 13d) = 32$

Remove the parentheses and simplify:

$a + 17d - a - 13d = 32$

Combine the terms:

$(a - a) + (17d - 13d) = 32$

$0 + 4d = 32$

$4d = 32$

Now, solve for the common difference $d$ by dividing both sides by 4:

$d = \frac{32}{4}$

$d = 8$

So, the common difference of the AP is 8.

Comparing this result with the given options, we find that 8 corresponds to option (A).

Final Answer:

The common difference of the AP is 8.

The correct option is (A) 8.

Given:

Let the first AP be denoted by AP$_1$ and the second AP be denoted by AP$_2$.

Both APs have the same common difference, let's call it $d$.

The first term of AP$_1$ is $a_1 = -1$.

The first term of AP$_2$ is $a'_1 = -8$.

To Find:

The difference between their 4th terms, i.e., $a_4 - a'_4$ or $a'_4 - a_4$. Let's find $a_4 - a'_4$.

Solution:

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

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

where $a$ is the first term and $d$ is the common difference.

For AP$_1$, the first term is $a_1 = -1$ and the common difference is $d$. The 4th term ($n=4$) is:

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

$a_4 = -1 + 3d$

For AP$_2$, the first term is $a'_1 = -8$ and the common difference is $d$. The 4th term ($n=4$) is:

$a'_4 = a'_1 + (4-1)d$

$a'_4 = -8 + 3d$

Now we find the difference between their 4th terms:

$a_4 - a'_4 = (-1 + 3d) - (-8 + 3d)$

Remove the parentheses:

$a_4 - a'_4 = -1 + 3d + 8 - 3d$

Combine the terms:

$a_4 - a'_4 = (-1 + 8) + (3d - 3d)$

$a_4 - a'_4 = 7 + 0$

$a_4 - a'_4 = 7$

Alternatively, the difference between the $n$-th terms of two APs with the same common difference is always equal to the difference between their first terms.

$a_n - a'_n = (a + (n-1)d) - (a' + (n-1)d) = a - a'$

For $n=4$, the difference is $a_4 - a'_4 = a_1 - a'_1 = -1 - (-8) = -1 + 8 = 7$.

The difference between their 4th terms is 7.

Comparing this result with the given options, we find that 7 corresponds to option (C).

Final Answer:

The difference between their 4th terms is 7.

The correct option is (C) 7.

Given:

In an Arithmetic Progression (AP), 7 times the 7th term is equal to 11 times the 11th term.

This can be written as: $7 \times a_7 = 11 \times a_{11}$

To Find:

The 18th term of the AP, i.e., $a_{18}$.

Solution:

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

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

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

Using this formula, we can write the 7th term ($a_7$) and the 11th term ($a_{11}$):

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

$a_{11} = a + (11-1)d = a + 10d$

According to the given condition, we have:

$7 \times a_7 = 11 \times a_{11}$

Substitute the expressions for $a_7$ and $a_{11}$ into this equation:

$7(a + 6d) = 11(a + 10d)$

Expand both sides of the equation:

$7a + 42d = 11a + 110d$

Rearrange the terms to one side to solve for the relationship between $a$ and $d$. Subtract $7a$ from both sides and subtract $42d$ from both sides:

$42d - 110d = 11a - 7a$

$-68d = 4a$

Divide both sides by 4 to find $a$ in terms of $d$:

$a = \frac{-68d}{4}$

$a = -17d$

Now we need to find the 18th term of the AP, $a_{18}$. Using the formula for the $n$-th term with $n=18$:

$a_{18} = a + (18-1)d$

$a_{18} = a + 17d$

Substitute the relation $a = -17d$ that we found into the expression for $a_{18}$:

$a_{18} = (-17d) + 17d$

$a_{18} = 0$

Thus, the 18th term of the AP is 0.

Comparing this result with the given options, we find that 0 corresponds to option (D).

Final Answer:

The 18th term of the AP is 0.

The correct option is (D) 0.

Given:

The given Arithmetic Progression (AP) is –11, –8, –5, ..., 49.

We need to find the 4th term from the end of this AP.

To Find:

The value of the 4th term counted from the last term of the AP.

Solution:

In the given AP: –11, –8, –5, ..., 49.

The first term is $a = -11$.

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

$d = -8 - (-11) = -8 + 11 = 3$

The last term of the AP is $l = 49$.

To find the $k$-th term from the end of an AP, we can consider the AP in reverse order. The reversed AP will have the last term of the original AP as its first term, and its common difference will be the negative of the common difference of the original AP.

The reversed AP starts with the first term $a' = 49$.

The common difference of the reversed AP is $d' = -d = -3$.

The 4th term from the end of the original AP is the same as the 4th term from the beginning of the reversed AP.

We need to find the 4th term ($n=4$) of the AP with first term $a' = 49$ and common difference $d' = -3$.

Using the formula for the $n$-th term of an AP, $a'_n = a' + (n-1)d'$:

$a'_4 = a' + (4-1)d'$

$a'_4 = 49 + (3)(-3)$

$a'_4 = 49 - 9$

$a'_4 = 40$

Thus, the 4th term from the end of the given AP is 40.

Alternate Method:

First, find the total number of terms ($n$) in the AP using the formula $a_n = a + (n-1)d$, where $a_n$ is the last term, 49.

$49 = -11 + (n-1)3$

$49 + 11 = 3(n-1)$

$60 = 3(n-1)$

$\frac{60}{3} = n-1$

$20 = n-1$

$n = 21$

The AP has 21 terms. The $k$-th term from the end of an AP with $n$ terms is the $(n-k+1)$-th term from the beginning.

The 4th term from the end ($k=4$) is the $(21-4+1)$-th term from the beginning.

Position from beginning = $21 - 4 + 1 = 18$.

So, we need to find the 18th term ($a_{18}$) of the original AP using $a = -11$ and $d = 3$.

$a_{18} = a + (18-1)d$

$a_{18} = -11 + (17)(3)$

$a_{18} = -11 + 51$

$a_{18} = 40$

Both methods give the same result.

Comparing this result with the given options, we find that 40 corresponds to option (B).

Final Answer:

The 4th term from the end of the AP is 40.

The correct option is (B) 40.

To Identify:

The famous mathematician associated with finding the sum of the first 100 natural numbers.

Solution:

The story of finding the sum of the first 100 natural numbers, which is $1 + 2 + 3 + ... + 100$, is famously associated with the German mathematician Carl Friedrich Gauss.

When he was a young schoolboy, his teacher asked the class to sum the numbers from 1 to 100, perhaps to keep them busy. While others added the numbers sequentially, Gauss quickly saw a pattern.

He realised that if you pair the numbers from the beginning and the end, their sum is always the same:

$1 + 100 = 101$

$2 + 99 = 101$

$3 + 98 = 101$

...and so on.

There are 100 numbers in total, so there are $\frac{100}{2} = 50$ such pairs.

The sum of each pair is 101.

Therefore, the total sum is $50 \times 101 = 5050$.

This method is a foundational concept that leads to the formula for the sum of an arithmetic series, which can be applied to the series of natural numbers $1, 2, ..., 100$ as an AP with $a=1$, $d=1$, and $n=100$. The sum formula is $S_n = \frac{n}{2}(a + a_n)$ or $S_n = \frac{n}{2}(2a + (n-1)d)$. Using the first version:

$S_{100} = \frac{100}{2}(1 + 100)$

$S_{100} = 50 \times 101$

$S_{100} = 5050$

This quick method is attributed to Carl Friedrich Gauss.

Final Answer:

The famous mathematician associated with finding the sum of the first 100 natural numbers is Gauss.

The correct option is (C) Gauss.

Given:

First term of the AP, $a = -5$.

Common difference, $d = 2$.

Number of terms, $n = 6$.

To Find:

The sum of the first 6 terms of the AP, denoted by $S_6$.

Solution:

The formula for the sum of the first $n$ terms of an Arithmetic Progression is given by:

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

Substitute the given values $a = -5$, $d = 2$, and $n = 6$ into this formula:

$S_6 = \frac{6}{2}(2 \times (-5) + (6-1) \times 2)$

Simplify the expression:

$S_6 = 3(-10 + (5) \times 2)$

$S_6 = 3(-10 + 10)$

$S_6 = 3(0)$

$S_6 = 0$

Thus, the sum of the first 6 terms of the AP is 0.

Alternatively, we can list the first 6 terms and sum them:

$a_1 = -5$

$a_2 = a_1 + d = -5 + 2 = -3$

$a_3 = a_2 + d = -3 + 2 = -1$

$a_4 = a_3 + d = -1 + 2 = 1$

$a_5 = a_4 + d = 1 + 2 = 3$

$a_6 = a_5 + d = 3 + 2 = 5$

The terms are –5, –3, –1, 1, 3, 5.

Sum of the terms: $S_6 = (-5) + (-3) + (-1) + 1 + 3 + 5$

$S_6 = -5 - 3 - 1 + 1 + 3 + 5$

$S_6 = (-5) + (-3+3) + (-1+1) + 5$

$S_6 = -5 + 0 + 0 + 5$

$S_6 = -5 + 5$

$S_6 = 0$

Both methods confirm that the sum of the first 6 terms is 0.

Comparing this result with the given options, we find that 0 corresponds to option (A).

Final Answer:

The sum of the first 6 terms is 0.

The correct option is (A) 0.

Given:

The given Arithmetic Progression (AP) is 10, 6, 2, ...

We need to find the sum of the first 16 terms of this AP.

To Find:

The sum of the first 16 terms, $S_{16}$.

Solution:

In the given AP: 10, 6, 2, ...

The first term is $a = 10$.

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

$d = a_2 - a_1 = 6 - 10 = -4$

$d = a_3 - a_2 = 2 - 6 = -4$

So, the common difference is $d = -4$.

We need to find the sum of the first 16 terms, which means $n = 16$.

The formula for the sum of the first $n$ terms of an Arithmetic Progression is given by:

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

Substitute the values $a = 10$, $d = -4$, and $n = 16$ into this formula:

$S_{16} = \frac{16}{2}(2 \times 10 + (16-1) \times (-4))$

Simplify the expression inside the parentheses:

$S_{16} = 8(20 + (15) \times (-4))$

$S_{16} = 8(20 - 60)$

$S_{16} = 8(-40)$

Perform the multiplication:

$S_{16} = -320$

Thus, the sum of the first 16 terms of the AP is -320.

Comparing this result with the given options, we find that -320 corresponds to option (A).

Final Answer:

The sum of the first 16 terms of the AP is -320.

The correct option is (A) –320.

Given:

In an Arithmetic Progression (AP), we are given:

First term, $a = 1$

The $n$-th term, $a_n = 20$

The sum of the first $n$ terms, $S_n = 399$

To Find:

The number of terms, $n$.

Solution:

The formula for the sum of the first $n$ terms of an Arithmetic Progression, when the first term ($a$) and the last term ($a_n$ or $l$) are known, is given by:

$S_n = \frac{n}{2}(a + a_n)$

Substitute the given values $a = 1$, $a_n = 20$, and $S_n = 399$ into this formula:

$399 = \frac{n}{2}(1 + 20)$

$399 = \frac{n}{2}(21)$

To solve for $n$, multiply both sides of the equation by 2:

$399 \times 2 = n \times 21$

$798 = 21n$

Now, divide both sides by 21:

$n = \frac{798}{21}$

Performing the division:

$n = 38$

So, the number of terms in the AP is 38.

We can verify this using the formula for the $n$-th term, $a_n = a + (n-1)d$. We know $a=1$, $a_{38}=20$, and $n=38$. We can find $d$ first: $20 = 1 + (38-1)d \implies 19 = 37d \implies d = \frac{19}{37}$. Then the sum $S_{38} = \frac{38}{2}(2 \times 1 + (38-1)\frac{19}{37}) = 19(2 + 37 \times \frac{19}{37}) = 19(2 + 19) = 19 \times 21 = 399$. This confirms our value of $n=38$ is correct.

Comparing this result with the given options, we find that 38 corresponds to option (C).

Final Answer:

The value of $n$ is 38.

The correct option is (C) 38.

Given:

The requirement to find the sum of the first five multiples of 3.

To Find:

The sum of the first five multiples of 3.

Solution:

The first five multiples of 3 are the numbers obtained by multiplying 3 by the first five positive integers, which are 1, 2, 3, 4, and 5.

The multiples are:

$3 \times 1 = 3$

$3 \times 2 = 6$

$3 \times 3 = 9$

$3 \times 4 = 12$

$3 \times 5 = 15$

The list of the first five multiples of 3 is 3, 6, 9, 12, 15.

This list forms an Arithmetic Progression (AP) because the difference between consecutive terms is constant.

The first term is $a = 3$.

The common difference is $d = 6 - 3 = 3$. (Also $9-6=3$, $12-9=3$, $15-12=3$).

We need to find the sum of the first 5 terms of this AP, so $n = 5$.

The formula for the sum of the first $n$ terms of an Arithmetic Progression is given by:

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

Substitute the values $a = 3$, $d = 3$, and $n = 5$ into the formula:

$S_5 = \frac{5}{2}(2 \times 3 + (5-1) \times 3)$

Simplify the expression inside the parentheses:

$S_5 = \frac{5}{2}(6 + (4) \times 3)$

$S_5 = \frac{5}{2}(6 + 12)$

$S_5 = \frac{5}{2}(18)$

Now, perform the multiplication and division:

$S_5 = 5 \times \frac{18}{2}$

$S_5 = 5 \times 9$

$S_5 = 45$

Thus, the sum of the first five multiples of 3 is 45.

Alternatively, we can directly sum the five multiples:

$S_5 = 3 + 6 + 9 + 12 + 15$

$S_5 = 9 + 9 + 12 + 15$

$S_5 = 18 + 12 + 15$

$S_5 = 30 + 15$

$S_5 = 45$

Both methods yield the same result.

Comparing this result with the given options, we find that 45 corresponds to option (A).

Final Answer:

The sum of the first five multiples of 3 is 45.

The correct option is (A) 45.

Given:

The given Arithmetic Progression (AP) is 10, 5, 0, –5, ...

The statement claims that the common difference $d$ is equal to 5.

To Justify:

Determine whether the statement "$d=5$" for the given AP is true or false.

Solution:

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by $d$.

In the given AP: 10, 5, 0, –5, ...

The terms are $a_1 = 10$, $a_2 = 5$, $a_3 = 0$, $a_4 = -5$, and so on.

To find the common difference $d$, we calculate the difference between any term and its preceding term.

Difference between the second term ($a_2$) and the first term ($a_1$):

$d = a_2 - a_1 = 5 - 10 = -5$

Let's verify the common difference using other consecutive terms:

Difference between the third term ($a_3$) and the second term ($a_2$):

$d = a_3 - a_2 = 0 - 5 = -5$

Difference between the fourth term ($a_4$) and the third term ($a_3$):

$d = a_4 - a_3 = -5 - 0 = -5$

Since the difference between consecutive terms is consistently -5, the common difference of the given AP is $d = -5$.

The given statement says that the common difference is 5.

Our calculation shows that the common difference is -5.

Therefore, the given statement is incorrect.

Conclusion:

The common difference of the AP: 10, 5, 0, –5, ... is -5, not 5.

Thus, the given statement is False.

Given:

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

Rate of compound interest, $R = 10\%$ per annum.

The sequence of amounts at the end of the first year, second year, third year, and so on.

To Justify:

Determine if the amounts at the end of consecutive years form an Arithmetic Progression (AP).

Solution:

For a sequence of numbers to form an Arithmetic Progression, the difference between any term and its preceding term must be constant. This constant difference is called the common difference.

The formula for the amount ($A_n$) after $n$ years when the interest is compounded annually is given by:

$A_n = P \left(1 + \frac{R}{100}\right)^n$

Given $P = 1000$ and $R = 10$, the formula becomes:

$A_n = 1000 \left(1 + \frac{10}{100}\right)^n = 1000 \left(1 + 0.1\right)^n = 1000 (1.1)^n$

Let's calculate the amount at the end of the first, second, and third years:

Amount at the end of the first year ($n=1$):

$A_1 = 1000 (1.1)^1 = 1000 \times 1.1 = 1100$

Amount at the end of the second year ($n=2$):

$A_2 = 1000 (1.1)^2 = 1000 \times 1.21 = 1210$

Amount at the end of the third year ($n=3$):

$A_3 = 1000 (1.1)^3 = 1000 \times 1.331 = 1331$

The sequence of amounts is 1100, 1210, 1331, ...

Now, let's find the difference between consecutive terms:

Difference between the second term and the first term:

$A_2 - A_1 = 1210 - 1100 = 110$

Difference between the third term and the second term:

$A_3 - A_2 = 1331 - 1210 = 121$

Since the difference between the first two pairs of consecutive terms is not constant ($110 \neq 121$), the amounts at the end of first year, second year, third year, and so on, do not form an Arithmetic Progression.

Conclusion:

The amounts at the end of consecutive years when interest is compounded do not have a constant difference. Therefore, they do not form an AP.

The given statement is False.

Given:

A statement claiming that the $n$-th term of an AP cannot be of the form $n^2 + 1$.

To Justify:

Verify whether the given statement is true or false by examining the properties of an AP.

Solution:

For a sequence to be an Arithmetic Progression (AP), the difference between any term and its preceding term must be a constant value. This constant difference is called the common difference, denoted by $d$.

Let's assume that the $n$-th term of a sequence is given by the formula $a_n = n^2 + 1$.

To check if this sequence is an AP, we need to find the difference between consecutive terms, i.e., $a_{n+1} - a_n$.

First, find the expression for the $(n+1)$-th term by replacing $n$ with $(n+1)$ in the formula for $a_n$:

$a_{n+1} = (n+1)^2 + 1$

Expand $(n+1)^2$ using the identity $(x+y)^2 = x^2 + 2xy + y^2$:

$a_{n+1} = (n^2 + 2n + 1) + 1$

$a_{n+1} = n^2 + 2n + 2$

Now, calculate the difference between the $(n+1)$-th term and the $n$-th term:

$a_{n+1} - a_n = (n^2 + 2n + 2) - (n^2 + 1)$

Remove the parentheses and simplify the expression:

$a_{n+1} - a_n = n^2 + 2n + 2 - n^2 - 1$

$a_{n+1} - a_n = (n^2 - n^2) + 2n + (2 - 1)$

$a_{n+1} - a_n = 0 + 2n + 1$

$a_{n+1} - a_n = 2n + 1$

The difference between consecutive terms is $2n + 1$. This difference depends on the value of $n$. For instance:

For $n=1$, the difference between the 2nd and 1st term is $2(1)+1 = 3$.

For $n=2$, the difference between the 3rd and 2nd term is $2(2)+1 = 5$.

For $n=3$, the difference between the 4th and 3rd term is $2(3)+1 = 7$.

Since the difference between consecutive terms is not constant (it varies with $n$), the sequence defined by $a_n = n^2 + 1$ does not have a common difference. Therefore, it is not an Arithmetic Progression.

The $n$-th term of an AP must be a linear expression in $n$, i.e., it must be of the form $a_n = An + B$, where $A$ and $B$ are constants ($A$ being the common difference and $B = a - d$). An expression involving $n^2$ is a quadratic expression, not linear.

Conclusion:

Since the difference between consecutive terms for a sequence with $n$-th term $a_n = n^2 + 1$ is $2n + 1$, which is not constant, the sequence is not an AP. Therefore, the statement that the $n$-th term of an AP cannot be $n^2 + 1$ is true.

The given statement is True.

A sequence forms an Arithmetic Progression (AP) if the difference between consecutive terms is constant. This constant difference is called the common difference ($d$).

(i) –1, –1, –1, –1, ...

Given: The sequence is –1, –1, –1, –1, ...

To Justify: Check if the difference between consecutive terms is constant.

Solution:

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

Difference between consecutive terms:

$a_2 - a_1 = -1 - (-1) = -1 + 1 = 0$

$a_3 - a_2 = -1 - (-1) = -1 + 1 = 0$

$a_4 - a_3 = -1 - (-1) = -1 + 1 = 0$

The difference between consecutive terms is consistently 0.

Conclusion: The given sequence forms an AP with a common difference $d = 0$.

(ii) 0, 2, 0, 2, ...

Given: The sequence is 0, 2, 0, 2, ...

To Justify: Check if the difference between consecutive terms is constant.

Solution:

Let the terms be $a_1 = 0$, $a_2 = 2$, $a_3 = 0$, $a_4 = 2$.

Difference between consecutive terms:

$a_2 - a_1 = 2 - 0 = 2$

$a_3 - a_2 = 0 - 2 = -2$

The difference between consecutive terms is not constant (2 $\neq$ -2).

Conclusion: The given sequence does not form an AP.

(iii) 1, 1, 2, 2, 3, 3,...

Given: The sequence is 1, 1, 2, 2, 3, 3,...

To Justify: Check if the difference between consecutive terms is constant.

Solution:

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

Difference between consecutive terms:

$a_2 - a_1 = 1 - 1 = 0$

$a_3 - a_2 = 2 - 1 = 1$

The difference between consecutive terms is not constant (0 $\neq$ 1).

Conclusion: The given sequence does not form an AP.

(iv) 11, 22, 33,…

Given: The sequence is 11, 22, 33,...

To Justify: Check if the difference between consecutive terms is constant.

Solution:

Let the terms be $a_1 = 11$, $a_2 = 22$, $a_3 = 33$.

Difference between consecutive terms:

$a_2 - a_1 = 22 - 11 = 11$

$a_3 - a_2 = 33 - 22 = 11$

Assuming the pattern continues, the difference between any consecutive terms will be 11.

Conclusion: The given sequence forms an AP with a common difference $d = 11$.

(v) $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{1}{4}$ ,…

Given: The sequence is $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$, ...

To Justify: Check if the difference between consecutive terms is constant.

Solution:

Let the terms be $a_1 = \frac{1}{2}$, $a_2 = \frac{1}{3}$, $a_3 = \frac{1}{4}$.

Difference between consecutive terms:

$a_2 - a_1 = \frac{1}{3} - \frac{1}{2} = \frac{2-3}{6} = -\frac{1}{6}$

$a_3 - a_2 = \frac{1}{4} - \frac{1}{3} = \frac{3-4}{12} = -\frac{1}{12}$

The difference between consecutive terms is not constant ($-\frac{1}{6} \neq -\frac{1}{12}$).

Conclusion: The given sequence does not form an AP.

(vi) 2, 2², 2³, 2⁴, …

Given: The sequence is 2, 2², 2³, 2⁴, ... which is 2, 4, 8, 16, ...

To Justify: Check if the difference between consecutive terms is constant.

Solution:

Let the terms be $a_1 = 2$, $a_2 = 4$, $a_3 = 8$, $a_4 = 16$.

Difference between consecutive terms:

$a_2 - a_1 = 4 - 2 = 2$

$a_3 - a_2 = 8 - 4 = 4$

The difference between consecutive terms is not constant (2 $\neq$ 4).

Conclusion: The given sequence does not form an AP.

(vii) $\sqrt{3}$ , $\sqrt{12}$ , $\sqrt{27}$ , $\sqrt{48}$ ,….

Given: The sequence is $\sqrt{3}$, $\sqrt{12}$, $\sqrt{27}$, $\sqrt{48}$, ...

To Justify: Check if the difference between consecutive terms is constant.

Solution:

First, simplify the terms by taking out perfect square factors from the square roots:

$a_1 = \sqrt{3}$

$a_2 = \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}$

$a_3 = \sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}$

$a_4 = \sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}$

The sequence can be written as $\sqrt{3}$, $2\sqrt{3}$, $3\sqrt{3}$, $4\sqrt{3}$, ...

Difference between consecutive terms:

$a_2 - a_1 = 2\sqrt{3} - \sqrt{3} = (2-1)\sqrt{3} = \sqrt{3}$

$a_3 - a_2 = 3\sqrt{3} - 2\sqrt{3} = (3-2)\sqrt{3} = \sqrt{3}$

$a_4 - a_3 = 4\sqrt{3} - 3\sqrt{3} = (4-3)\sqrt{3} = \sqrt{3}$

The difference between consecutive terms is consistently $\sqrt{3}$.

Conclusion: The given sequence forms an AP with a common difference $d = \sqrt{3}$.

Given:

The sequence is –1, $-\frac{3}{2}$, -2, $\frac{5}{2}$, ...

The statement claims that this sequence forms an AP because $a_2 – a_1 = a_3 – a_2$.

To Justify:

Determine if the statement is true or false based on the definition of an Arithmetic Progression (AP).

Solution:

For a sequence to be an Arithmetic Progression (AP), the difference between any term and its preceding term must be constant throughout the sequence. This constant difference is called the common difference ($d$).

The given sequence is –1, $-\frac{3}{2}$, -2, $\frac{5}{2}$, ...

Let the terms be $a_1 = -1$, $a_2 = -\frac{3}{2}$, $a_3 = -2$, $a_4 = \frac{5}{2}$.

First, let's calculate the difference between the second term and the first term ($a_2 - a_1$):

$a_2 - a_1 = -\frac{3}{2} - (-1) = -\frac{3}{2} + 1 = -\frac{3}{2} + \frac{2}{2} = -\frac{1}{2}$

Next, let's calculate the difference between the third term and the second term ($a_3 - a_2$):

$a_3 - a_2 = -2 - (-\frac{3}{2}) = -2 + \frac{3}{2} = -\frac{4}{2} + \frac{3}{2} = -\frac{1}{2}$

Indeed, we find that $a_2 - a_1 = a_3 - a_2 = -\frac{1}{2}$. This part of the statement is true based on the first three terms.

However, for the sequence to be an AP, the difference must be constant for *all* consecutive terms. Let's calculate the difference between the fourth term and the third term ($a_4 - a_3$):

$a_4 - a_3 = \frac{5}{2} - (-2) = \frac{5}{2} + 2 = \frac{5}{2} + \frac{4}{2} = \frac{9}{2}$

The difference between the third and fourth terms is $\frac{9}{2}$.

We see that $a_2 - a_1 = -\frac{1}{2}$ and $a_3 - a_2 = -\frac{1}{2}$, but $a_4 - a_3 = \frac{9}{2}$.

Since the difference between consecutive terms is not constant (e.g., $a_3 - a_2 \neq a_4 - a_3$), the sequence –1, $-\frac{3}{2}$, -2, $\frac{5}{2}$, ... does not have a common difference and therefore is not an Arithmetic Progression.

The condition $a_2 - a_1 = a_3 - a_2$ is necessary for the first three terms to be in AP, but it is not sufficient to prove that the entire sequence is an AP, especially if there are more terms given or implied by the "...". For a sequence to be an AP, the difference between any term and its preceding term must be the same value.

Conclusion:

While it is true that the difference between the first two pairs of terms is equal ($a_2 – a_1 = a_3 – a_2 = -\frac{1}{2}$), the difference between the subsequent pair of terms ($a_4 - a_3 = \frac{9}{2}$) is different. Since the difference between consecutive terms is not constant throughout the sequence, it does not form an AP.

Therefore, the statement that the sequence forms an AP based solely on $a_2 – a_1 = a_3 – a_2$ is False.

Given:

The Arithmetic Progression (AP) is –3, –7, –11, ...

To Justify:

Determine if the difference $a_{30} – a_{20}$ can be found directly without calculating $a_{30}$ and $a_{20}$ separately, and provide reasons.

Solution:

Yes, we can find the difference $a_{30} – a_{20}$ directly without calculating the values of $a_{30}$ and $a_{20}$ individually.

Reason:

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

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

where $a$ is the first term and $d$ is the common difference.

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

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

The difference between the $m$-th term and the $n$-th term ($m > n$) is:

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

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

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

$a_m - a_n = (m - 1 - n + 1)d$

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

This formula shows that the difference between any two terms of an AP depends only on the common difference ($d$) and the difference in their positions ($m-n$), not on the first term ($a$) or the specific values of $m$ and $n$.

In this question, we need to find $a_{30} - a_{20}$. Here, $m=30$ and $n=20$.

So, $a_{30} - a_{20} = (30 - 20)d = 10d$.

To find the value of $a_{30} - a_{20}$, we only need to know the common difference $d$ of the given AP.

From the given AP: –3, –7, –11, ...

The first term is $a = -3$.

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

$d = -7 - (-3) = -7 + 3 = -4$

Now, we can calculate the difference $a_{30} - a_{20}$ using the formula $10d$:

$a_{30} - a_{20} = 10 \times (-4)$

$a_{30} - a_{20} = -40$

Thus, the difference between the 30th term and the 20th term is -40, and we were able to find this value directly using the common difference, without calculating $a_{30}$ and $a_{20}$ separately.

Conclusion:

Yes, it is true that we can find $a_{30} – a_{20}$ directly without finding $a_{30}$ and $a_{20}$ separately because the difference between the $m$-th and $n$-th terms of an AP is simply $(m-n)d$.

$a_{30} - a_{20} = (30-20)d = 10d = 10 \times (-4) = -40$.

Two Arithmetic Progressions (APs) with the same common difference, $d$.

First term of the first AP, $a_1 = 2$.

First term of the second AP, $a'_1 = 7$.

Why?

The reason the difference between any two corresponding terms of these two APs is the same is because the terms involving the common difference cancel out when calculating the difference between corresponding terms, leaving only the difference between the first terms.

Justification:

Let the common difference of both APs be $d$.

Let the first term of the first AP be $a = 2$.

Let the first term of the second AP be $a' = 7$.

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

For the first AP, the $n$-th term is:

$a_n = 2 + (n-1)d$

For the second AP, the $n$-th term is:

$a'_n = 7 + (n-1)d$

Now, let's find the difference between their corresponding $n$-th terms, $a'_n - a_n$:

$a'_n - a_n = [7 + (n-1)d] - [2 + (n-1)d]$

Remove the parentheses:

$a'_n - a_n = 7 + (n-1)d - 2 - (n-1)d$

Combine the terms:

$a'_n - a_n = (7 - 2) + [(n-1)d - (n-1)d]$

$a'_n - a_n = 5 + 0$

$a'_n - a_n = 5$

The difference between the $n$-th term of the second AP and the $n$-th term of the first AP is a constant value, 5. This difference is independent of $n$, the term number.

Therefore, the difference between their 10th terms (when $n=10$) will be $a'_{10} - a_{10} = 5$.

The difference between their 21st terms (when $n=21$) will be $a'_{21} - a_{21} = 5$.

The difference between any two corresponding terms (for any value of $n$) will be 5.

This constant difference (5) is simply the difference between the first terms of the two APs ($a'_1 - a_1 = 7 - 2 = 5$).

Conclusion:

The difference between any two corresponding terms of the two APs is constant because the component depending on the term number ($n$) and the common difference ($d$) is identical for both APs and thus cancels out when the difference is calculated, leaving only the difference between their first terms.

Given:

The given Arithmetic Progression (AP) is 31, 28, 25, ...

We need to determine if 0 is a term in this AP.

To Justify:

Determine whether there exists a positive integer $n$ such that the $n$-th term of the AP is 0.

Solution:

In the given AP: 31, 28, 25, ...

The first term is $a = 31$.

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

$d = a_2 - a_1 = 28 - 31 = -3$

$d = a_3 - a_2 = 25 - 28 = -3$

The common difference is $d = -3$.

Let's assume that 0 is the $n$-th term of this AP for some positive integer $n$.

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

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

Set $a_n = 0$ and substitute the values $a = 31$ and $d = -3$ into the formula:

$0 = 31 + (n-1)(-3)$

Now, we solve this equation for $n$.

Subtract 31 from both sides:

$-31 = (n-1)(-3)$

Divide both sides by -3:

$\frac{-31}{-3} = n-1$

$\frac{31}{3} = n-1$

Add 1 to both sides to isolate $n$:

$n = \frac{31}{3} + 1$

$n = \frac{31}{3} + \frac{3}{3}$

$n = \frac{31 + 3}{3}$

$n = \frac{34}{3}$

For a number to be a term in an AP, its term number ($n$) must be a positive integer (1, 2, 3, ...). In this case, we found $n = \frac{34}{3}$.

Since $\frac{34}{3}$ is not an integer, 0 is not a term in the given Arithmetic Progression.

Conclusion:

When we assume that 0 is the $n$-th term and solve for $n$, we get $n = \frac{34}{3}$. Since the term number must be a positive integer, and $\frac{34}{3}$ is a fraction, 0 is not a term of the AP: 31, 28, 25, ...

Thus, it is False to say that 0 is a term of the given AP.

Given:

Taxi fare structure: Rs 15 for the first km and Rs 8 for each additional km.

Statement: The total fare after each km is 15, 8, 8, 8, ... and this does not form an AP.

To Justify:

Determine if the statement is true or false by calculating the actual sequence of total fares and checking if it forms an AP.

Solution:

We need to calculate the total taxi fare after each successive kilometer to form the sequence of "total fare after each km".

Total fare after 1 km (1st term):

$a_1 = \textsf{₹}15$ (Fare for the first km)

Total fare after 2 km (2nd term):

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

Total fare after 3 km (3rd term):

$a_3 = \textsf{₹}23 \text{ (for 2 km)} + \textsf{₹}8 \text{ (for 3rd km)} = \textsf{₹}31$

Total fare after 4 km (4th term):

$a_4 = \textsf{₹}31 \text{ (for 3 km)} + \textsf{₹}8 \text{ (for 4th km)} = \textsf{₹}39$

The sequence representing the total taxi fare after each kilometer is 15, 23, 31, 39, ...

Now, let's check if this sequence forms an Arithmetic Progression (AP) by finding the difference between consecutive terms:

Difference between the second term and the first term:

$a_2 - a_1 = 23 - 15 = 8$

Difference between the third term and the second term:

$a_3 - a_2 = 31 - 23 = 8$

Difference between the fourth term and the third term:

$a_4 - a_3 = 39 - 31 = 8$

The difference between consecutive terms is constant, and the common difference is $d = 8$.

Therefore, the sequence of total taxi fares after each kilometer, which is 15, 23, 31, 39, ..., forms an AP with the first term $a=15$ and common difference $d=8$.

The statement claims that the sequence of total fares is 15, 8, 8, 8, ... This sequence is incorrect. The correct sequence of total fares is 15, 23, 31, 39, ...

The statement further claims that the sequence 15, 8, 8, 8, ... does not form an AP. While it is true that the sequence 15, 8, 8, 8, ... does not form an AP (as $8-15 = -7$ and $8-8=0$), the premise that this is the sequence of total fares is false.

Conclusion:

The actual sequence of total taxi fares after each km is 15, 23, 31, 39, ..., which forms an AP with a common difference of 8. The sequence presented in the statement (15, 8, 8, 8, ...) is not the correct sequence of total fares.

Therefore, the statement is False.

For a list of numbers to form an Arithmetic Progression (AP), the difference between any term and its preceding term must be constant. This constant difference is called the common difference.

(i) The fee charged from a student every month by a school for the whole session, when the monthly fee is Rs 400.

List of numbers: The fee charged each month is a constant amount.

Fee in month 1 = $\textsf{₹}400$

Fee in month 2 = $\textsf{₹}400$

Fee in month 3 = $\textsf{₹}400$

... and so on.

The list of fees is 400, 400, 400, ...

Check for AP:

Difference between 2nd term and 1st term = $400 - 400 = 0$

Difference between 3rd term and 2nd term = $400 - 400 = 0$

The difference between consecutive terms is constant (0).

Conclusion: Yes, this situation forms an AP with the first term $a = 400$ and common difference $d = 0$.

(ii) The fee charged every month by a school from Classes I to XII, when the monthly fee for Class I is Rs 250, and it increases by Rs 50 for the next higher class.

List of numbers: The fee for each class level.

Fee for Class I = $\textsf{₹}250$ ($a_1$)

Fee for Class II = Fee for Class I + Increase = $\textsf{₹}250 + \textsf{₹}50 = \textsf{₹}300$ ($a_2$)

Fee for Class III = Fee for Class II + Increase = $\textsf{₹}300 + \textsf{₹}50 = \textsf{₹}350$ ($a_3$)

Fee for Class IV = Fee for Class III + Increase = $\textsf{₹}350 + \textsf{₹}50 = \textsf{₹}400$ ($a_4$)

... and so on, up to Class XII.

The list of fees is 250, 300, 350, 400, ...

Check for AP:

Difference between 2nd term and 1st term = $300 - 250 = 50$

Difference between 3rd term and 2nd term = $350 - 300 = 50$

Difference between 4th term and 3rd term = $400 - 350 = 50$

The difference between consecutive terms is constant (50).

Conclusion: Yes, this situation forms an AP with the first term $a = 250$ and common difference $d = 50$.

(iii) The amount of money in the account of Varun at the end of every year when Rs 1000 is deposited at simple interest of 10% per annum.

List of numbers: The total amount at the end of each year.

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

Annual simple interest = $\frac{P \times R \times 1}{100} = \frac{1000 \times 10 \times 1}{100} = \textsf{₹}100$.

Amount at the end of Year 1 = Principal + Interest for Year 1 = $1000 + 100 = \textsf{₹}1100$ ($A_1$)

Amount at the end of Year 2 = Amount at Year 1 + Interest for Year 2 = $1100 + 100 = \textsf{₹}1200$ ($A_2$)

Amount at the end of Year 3 = Amount at Year 2 + Interest for Year 3 = $1200 + 100 = \textsf{₹}1300$ ($A_3$)

Amount at the end of Year 4 = Amount at Year 3 + Interest for Year 4 = $1300 + 100 = \textsf{₹}1400$ ($A_4$)

... and so on.

The list of amounts is 1100, 1200, 1300, 1400, ...

Check for AP:

Difference between 2nd term and 1st term = $1200 - 1100 = 100$

Difference between 3rd term and 2nd term = $1300 - 1200 = 100$

Difference between 4th term and 3rd term = $1400 - 1300 = 100$

The difference between consecutive terms is constant (100).

Conclusion: Yes, this situation forms an AP with the first term $a = 1100$ and common difference $d = 100$.

(iv) The number of bacteria in a certain food item after each second, when they double in every second.

List of numbers: The number of bacteria after each second. Let the initial number of bacteria (at time = 0) be $N_0$.

Number of bacteria after 1st second = $N_0 \times 2 = 2N_0$ ($B_1$)

Number of bacteria after 2nd second = $2N_0 \times 2 = 4N_0$ ($B_2$)

Number of bacteria after 3rd second = $4N_0 \times 2 = 8N_0$ ($B_3$)

Number of bacteria after 4th second = $8N_0 \times 2 = 16N_0$ ($B_4$)

... and so on.

The list of numbers of bacteria is $2N_0, 4N_0, 8N_0, 16N_0, ...$ (assuming $N_0 > 0$)

Check for AP:

Difference between 2nd term and 1st term = $4N_0 - 2N_0 = 2N_0$

Difference between 3rd term and 2nd term = $8N_0 - 4N_0 = 4N_0$

Difference between 4th term and 3rd term = $16N_0 - 8N_0 = 8N_0$

The difference between consecutive terms is not constant ($2N_0 \neq 4N_0 \neq 8N_0$, assuming $N_0 \neq 0$). This sequence grows by a constant multiplicative factor (2), which means it is a Geometric Progression, not an Arithmetic Progression.

Conclusion: No, this situation does not form an AP.

For an expression to be the $n$-th term of an Arithmetic Progression (AP), the difference between any term and its preceding term must be a constant value. This constant difference is the common difference ($d$). If the $n$-th term is given by $a_n$, then the sequence is an AP if and only if $a_{n+1} - a_n$ is a constant for all positive integers $n$. The general form of the $n$-th term of an AP is a linear expression in $n$, specifically $a_n = An + B$, where $A$ and $B$ are constants.

(i) $a_n = 2n – 3$

Given: The $n$-th term is given by $a_n = 2n - 3$.

To Justify: Check if the difference $a_{n+1} - a_n$ is constant.

Solution:

Find the $(n+1)$-th term by substituting $(n+1)$ for $n$ in the expression for $a_n$:

$a_{n+1} = 2(n+1) - 3 = 2n + 2 - 3 = 2n - 1$

Now, find the difference between the $(n+1)$-th term and the $n$-th term:

$a_{n+1} - a_n = (2n - 1) - (2n - 3)$

$a_{n+1} - a_n = 2n - 1 - 2n + 3$

$a_{n+1} - a_n = (2n - 2n) + (-1 + 3)$

$a_{n+1} - a_n = 0 + 2 = 2$

The difference $a_{n+1} - a_n$ is 2, which is a constant value. This constant value is the common difference $d$.

Also, the expression $a_n = 2n - 3$ is a linear expression in $n$ (of the form $An+B$ with $A=2$ and $B=-3$).

Conclusion: Yes, $a_n = 2n - 3$ can be the $n$-th term of an AP. The statement is True.

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

Given: The $n$-th term is given by $a_n = 3n^2 + 5$.

To Justify: Check if the difference $a_{n+1} - a_n$ is constant.

Solution:

Find the $(n+1)$-th term:

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

$a_{n+1} = 3n^2 + 6n + 3 + 5 = 3n^2 + 6n + 8$

Now, find the difference between the $(n+1)$-th term and the $n$-th term:

$a_{n+1} - a_n = (3n^2 + 6n + 8) - (3n^2 + 5)$

$a_{n+1} - a_n = 3n^2 + 6n + 8 - 3n^2 - 5$

$a_{n+1} - a_n = (3n^2 - 3n^2) + 6n + (8 - 5)$

$a_{n+1} - a_n = 0 + 6n + 3 = 6n + 3$

The difference $a_{n+1} - a_n = 6n + 3$. This expression depends on $n$, so it is not a constant value. For example, when $n=1$, the difference is $6(1)+3=9$; when $n=2$, the difference is $6(2)+3=15$.

Also, the expression $a_n = 3n^2 + 5$ is a quadratic expression in $n$, not a linear one.

Conclusion: No, $a_n = 3n^2 + 5$ cannot be the $n$-th term of an AP because the difference between consecutive terms is not constant. The statement is True (that it cannot be the nth term of an AP).

(iii) $a_n = 1 + n + n^2$

Given: The $n$-th term is given by $a_n = 1 + n + n^2$.

To Justify: Check if the difference $a_{n+1} - a_n$ is constant.

Solution:

Find the $(n+1)$-th term:

$a_{n+1} = 1 + (n+1) + (n+1)^2$

$a_{n+1} = 1 + n + 1 + (n^2 + 2n + 1)$

$a_{n+1} = n^2 + 3n + 3$

Now, find the difference between the $(n+1)$-th term and the $n$-th term:

$a_{n+1} - a_n = (n^2 + 3n + 3) - (1 + n + n^2)$

$a_{n+1} - a_n = n^2 + 3n + 3 - 1 - n - n^2$

$a_{n+1} - a_n = (n^2 - n^2) + (3n - n) + (3 - 1)$

$a_{n+1} - a_n = 0 + 2n + 2 = 2n + 2$

The difference $a_{n+1} - a_n = 2n + 2$. This expression depends on $n$, so it is not a constant value. For example, when $n=1$, the difference is $2(1)+2=4$; when $n=2$, the difference is $2(2)+2=6$.

Also, the expression $a_n = 1 + n + n^2$ is a quadratic expression in $n$, not a linear one.

Conclusion: No, $a_n = 1 + n + n^2$ cannot be the $n$-th term of an AP because the difference between consecutive terms is not constant. The statement is True (that it cannot be the nth term of an AP).

Given:

The three numbers $n – 2$, $4n – 1$, and $5n + 2$ are in Arithmetic Progression (AP).

To Find:

The value of $n$.

Solution:

If three numbers, say $a$, $b$, and $c$, are in AP, then the difference between consecutive terms is constant. This means the difference between the second term and the first term is equal to the difference between the third term and the second term.

Mathematically, if $a_1, a_2, a_3$ are in AP, then $a_2 - a_1 = a_3 - a_2$.

In this problem, the given terms are:

First term, $a_1 = n - 2$

Second term, $a_2 = 4n - 1$

Third term, $a_3 = 5n + 2$

Since these terms are in AP, the common difference must be the same between consecutive terms. Therefore, we have:

Substitute the given expressions for the terms into Equation (i):

$(4n - 1) - (n - 2) = (5n + 2) - (4n - 1)$

Remove the parentheses, remembering to change the signs for the terms being subtracted:

$4n - 1 - n + 2 = 5n + 2 - 4n + 1$

Combine like terms on both sides of the equation:

$(4n - n) + (-1 + 2) = (5n - 4n) + (2 + 1)$

$3n + 1 = n + 3$

Now, we need to solve this linear equation for $n$. Subtract $n$ from both sides:

$3n - n + 1 = 3$

$2n + 1 = 3$

Subtract 1 from both sides:

$2n = 3 - 1$

$2n = 2$

Divide both sides by 2:

$n = \frac{2}{2}$

$n = 1$

So, the value of $n$ is 1.

Let's check the terms with $n=1$:

$a_1 = 1 - 2 = -1$

$a_2 = 4(1) - 1 = 4 - 1 = 3$

$a_3 = 5(1) + 2 = 5 + 2 = 7$

The sequence is -1, 3, 7. Let's check the differences:

$a_2 - a_1 = 3 - (-1) = 3 + 1 = 4$

$a_3 - a_2 = 7 - 3 = 4$

Since the difference is constant (4), the terms are indeed in AP when $n=1$.

Final Answer:

The value of $n$ is 1.

Given:

The given Arithmetic Progression (AP) is –11, –7, –3,..., 49.

To Find:

The value of the middle most term(s) of the AP.

Solution:

In the given AP: –11, –7, –3,..., 49.

The first term is $a = -11$.

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

$d = -7 - (-11) = -7 + 11 = 4$

$d = -3 - (-7) = -3 + 7 = 4$

So, the common difference is $d = 4$.

The last term is $a_n = 49$.

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

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

Substitute the values $a_n = 49$, $a = -11$, and $d = 4$ to find the number of terms, $n$:

$49 = -11 + (n-1)4$

Add 11 to both sides:

$49 + 11 = (n-1)4$

$60 = (n-1)4$

Divide both sides by 4:

$\frac{60}{4} = n-1$

$15 = n-1$

Add 1 to both sides:

$n = 15 + 1 = 16$

The number of terms in the AP is $n=16$. Since $n$ is an even number, there are two middle terms.

The positions of the middle terms are $\frac{n}{2}$ and $\frac{n}{2} + 1$.

The first middle term is the $\frac{16}{2} = 8$-th term.

The second middle term is the $\frac{16}{2} + 1 = 8 + 1 = 9$-th term.

Calculate the 8th term ($a_8$) using the formula $a_n = a + (n-1)d$:

$a_8 = a + (8-1)d = a + 7d$

$a_8 = -11 + 7(4)$

$a_8 = -11 + 28$

$a_8 = 17$

Calculate the 9th term ($a_9$) using the formula $a_n = a + (n-1)d$:

$a_9 = a + (9-1)d = a + 8d$

$a_9 = -11 + 8(4)$

$a_9 = -11 + 32$

$a_9 = 21$

The middle terms of the AP are 17 and 21.

Final Answer:

The middle most terms of the AP are 17 and 21.

Given:

The sum of the first three terms of an AP is 33.

The product of the first and the third term exceeds the second term by 29.

To Find:

The Arithmetic Progression (AP).

Solution:

Let the first three terms of the AP be represented as $a-d$, $a$, and $a+d$, where $a$ is the second term and $d$ is the common difference. Choosing these terms simplifies calculations involving their sum.

According to the first condition, the sum of the first three terms is 33:

$(a-d) + a + (a+d) = 33$

Combine the terms:

$a - d + a + a + d = 33$

$3a = 33$

Divide both sides by 3:

From Equation (i), we find the value of $a$:

$a = 11$

So, the second term of the AP is 11.

According to the second condition, the product of the first term and the third term exceeds the second term by 29. This means the product of the first and third term is equal to the second term plus 29.

$(a-d)(a+d) = a + 29$

Using the algebraic identity $(x-y)(x+y) = x^2 - y^2$, we can simplify the left side:

$a^2 - d^2 = a + 29$

Now, substitute the value of $a = 11$ (obtained from Equation (i)) into Equation (ii):

$11^2 - d^2 = 11 + 29$

Calculate the square of 11 and the sum on the right side:

$121 - d^2 = 40$

Rearrange the equation to solve for $d^2$. Subtract $d^2$ from both sides and subtract 40 from both sides:

$121 - 40 = d^2$

$81 = d^2$

Take the square root of both sides to find the value(s) of $d$:

$d = \pm \sqrt{81}$

$d = \pm 9$

We have two possible values for the common difference, $d=9$ and $d=-9$. Each value corresponds to a possible AP that satisfies the given conditions.

Case 1: $a = 11$ and $d = 9$

The first three terms are:

First term ($a-d$) = $11 - 9 = 2$

Second term ($a$) = $11$

Third term ($a+d$) = $11 + 9 = 20$

The AP is 2, 11, 20, ... (with first term 2 and common difference 9).

Case 2: $a = 11$ and $d = -9$

The first three terms are:

First term ($a-d$) = $11 - (-9) = 11 + 9 = 20$

Second term ($a$) = $11$

Third term ($a+d$) = $11 + (-9) = 11 - 9 = 2$

The AP is 20, 11, 2, ... (with first term 20 and common difference -9).

Both APs satisfy the given conditions (sum of first three terms is $2+11+20 = 33$ and $20+11+2=33$; product of first and third minus second term is $(2)(20) - 11 = 40 - 11 = 29$ and $(20)(2) - 11 = 40 - 11 = 29$).

Final Answer:

The Arithmetic Progression can be 2, 11, 20, ... or 20, 11, 2, ...

We need to find the common difference for each description in Column A and match it with the corresponding value in Column B.

The common difference ($d$) of an AP is the difference between any term and its preceding term, i.e., $d = a_n - a_{n-1}$. For an AP with first term $a$ and common difference $d$, the $n$-th term is given by $a_n = a + (n-1)d$.

(A₁) 2, – 2, – 6, –10,...

This is a sequence of terms. The first term $a_1 = 2$. The second term $a_2 = -2$.

The common difference $d = a_2 - a_1 = -2 - 2 = -4$.

(A₂) a = –18, n = 10, a_n = 0

We are given the first term $a = -18$, the number of terms $n = 10$, and the $n$-th term $a_{10} = 0$.

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

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

$0 = -18 + 9d$

Add 18 to both sides:

$18 = 9d$

Divide by 9:

$d = \frac{18}{9} = 2$

(A₃) a = 0, a₁₀ = 6

We are given the first term $a = 0$ and the 10th term $a_{10} = 6$.

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

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

$6 = 0 + 9d$

$6 = 9d$

Divide by 9:

$d = \frac{6}{9} = \frac{2}{3}$

(A₄) a₂ = 13, a₄ =3

We are given the 2nd term $a_2 = 13$ and the 4th term $a_4 = 3$.

Using the property that the difference between the $m$-th term and the $n$-th term is $(m-n)d$, we have:

$a_4 - a_2 = (4-2)d$

$3 - 13 = 2d$

$-10 = 2d$

Divide by 2:

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

Now, let's match the calculated common differences with the options in Column B:

(A₁) has $d = -4$, which matches (B₄).

(A₂) has $d = 2$, which matches (B₅).

(A₃) has $d = \frac{2}{3}$, which matches (B₁).

(A₄) has $d = -5$, which matches (B₂).

Matches:

(A₁) $\leftrightarrow$ (B₄)

(A₂) $\leftrightarrow$ (B₅)

(A₃) $\leftrightarrow$ (B₁)

(A₄) $\leftrightarrow$ (B₂)

For a sequence to be an Arithmetic Progression (AP), the difference between any term and its preceding term must be a constant value. This constant value is the common difference ($d$). If the sequence is an AP with first term $a_1$ and common difference $d$, the terms are $a_1, a_1+d, a_1+2d, a_1+3d, ...$ The $n$-th term is $a_n = a_1 + (n-1)d$.

(i) 0 , $\frac{1}{4}$ , $\frac{1}{2}$ , $\frac{3}{4}$ , …

Given: The sequence is 0, $\frac{1}{4}$, $\frac{1}{2}$, $\frac{3}{4}$, ...

Verification:

Calculate the difference between consecutive terms:

$a_2 - a_1 = \frac{1}{4} - 0 = \frac{1}{4}$

$a_3 - a_2 = \frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}$

$a_4 - a_3 = \frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4}$

The difference between consecutive terms is constant, $d = \frac{1}{4}$.

Conclusion: The sequence is an AP with $a_1 = 0$ and $d = \frac{1}{4}$.

Next three terms: The given terms are $a_1, a_2, a_3, a_4$. We need $a_5, a_6, a_7$.

$a_5 = a_4 + d = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1$

$a_6 = a_5 + d = 1 + \frac{1}{4} = \frac{5}{4}$

$a_7 = a_6 + d = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2}$

The next three terms are 1, $\frac{5}{4}$, $\frac{3}{2}$.

(ii) 5 , $\frac{14}{3}$ , $\frac{13}{3}$ , 4 , ...

Given: The sequence is 5, $\frac{14}{3}$, $\frac{13}{3}$, 4, ...

Verification:

Calculate the difference between consecutive terms:

$a_2 - a_1 = \frac{14}{3} - 5 = \frac{14}{3} - \frac{15}{3} = -\frac{1}{3}$

$a_3 - a_2 = \frac{13}{3} - \frac{14}{3} = -\frac{1}{3}$

$a_4 - a_3 = 4 - \frac{13}{3} = \frac{12}{3} - \frac{13}{3} = -\frac{1}{3}$

The difference between consecutive terms is constant, $d = -\frac{1}{3}$.

Conclusion: The sequence is an AP with $a_1 = 5$ and $d = -\frac{1}{3}$.

Next three terms: The given terms are $a_1, a_2, a_3, a_4$. We need $a_5, a_6, a_7$.

$a_5 = a_4 + d = 4 + (-\frac{1}{3}) = \frac{12}{3} - \frac{1}{3} = \frac{11}{3}$

$a_6 = a_5 + d = \frac{11}{3} + (-\frac{1}{3}) = \frac{10}{3}$

$a_7 = a_6 + d = \frac{10}{3} + (-\frac{1}{3}) = \frac{9}{3} = 3$

The next three terms are $\frac{11}{3}$, $\frac{10}{3}$, 3.

(iii) $\sqrt{3}$ , 2$\sqrt{3}$ , 3$\sqrt{3}$ , …

Given: The sequence is $\sqrt{3}$, $2\sqrt{3}$, $3\sqrt{3}$, ...

Verification:

Calculate the difference between consecutive terms:

$a_2 - a_1 = 2\sqrt{3} - \sqrt{3} = (2-1)\sqrt{3} = \sqrt{3}$

$a_3 - a_2 = 3\sqrt{3} - 2\sqrt{3} = (3-2)\sqrt{3} = \sqrt{3}$

The difference between consecutive terms is constant, $d = \sqrt{3}$. Assuming the pattern continues.

Conclusion: The sequence is an AP with $a_1 = \sqrt{3}$ and $d = \sqrt{3}$.

Next three terms: The given terms are $a_1, a_2, a_3$. We need $a_4, a_5, a_6$.

$a_4 = a_3 + d = 3\sqrt{3} + \sqrt{3} = 4\sqrt{3}$

$a_5 = a_4 + d = 4\sqrt{3} + \sqrt{3} = 5\sqrt{3}$

$a_6 = a_5 + d = 5\sqrt{3} + \sqrt{3} = 6\sqrt{3}$

The next three terms are $4\sqrt{3}$, $5\sqrt{3}$, $6\sqrt{3}$.

(iv) a + b, (a + 1) + b, (a + 1) + (b + 1), ...

Given: The sequence is $a+b$, $(a+1)+b$, $(a+1)+(b+1)$, ...

Simplify the terms:

$a_1 = a+b$

$a_2 = a+1+b$

$a_3 = a+1+b+1 = a+b+2$

The sequence is $a+b$, $a+b+1$, $a+b+2$, ...

Verification:

Calculate the difference between consecutive terms:

$a_2 - a_1 = (a+b+1) - (a+b) = a+b+1 - a - b = 1$

$a_3 - a_2 = (a+b+2) - (a+b+1) = a+b+2 - a - b - 1 = 1$

The difference between consecutive terms is constant, $d = 1$. Assuming 'a' and 'b' are constants.

Conclusion: The sequence is an AP with $a_1 = a+b$ and $d = 1$.

Next three terms: The given terms are $a_1, a_2, a_3$. We need $a_4, a_5, a_6$.

$a_4 = a_3 + d = (a+b+2) + 1 = a+b+3$

$a_5 = a_4 + d = (a+b+3) + 1 = a+b+4$

$a_6 = a_5 + d = (a+b+4) + 1 = a+b+5$

The next three terms are $a+b+3$, $a+b+4$, $a+b+5$.

(v) a , 2a + 1 , 3a + 2 , 4a + 3 ,...

Given: The sequence is $a$, $2a+1$, $3a+2$, $4a+3$, ...

Verification:

Calculate the difference between consecutive terms:

$a_2 - a_1 = (2a+1) - a = 2a + 1 - a = a + 1$

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

$a_4 - a_3 = (4a+3) - (3a+2) = 4a + 3 - 3a - 2 = a + 1$

The difference between consecutive terms is constant, $d = a+1$. Assuming 'a' is a constant.

Conclusion: The sequence is an AP with $a_1 = a$ and common difference $d = a+1$.

Next three terms: The given terms are $a_1, a_2, a_3, a_4$. We need $a_5, a_6, a_7$.

$a_5 = a_4 + d = (4a+3) + (a+1) = 5a+4$

$a_6 = a_5 + d = (5a+4) + (a+1) = 6a+5$

$a_7 = a_6 + d = (6a+5) + (a+1) = 7a+6$

The next three terms are $5a+4$, $6a+5$, $7a+6$.

To find the first three terms of an AP given the first term ($a$) and the common difference ($d$), we use the definitions:

First term: $a_1 = a$

Second term: $a_2 = a_1 + d = a + d$

Third term: $a_3 = a_2 + d = a + 2d$

(i) a = $\frac{1}{2}$ , d = $-\frac{1}{6}$

Given: $a = \frac{1}{2}$, $d = -\frac{1}{6}$

To Find: The first three terms ($a_1, a_2, a_3$).

Solution:

$a_1 = a = \frac{1}{2}$

$a_2 = a + d = \frac{1}{2} + (-\frac{1}{6}) = \frac{1}{2} - \frac{1}{6}$

To subtract the fractions, find a common denominator, which is 6.

$a_2 = \frac{1 \times 3}{2 \times 3} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{3-1}{6} = \frac{2}{6} = \frac{1}{3}$

$a_3 = a + 2d = \frac{1}{2} + 2(-\frac{1}{6}) = \frac{1}{2} - \frac{2}{6}$

Simplify the second term: $\frac{2}{6} = \frac{1}{3}$.

$a_3 = \frac{1}{2} - \frac{1}{3}$

To subtract the fractions, find a common denominator, which is 6.

$a_3 = \frac{1 \times 3}{2 \times 3} - \frac{1 \times 2}{3 \times 2} = \frac{3}{6} - \frac{2}{6} = \frac{3-2}{6} = \frac{1}{6}$

The first three terms are $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{6}$.

(ii) a = –5 , d = –3

Given: $a = -5$, $d = -3$

To Find: The first three terms ($a_1, a_2, a_3$).

Solution:

$a_1 = a = -5$

$a_2 = a + d = -5 + (-3) = -5 - 3 = -8$

$a_3 = a + 2d = -5 + 2(-3) = -5 - 6 = -11$

The first three terms are -5, -8, -11.

(iii) a = $\sqrt{2}$ , d = $\frac{1}{\sqrt{2}}$

Given: $a = \sqrt{2}$, $d = \frac{1}{\sqrt{2}}$

To Find: The first three terms ($a_1, a_2, a_3$).

Solution:

$a_1 = a = \sqrt{2}$

$a_2 = a + d = \sqrt{2} + \frac{1}{\sqrt{2}}$

To add the terms, find a common denominator, which is $\sqrt{2}$.

$a_2 = \frac{\sqrt{2} \times \sqrt{2}}{1 \times \sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2+1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$

This can also be rationalized: $a_2 = \frac{3}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$.

$a_3 = a + 2d = \sqrt{2} + 2(\frac{1}{\sqrt{2}}) = \sqrt{2} + \frac{2}{\sqrt{2}}$

Simplify the second term: $\frac{2}{\sqrt{2}} = \frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}} = \sqrt{2}$.

$a_3 = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$

Alternatively, using $a_3 = a_2 + d$:

$a_3 = \frac{3}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{3+1}{\sqrt{2}} = \frac{4}{\sqrt{2}}$

Rationalize: $a_3 = \frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.

The first three terms are $\sqrt{2}$, $\frac{3}{\sqrt{2}}$ (or $\frac{3\sqrt{2}}{2}$), $2\sqrt{2}$.

Given:

The numbers a, 7, b, 23, and c are in Arithmetic Progression (AP).

To Find:

The values of a, b, and c.

Solution:

If a sequence of numbers is in an Arithmetic Progression, the difference between any term and its preceding term is constant. This constant difference is called the common difference, denoted by $d$.

Let the given terms be $a_1, a_2, a_3, a_4, a_5$.

$a_1 = a$

$a_2 = 7$

$a_3 = b$

$a_4 = 23$

$a_5 = c$

Since these terms form an AP, the common difference $d$ must be the same between consecutive terms:

$a_2 - a_1 = d \implies 7 - a = d$

$a_3 - a_2 = d \implies b - 7 = d$

$a_4 - a_3 = d \implies 23 - b = d$

$a_5 - a_4 = d \implies c - 23 = d$

From the definition of an AP, for any three consecutive terms $x, y, z$, we have $y - x = z - y$, which implies $2y = x + z$, or $y = \frac{x+z}{2}$. This means the middle term is the arithmetic mean of the terms surrounding it.

Considering the terms $a_2, a_3, a_4$, which are 7, b, and 23, since they are consecutive terms in the AP, the middle term b must be the average of 7 and 23.

$a_3 = \frac{a_2 + a_4}{2}$

$b = \frac{7 + 23}{2}$

$b = \frac{30}{2}$

Now that we have the value of b, we can find the common difference $d$ using the differences involving b:

$d = a_3 - a_2 = b - 7 = 15 - 7 = 8$

Alternatively, using $a_4 - a_3 = 23 - b$:

$d = 23 - 15 = 8$

The common difference of the AP is $d = 8$.

Now we can find the values of a and c using the common difference.

Using $a_2 - a_1 = d$:

$7 - a = 8$

Subtract 7 from both sides:

$-a = 8 - 7$

$-a = 1$

Multiply by -1:

Using $a_5 - a_4 = d$:

$c - 23 = 8$

Add 23 to both sides:

From Equation (iii):

c = 31

So, the values are a = -1, b = 15, and c = 31. The AP is -1, 7, 15, 23, 31.

Final Answer:

The values are a = -1, b = 15, and c = 31.

Given:

In an Arithmetic Progression (AP):

The fifth term ($a_5$) is 19.

The difference of the eighth term from the thirteenth term is 20, which means $a_{13} - a_8 = 20$.

To Find:

Determine the Arithmetic Progression (AP). This means finding the first term ($a$) and the common difference ($d$).

Solution:

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

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

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

Using the first given condition, the fifth term ($a_5$) is 19. For $n=5$, we have:

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

$19 = a + 4d$

Using the second given condition, the difference between the thirteenth term ($a_{13}$) and the eighth term ($a_8$) is 20.

$a_{13} - a_8 = 20$

First, express $a_{13}$ and $a_8$ using the formula for the $n$-th term:

$a_{13} = a + (13-1)d = a + 12d$

$a_8 = a + (8-1)d = a + 7d$

Substitute these expressions into the second condition:

$(a + 12d) - (a + 7d) = 20$

Simplify the equation:

$a + 12d - a - 7d = 20$

$(a - a) + (12d - 7d) = 20$

$0 + 5d = 20$

From Equation (2), solve for $d$:

$d = \frac{20}{5}$

$d = 4$

Now that we have the common difference $d = 4$, substitute this value into Equation (1) to find the first term $a$:

$a + 4(4) = 19$

$a + 16 = 19$

Subtract 16 from both sides:

$a = 19 - 16$

$a = 3$

The first term of the AP is $a = 3$ and the common difference is $d = 4$.

The AP is formed by starting with the first term and repeatedly adding the common difference. The terms are $a, a+d, a+2d, a+3d, ...$

The first few terms are:

$a_1 = a = 3$

$a_2 = a + d = 3 + 4 = 7$

$a_3 = a + 2d = 3 + 2(4) = 3 + 8 = 11$

$a_4 = a + 3d = 3 + 3(4) = 3 + 12 = 15$

$a_5 = a + 4d = 3 + 4(4) = 3 + 16 = 19$ (This matches the given fifth term)

The AP is 3, 7, 11, 15, 19, ...

Final Answer:

The Arithmetic Progression is 3, 7, 11, 15, ...

Given:

In an Arithmetic Progression (AP):

The 26th term, $a_{26} = 0$.

The 11th term, $a_{11} = 3$.

The last term, $a_n = -\frac{1}{5}$.

To Find:

The common difference ($d$) and the number of terms ($n$) in the AP.

Solution:

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

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

$a_k = a + (k-1)d$

Using the given information about the 26th term ($a_{26} = 0$):

$a_{26} = a + (26-1)d$

Using the given information about the 11th term ($a_{11} = 3$):

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

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

From (1): $a + 25d = 0$

From (2): $a + 10d = 3$

Subtract Equation (2) from Equation (1):

$(a + 25d) - (a + 10d) = 0 - 3$

$a + 25d - a - 10d = -3$

$15d = -3$

Solve for $d$ by dividing both sides by 15:

$d = \frac{-3}{15} = -\frac{1}{5}$

So, the common difference is $d = -\frac{1}{5}$.

Now substitute the value of $d = -\frac{1}{5}$ into either Equation (1) or (2) to find the first term $a$. Let's use Equation (1):

$a + 25(-\frac{1}{5}) = 0$

$a + \cancel{25}^{5} \times (-\frac{1}{\cancel{5}_{1}}) = 0$

$a - 5 = 0$

Add 5 to both sides:

$a = 5$

The first term of the AP is $a = 5$ and the common difference is $d = -\frac{1}{5}$.

We are given that the last term of the AP is $a_n = -\frac{1}{5}$. We use the formula $a_n = a + (n-1)d$ to find the number of terms, $n$.

Substitute $a_n = -\frac{1}{5}$, $a = 5$, and $d = -\frac{1}{5}$ into the formula:

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

Subtract 5 from both sides:

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

$-\frac{1}{5} - \frac{25}{5} = (n-1)(-\frac{1}{5})$

$-\frac{26}{5} = (n-1)(-\frac{1}{5})$

Multiply both sides by -5:

$-\frac{26}{5} \times (-5) = (n-1)(-\frac{1}{5}) \times (-5)$

$\cancel{-}\frac{26}{\cancel{5}} \times \cancel{-}\cancel{5} = (n-1) \times \cancel{-}\frac{1}{\cancel{5}} \times \cancel{-}\cancel{5}$

$26 = n-1$

Add 1 to both sides:

$n = 26 + 1$

$n = 27$

The number of terms in the AP is 27.

Final Answer:

The common difference is $-\frac{1}{5}$ and the number of terms is 27.

Given:

In an Arithmetic Progression (AP):

The sum of the 5th term ($a_5$) and the 7th term ($a_7$) is 52, which means $a_5 + a_7 = 52$.

The 10th term ($a_{10}$) is 46.

To Find:

Determine the Arithmetic Progression (AP). This means finding the first term ($a$) and the common difference ($d$).

Solution:

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

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

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

Using the given information about the 10th term ($a_{10} = 46$):

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

Using the given information about the sum of the 5th and 7th terms ($a_5 + a_7 = 52$):

First, express $a_5$ and $a_7$ using the formula for the $n$-th term:

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

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

Substitute these expressions into the sum equation:

$(a + 4d) + (a + 6d) = 52$

Combine like terms:

$(a + a) + (4d + 6d) = 52$

We can simplify Equation (2) by dividing all terms by 2:

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

From (1): $a + 9d = 46$

From (3): $a + 5d = 26$

Subtract Equation (3) from Equation (1):

$(a + 9d) - (a + 5d) = 46 - 26$

$a + 9d - a - 5d = 20$

$4d = 20$

Solve for $d$ by dividing both sides by 4:

$d = \frac{20}{4} = 5$

So, the common difference is $d = 5$.

Now substitute the value of $d = 5$ into either Equation (1) or (3) to find the first term $a$. Let's use Equation (3):

$a + 5(5) = 26$

$a + 25 = 26$

Subtract 25 from both sides:

$a = 26 - 25$

$a = 1$

The first term of the AP is $a = 1$ and the common difference is $d = 5$.

The AP is formed by starting with the first term and repeatedly adding the common difference. The terms are $a, a+d, a+2d, a+3d, ...$

The first few terms are:

$a_1 = a = 1$

$a_2 = a + d = 1 + 5 = 6$

$a_3 = a + 2d = 1 + 2(5) = 1 + 10 = 11$

$a_4 = a + 3d = 1 + 3(5) = 1 + 15 = 16$

$a_5 = a + 4d = 1 + 4(5) = 1 + 20 = 21$

$a_6 = a + 5d = 1 + 5(5) = 1 + 25 = 26$

$a_7 = a + 6d = 1 + 6(5) = 1 + 30 = 31$

$a_{10} = a + 9d = 1 + 9(5) = 1 + 45 = 46$ (This matches the given 10th term)

Check the sum of the 5th and 7th terms: $a_5 + a_7 = 21 + 31 = 52$ (This matches the given condition)

The AP is 1, 6, 11, 16, 21, ...

Final Answer:

The Arithmetic Progression is 1, 6, 11, 16, ...

Given:

In an Arithmetic Progression (AP):

The 7th term ($a_7$) is 24 less than the 11th term ($a_{11}$), which means $a_7 = a_{11} - 24$. This can be rewritten as $a_{11} - a_7 = 24$.

The first term ($a$) is 12.

To Find:

The 20th term ($a_{20}$) of the AP.

Solution:

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

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

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

We are given that $a_{11} - a_7 = 24$.

Using the formula for the $n$-th term, we can write expressions for $a_{11}$ and $a_7$:

$a_{11} = a + (11-1)d = a + 10d$

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

Substitute these expressions into the given difference equation:

$(a + 10d) - (a + 6d) = 24$

Simplify the left side of the equation:

$a + 10d - a - 6d = 24$

$(a - a) + (10d - 6d) = 24$

$0 + 4d = 24$

Solve for the common difference $d$ by dividing both sides by 4:

$d = \frac{24}{4}$

$d = 6$

So, the common difference of the AP is 6.

We are given that the first term is $a = 12$.

Now we need to find the 20th term ($a_{20}$) of the AP using the formula $a_n = a + (n-1)d$ with $n=20$, $a=12$, and $d=6$.

$a_{20} = a + (20-1)d$

$a_{20} = a + 19d$

Substitute the values of $a$ and $d$:

$a_{20} = 12 + 19 \times 6$

Perform the multiplication:

$19 \times 6 = 114$

Now, perform the addition:

$a_{20} = 12 + 114$

$a_{20} = 126$

The 20th term of the AP is 126.

Final Answer:

The 20th term of the AP is 126.

Given:

The 9th term of an Arithmetic Progression (AP) is zero, i.e., $a_9 = 0$.

To Prove:

The 29th term of the AP is twice its 19th term, i.e., $a_{29} = 2 \times a_{19}$.

Proof:

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

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

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

We are given that the 9th term is zero. Using the formula with $n=9$:

Substitute the given value $a_9 = 0$ into Equation (1):

$0 = a + 8d$

This gives us a relationship between the first term and the common difference:

Now, let's express the 19th term ($a_{19}$) using the formula for the $n$-th term with $n=19$:

$a_{19} = a + (19-1)d$

Substitute the relationship $a = -8d$ (from Equation (2)) into Equation (3):

$a_{19} = (-8d) + 18d$

$a_{19} = 10d$

Next, let's express the 29th term ($a_{29}$) using the formula for the $n$-th term with $n=29$:

$a_{29} = a + (29-1)d$

Substitute the relationship $a = -8d$ (from Equation (2)) into Equation (5):

$a_{29} = (-8d) + 28d$

$a_{29} = 20d$

Now, we need to show that $a_{29} = 2 \times a_{19}$.

From Equation (4), we have $a_{19} = 10d$.

From Equation (6), we have $a_{29} = 20d$.

We can write $20d$ as $2 \times (10d)$.

So, $a_{29} = 2 \times a_{19}$.

This matches the statement we needed to prove.

Conclusion:

Since we have shown that $a_{29} = 20d$ and $a_{19} = 10d$, it follows that $a_{29} = 2 \times a_{19}$.

Hence, if the 9th term of an AP is zero, its 29th term is twice its 19th term. Proved.

Given:

The given Arithmetic Progression (AP) is 7, 10, 13, ...

We need to determine if the number 55 is a term of this AP.

To Find:

Whether 55 is a term of the AP. If it is, find its term number ($n$).

Solution:

In the given AP: 7, 10, 13, ...

The first term is $a = 7$.

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

$d = a_2 - a_1 = 10 - 7 = 3$

$d = a_3 - a_2 = 13 - 10 = 3$

So, the common difference is $d = 3$.

To determine if 55 is a term of this AP, we assume that 55 is the $n$-th term ($a_n$) for some positive integer $n$.

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

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

Set $a_n = 55$ and substitute the values $a = 7$ and $d = 3$ into the formula:

$55 = 7 + (n-1)3$

Now, we solve this equation for $n$.

Subtract 7 from both sides:

$55 - 7 = (n-1)3$

$48 = (n-1)3$

Divide both sides by 3:

$\frac{48}{3} = n-1$

$16 = n-1$

Add 1 to both sides to isolate $n$:

$n = 16 + 1$

$n = 17$

Since the value of $n$ we obtained is a positive integer (17), it means that 55 is indeed a term of the given AP, and it is the 17th term.

Conclusion:

Yes, 55 is a term of the AP: 7, 10, 13, ... It is the 17th term.

Given:

The three expressions $k^2 + 4k + 8$, $2k^2 + 3k + 6$, and $3k^2 + 4k + 4$ are consecutive terms of an Arithmetic Progression (AP).

To Find:

The value of $k$.

Solution:

Let the three consecutive terms of the AP be $a_1$, $a_2$, and $a_3$.

$a_1 = k^2 + 4k + 8$

$a_2 = 2k^2 + 3k + 6$

$a_3 = 3k^2 + 4k + 4$

For three terms to be consecutive terms in an Arithmetic Progression, the difference between the second term and the first term must be equal to the difference between the third term and the second term.

$a_2 - a_1 = a_3 - a_2$

This property can also be expressed as $2a_2 = a_1 + a_3$. Let's use this property.

Now, simplify both sides of Equation (i):

Left side:

$2(2k^2 + 3k + 6) = 4k^2 + 6k + 12$

Right side:

$(k^2 + 4k + 8) + (3k^2 + 4k + 4) = (k^2 + 3k^2) + (4k + 4k) + (8 + 4)$

$= 4k^2 + 8k + 12$

Equate the simplified left and right sides:

$4k^2 + 6k + 12 = 4k^2 + 8k + 12$

Subtract $4k^2$ from both sides:

$6k + 12 = 8k + 12$

Subtract 12 from both sides:

$6k = 8k$

Subtract $6k$ from both sides:

$0 = 8k - 6k$

$0 = 2k$

Divide both sides by 2:

$k = 0$

The value of $k$ is 0.

Let's verify the terms when $k=0$:

$a_1 = (0)^2 + 4(0) + 8 = 0 + 0 + 8 = 8$

$a_2 = 2(0)^2 + 3(0) + 6 = 0 + 0 + 6 = 6$

$a_3 = 3(0)^2 + 4(0) + 4 = 0 + 0 + 4 = 4$

The terms are 8, 6, 4. The difference between consecutive terms is $6-8=-2$ and $4-6=-2$. Since the difference is constant, these terms form an AP.

Final Answer:

The value of $k$ is 0.

Given:

The number 207 is split into three parts.

These three parts are in Arithmetic Progression (AP).

The sum of the three parts is 207.

The product of the two smaller parts is 4623.

To Find:

The three parts.

Solution:

Let the three parts in AP be $a-d$, $a$, and $a+d$, where $a$ is the middle term and $d$ is the common difference.

According to the first condition, the sum of the three parts is 207:

$(a-d) + a + (a+d) = 207$

Combine the terms:

$3a = 207$

Divide both sides by 3 to find the value of $a$:

From Equation (i), we get:

$a = 69$

So, the middle part is 69. The three parts of the AP are $69-d$, 69, and $69+d$.

The "two smaller parts" depend on the sign of the common difference $d$.

If $d > 0$, the terms are in increasing order: $69-d < 69 < 69+d$. The two smaller parts are $69-d$ and 69.

If $d < 0$, let $d = -|d|$ where $|d|>0$. The terms are $69+|d|$, 69, $69-|d|$. The terms are in decreasing order: $69+|d| > 69 > 69-|d|$. The two smaller parts are $69+d$ (which is $69-|d|$) and 69.

If $d=0$, the terms are 69, 69, 69. The two smaller parts are 69 and 69. Their product is $69 \times 69 = 4761$, which is not 4623. So $d \neq 0$.

In both cases where $d \neq 0$, the set of the two smaller parts is $\{\min(69-d, 69+d), 69\}$. Their product is given as 4623.

$\min(69-d, 69+d) \times 69 = 4623$

Divide both sides by 69:

Perform the division $\frac{4623}{69}$: $\begin{array}{r} 67 \\ 69{\overline{\smash{\big)}\,4623}} \\ \underline{-414\downarrow} \\ 483 \\ \underline{-483} \\ 0 \end{array}$ So, $\frac{4623}{69} = 67$.

Substitute this back into Equation (ii):

$\min(69-d, 69+d) = 67$

This means the smaller of the two outer terms ($69-d$ and $69+d$) must be 67.

Case A: Assume $69-d$ is the smaller term. This happens when $69-d \leq 69+d$, which implies $-d \leq d$, or $0 \leq 2d$, so $d \geq 0$.

$69-d = 67$

$d = 69 - 67$

$d = 2$

Since $d=2$ is $\geq 0$, this case is consistent. The three parts are:

$69-d = 69 - 2 = 67$

$a = 69$

$69+d = 69 + 2 = 71$

The parts are 67, 69, 71.

Case B: Assume $69+d$ is the smaller term. This happens when $69+d \leq 69-d$, which implies $d \leq -d$, or $2d \leq 0$, so $d \leq 0$.

$69+d = 67$

$d = 67 - 69$

$d = -2$

Since $d=-2$ is $\leq 0$, this case is consistent. The three parts are:

$69-d = 69 - (-2) = 69 + 2 = 71$

$a = 69$

$69+d = 69 + (-2) = 69 - 2 = 67$

The parts are 71, 69, 67.

In both valid cases ($d=2$ and $d=-2$), the set of three parts is {67, 69, 71}. These are the three parts that form an AP (67, 69, 71 or 71, 69, 67). The two smaller parts are 67 and 69, and their product is $67 \times 69 = 4623$, which matches the given condition.

Final Answer:

The three parts are 67, 69, and 71.

Given:

The angles of a triangle are in Arithmetic Progression (AP).

The greatest angle is twice the least angle.

To Find:

All the angles of the triangle.

Solution:

Let the three angles of the triangle in AP be $a - d$, $a$, and $a + d$, where $a$ is the middle angle and $d$ is the common difference. Assume $d > 0$, so $a - d$ is the least angle and $a + d$ is the greatest angle.

The sum of the angles in a triangle is $180^\circ$.

So,

Simplifying equation (i):

Dividing by 3:

Now, we use the second given condition: the greatest angle is twice the least angle.

According to the condition:

Expanding equation (iii):

Rearranging the terms to solve for $d$:

Substitute the value of $a$ from equation (ii) into equation (iv):

Dividing by 3:

Now we find the three angles using the values of $a$ and $d$ from (ii) and (v).

The angles are $a - d$, $a$, and $a + d$.

Verification:

The angles are $40^\circ$, $60^\circ$, and $80^\circ$.

Are they in AP? $60 - 40 = 20$, $80 - 60 = 20$. Yes, the common difference is $20^\circ$.

Is the sum $180^\circ$? $40^\circ + 60^\circ + 80^\circ = 180^\circ$. Yes.

Is the greatest angle twice the least? $80^\circ = 2 \times 40^\circ$. Yes, $80^\circ = 80^\circ$.

The angles of the triangle are $\mathbf{40^\circ}$, $\mathbf{60^\circ}$, and $\mathbf{80^\circ}$.

Given:

First AP: 9, 7, 5, ...

Second AP: 24, 21, 18, ...

The nth term of the first AP is equal to the nth term of the second AP.

To Find:

The value of $n$.

The value of the nth term.

Solution:

Let the first AP be AP$_1$ and the second AP be AP$_2$.

For AP$_1$: 9, 7, 5, ...

First term, $a_1 = 9$

Common difference, $d_1 = 7 - 9 = -2$

The nth term of AP$_1$ is given by the formula $a_n = a_1 + (n-1)d_1$.

For AP$_2$: 24, 21, 18, ...

First term, $a_2 = 24$

Common difference, $d_2 = 21 - 24 = -3$

The nth term of AP$_2$ is given by the formula $a_n = a_2 + (n-1)d_2$.

Given that the nth terms of the two APs are the same, we equate the expressions from (i) and (ii).

Now, solve for $n$:

So, the value of $n$ is 16.

Now, we find the value of the 16th term using either equation (i) or (ii).

Using equation (i):

Using equation (ii):

Both methods give the same value for the 16th term.

The value of $n$ is $\mathbf{16}$.

The value of the term is $\mathbf{-21}$.

Given:

In an AP, the sum of the 3rd and 8th terms is 7.

In the same AP, the sum of the 7th and 14th terms is -3.

To Find:

The 10th term of the AP.

Solution:

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

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

According to the first condition:

According to the second condition:

We now have a system of two linear equations with two variables, $a$ and $d$:

$2a + 9d = 7$ (i)

$2a + 19d = -3$ (ii)

Subtract equation (i) from equation (ii):

Substitute the value of $d = -1$ into equation (i):

Now we need to find the 10th term, $a_{10}$.

Substitute the values of $a=8$ and $d=-1$:

The 10th term of the AP is $\mathbf{-1}$.

Given:

The Arithmetic Progression (AP) is –2, –4, –6,..., –100.

To Find:

The 12th term from the end of the given AP.

Solution:

The given AP is –2, –4, –6,..., –100.

The first term is $a = -2$.

The common difference is $d = -4 - (-2) = -4 + 2 = -2$.

The last term is $l = -100$.

To find the 12th term from the end of the AP, we can consider the AP in reverse order.

The reversed AP will start with the last term of the original AP and its common difference will be the negative of the common difference of the original AP.

Reversed AP: –100, ..., –6, –4, –2.

First term of the reversed AP, $a' = -100$.

Common difference of the reversed AP, $d' = -d = -(-2) = 2$.

We need to find the 12th term of this reversed AP.

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

Using this formula for the reversed AP, the 12th term ($a'_{12}$) is:

Substitute the values $a' = -100$ and $d' = 2$:

The 12th term from the end of the AP is $\mathbf{-78}$.

Alternate Solution:

First, find the total number of terms in the AP.

Let the number of terms be $n$. The last term is $a_n = -100$.

The total number of terms is 50.

The 12th term from the end is the $(n - 12 + 1)$th term from the beginning.

So, we need to find the 39th term ($a_{39}$) of the original AP.

Substitute $a = -2$ and $d = -2$:

Both methods yield the same result.

Given:

The Arithmetic Progression (AP) is 53, 48, 43, ...

To Find:

The first term of the AP which is negative.

Solution:

The given AP is 53, 48, 43, ...

The first term is $a = 53$.

The common difference is $d = 48 - 53 = -5$.

We are looking for the first term $a_n$ such that $a_n < 0$.

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

We need to find the value of $n$ for which $a_n < 0$.

Substitute the values of $a$ and $d$:

Now, we solve the inequality for $n$:

Divide both sides by -5 and reverse the inequality sign:

Calculate the value of $\frac{58}{5}$:

Since $n$ represents the term number, it must be a positive integer.

The smallest integer value of $n$ that is greater than 11.6 is 12.

Therefore, the 12th term is the first term that is negative.

We can verify this by finding the 11th and 12th terms:

Since $a_{11} = 3$ (positive) and $a_{12} = -2$ (negative), the 12th term is indeed the first negative term.

The first negative term is the 12th term.

Solution:

Let the required number be $N$. When $N$ is divided by 4, it leaves a remainder of 3. This means $N$ can be expressed in the form:

$N = 4k + 3$, where $k$ is an integer.

The numbers lie between 10 and 300, which means the range for $N$ is $10 < N < 300$.

Substituting the expression for $N$, we get the inequality:

$10 < 4k + 3 < 300$

To find the possible integer values of $k$, we first isolate $4k$ by subtracting 3 from all parts of the inequality:

$10 - 3 < 4k + 3 - 3 < 300 - 3$

$7 < 4k < 297$

Next, we divide all parts of the inequality by 4:

Simplifying the inequality (i), we get:

Since $k$ must be an integer, the possible integer values for $k$ that satisfy inequality (ii) are the integers strictly greater than 1.75 and strictly less than 74.25. These values are $2, 3, 4, \dots, 73, 74$.

To find the number of integers in this sequence, we subtract the first integer from the last integer and add 1:

Number of values of $k = (\text{Last integer}) - (\text{First integer}) + 1$

Number of values of $k = 74 - 2 + 1$

Number of values of $k = 72 + 1$

Number of values of $k = 73$

Each integer value of $k$ corresponds to a unique number $N = 4k+3$ that lies between 10 and 300 and leaves a remainder of 3 when divided by 4.

Therefore, there are 73 such numbers.

The final answer is 73.

Given:

The given Arithmetic Progression (AP) is $-\frac{4}{3}$, -1, $-\frac{2}{3}$, …, $4\frac{1}{3}$.

To Find:

The sum of the two middle most terms of the given AP.

Solution:

The first term of the AP is $a = -\frac{4}{3}$.

The last term of the AP is $l = 4\frac{1}{3} = \frac{13}{3}$.

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

$d = (-1) - \left(-\frac{4}{3}\right) = -1 + \frac{4}{3} = \frac{-3+4}{3} = \frac{1}{3}$

Alternatively, $d = \left(-\frac{2}{3}\right) - (-1) = -\frac{2}{3} + 1 = \frac{-2+3}{3} = \frac{1}{3}$.

Let $n$ be the number of terms in the AP. The $n$-th term $a_n$ is given by the formula $a_n = a + (n-1)d$.

The last term is $a_n = \frac{13}{3}$. So, we have:

Multiply both sides of equation (i) by 3 to clear the denominators:

$3 \times \frac{13}{3} = 3 \times \left(-\frac{4}{3}\right) + 3 \times (n-1)\frac{1}{3}$

$13 = -4 + (n-1)$

$13 = -4 + n - 1$

$13 = n - 5$}

Adding 5 to both sides, we get:

$n = 13 + 5 = 18$

So, there are 18 terms in the AP.

Since the number of terms ($n=18$) is even, the two middle most terms are the $(\frac{n}{2})$-th term and the $(\frac{n}{2}+1)$-th term.

The middle terms are the $(\frac{18}{2})$-th term and the $(\frac{18}{2}+1)$-th term, which are the 9th term and the 10th term.

Now, we find the value of the 9th term ($a_9$) using the formula $a_n = a + (n-1)d$ with $n=9$:

$a_9 = a + (9-1)d = a + 8d$

$a_9 = -\frac{4}{3} + 8 \times \frac{1}{3} = -\frac{4}{3} + \frac{8}{3} = \frac{-4+8}{3} = \frac{4}{3}$

Next, we find the value of the 10th term ($a_{10}$) using the formula $a_n = a + (n-1)d$ with $n=10$:

$a_{10} = a + (10-1)d = a + 9d$}

$a_{10} = -\frac{4}{3} + 9 \times \frac{1}{3} = -\frac{4}{3} + 3 = \frac{-4+9}{3} = \frac{5}{3}$

The sum of the two middle most terms is the sum of the 9th term and the 10th term:

Sum $= a_9 + a_{10} = \frac{4}{3} + \frac{5}{3}$

Sum $= \frac{4+5}{3} = \frac{9}{3} = 3$

The sum of the two middle most terms is 3.

The final answer is 3.

Given:

First term of the AP, $a = -5$.

Last term of the AP, $l = 45$.

Sum of the terms of the AP, $S_n = 120$.

To Find:

The number of terms ($n$) and the common difference ($d$).

Solution:

We are given the first term ($a$), the last term ($l$), and the sum of the terms ($S_n$). We can use the formula for the sum of an AP when the first and last terms are known:

$S_n = \frac{n}{2}(a+l)$

Substitute the given values into the formula:

Simplify equation (i):

$120 = \frac{n}{2}(40)$

$120 = n \times \frac{40}{2}$

$120 = 20n$

To find $n$, divide both sides by 20:

$n = \frac{120}{20}$

$n = 6$

So, the number of terms in the AP is 6.

Now that we have the number of terms ($n$), the first term ($a$), and the last term ($l$), we can find the common difference ($d$) using the formula for the $n$-th term (which is the last term in this case):

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

Substitute the known values $l=45$, $a=-5$, and $n=6$ into the formula:

Simplify equation (ii):

$45 = -5 + 5d$

Add 5 to both sides of the equation:

$45 + 5 = 5d$

$50 = 5d$

To find $d$, divide both sides by 5:

$d = \frac{50}{5}$

$d = 10$

So, the common difference is 10.

The number of terms is 6 and the common difference is 10.

The final answer is Number of terms = 6, Common difference = 10.

Solution (i):

The given series is 1 + (–2) + (–5) + (–8) + ... + (–236).

This is an Arithmetic Progression (AP) with:

First term, $a = 1$.

Common difference, $d = (-2) - 1 = -3$.

Let the last term be $a_n = -236$.

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

Substituting the given values, we have:

From equation (i), we get:

-236 - 1 = (n-1)(-3)

-237 = -3(n-1)

Divide both sides by -3:

$\frac{-237}{-3} = n-1$

$79 = n-1$

Add 1 to both sides:

$n = 79 + 1 = 80$

So, there are 80 terms in the AP.

Now, we need to find the sum of these 80 terms. The formula for the sum of an AP ($S_n$) when the first term ($a$) and the last term ($l$ or $a_n$) are known is $S_n = \frac{n}{2}(a+l)$.

Here, $n=80$, $a=1$, and $l=-236$.

$S_{80} = \frac{80}{2}(1 + (-236))$

$S_{80} = 40(1 - 236)$

$S_{80} = 40(-235)$

Calculating the product:

$40 \times (-235) = -9400$

The sum of the series is -9400.

The final answer for part (i) is -9400.

Solution (ii):

The given series is $4 - \frac{1}{n}$ + $4 - \frac{2}{n}$ + $4 - \frac{3}{n}$ + ... upto n terms.

This is an Arithmetic Progression (AP) with:

First term, $a = 4 - \frac{1}{n}$.

Number of terms is $n$.

Common difference, $d = \left(4 - \frac{2}{n}\right) - \left(4 - \frac{1}{n}\right)$

$d = 4 - \frac{2}{n} - 4 + \frac{1}{n}$

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

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

Substitute the values of $a$, $n$, and $d$ into the formula:

From equation (ii), we simplify the expression inside the parenthesis:

$2\left(4 - \frac{1}{n}\right) = 8 - \frac{2}{n}$

$(n-1)\left(-\frac{1}{n}\right) = -\frac{n}{n} + \frac{1}{n} = -1 + \frac{1}{n}$

So, the expression inside the parenthesis becomes:

$\left(8 - \frac{2}{n}\right) + \left(-1 + \frac{1}{n}\right) = 8 - \frac{2}{n} - 1 + \frac{1}{n}$

$= (8-1) + \left(-\frac{2}{n} + \frac{1}{n}\right)$

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

Now, substitute this back into the sum formula:

$S_n = \frac{n}{2}\left(7 - \frac{1}{n}\right)$

$S_n = \frac{n}{2} \times 7 - \frac{n}{2} \times \frac{1}{n}$

$S_n = \frac{7n}{2} - \frac{\cancel{n}}{2} \times \frac{1}{\cancel{n}}$

$S_n = \frac{7n}{2} - \frac{1}{2}$

$S_n = \frac{7n - 1}{2}$

The sum of the series is $\frac{7n - 1}{2}$.

The final answer for part (ii) is $\frac{7n - 1}{2}$.

Solution (iii):

The given series is $\frac{a \;-\; b}{a \;+\; b}$ + $\frac{3a \;-\; 2b}{a \;+\; b}$ + $\frac{5a \;-\; 3b}{a \;+\; b}$ + … to 11 terms.

This is an Arithmetic Progression (AP) with:

First term, $a_{term} = \frac{a \;-\; b}{a \;+\; b}$.

Number of terms, $n = 11$.

Common difference, $d = \frac{3a \;-\; 2b}{a \;+\; b} - \frac{a \;-\; b}{a \;+\; b}$

$d = \frac{(3a - 2b) - (a - b)}{a \;+\; b}$

$d = \frac{3a - 2b - a + b}{a \;+\; b}$

$d = \frac{2a - b}{a \;+\; b}$

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

Substitute the values of $a_{term}$, $n$, and $d$ into the formula:

From equation (iii), we simplify the expression inside the parenthesis:

$2\left(\frac{a \;-\; b}{a \;+\; b}\right) = \frac{2(a \;-\; b)}{a \;+\; b}$

$(11-1)\left(\frac{2a \;-\; b}{a \;+\; b}\right) = 10\left(\frac{2a \;-\; b}{a \;+\; b}\right) = \frac{10(2a \;-\; b)}{a \;+\; b}$

So, the expression inside the parenthesis becomes:

$\frac{2(a \;-\; b)}{a \;+\; b} + \frac{10(2a \;-\; b)}{a \;+\; b}$

$= \frac{2(a \;-\; b) + 10(2a \;-\; b)}{a \;+\; b}$

$= \frac{2a - 2b + 20a - 10b}{a \;+\; b}$

$= \frac{(2a + 20a) + (-2b - 10b)}{a \;+\; b}$

$= \frac{22a - 12b}{a \;+\; b}$

Now, substitute this back into the sum formula:

$S_{11} = \frac{11}{2}\left(\frac{22a - 12b}{a \;+\; b}\right)$

$S_{11} = 11 \times \frac{22a - 12b}{2(a \;+\; b)}$

$S_{11} = 11 \times \frac{\cancel{2}(11a - 6b)}{\cancel{2}(a \;+\; b)}$

$S_{11} = \frac{11(11a - 6b)}{a \;+\; b}$

$S_{11} = \frac{121a - 66b}{a \;+\; b}$

The sum of the series to 11 terms is $\frac{121a - 66b}{a \;+\; b}$.

The final answer for part (iii) is $\frac{121a - 66b}{a \;+\; b}$.

Given:

The given Arithmetic Progression (AP) is –2, –7, –12,...

The term we are looking for is –77.

To Find:

1. Which term in the AP is –77?

2. The sum of the AP up to the term –77.

Solution:

The first term of the AP is $a = -2$.

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

$d = -7 + 2 = -5$.

Let –77 be the $n$-th term of the AP. The formula for the $n$-th term of an AP is $a_n = a + (n-1)d$.

Substitute the given values $a_n = -77$, $a = -2$, and $d = -5$ into the formula:

From equation (i), we solve for $n$:

-77 + 2 = (n-1)(-5)

-75 = -5(n-1)

Divide both sides by -5:

From equation (ii), we get:

$15 = n-1$

$n = 15 + 1$

$n = 16$

Therefore, –77 is the 16th term of the AP.

Now, we need to find the sum of the AP up to the term –77, which means the sum of the first 16 terms. The formula for the sum of the first $n$ terms of an AP ($S_n$) when the first term ($a$) and the last term ($l$ or $a_n$) are known is $S_n = \frac{n}{2}(a+l)$.

Here, $n=16$, $a=-2$, and the last term is $l = -77$.

Substitute these values into the sum formula:

From equation (iii), we calculate the sum:

$S_{16} = 8(-2 - 77)$

$S_{16} = 8(-79)$

$S_{16} = -632$

The sum of the AP up to the term –77 is –632.

The final answer is The 16th term is –77 and the sum of the AP upto this term is –632.

Given:

The $n$-th term of the sequence is given by the formula $a_n = 3 - 4n$.

To Show:

That the sequence $a_1, a_2, a_3, \dots$ forms an Arithmetic Progression (AP).

To Find:

The sum of the first 20 terms ($S_{20}$) of this AP.

Solution:

To show that the sequence $a_n$ forms an AP, we need to prove that the difference between any term and its preceding term is a constant value.

Let's find the difference between the $(n+1)$-th term and the $n$-th term, $a_{n+1} - a_n$.

First, find the expression for $a_{n+1}$ by replacing $n$ with $(n+1)$ in the formula $a_n = 3 - 4n$:

$a_{n+1} = 3 - 4(n+1)$

$a_{n+1} = 3 - 4n - 4$

$a_{n+1} = -1 - 4n$

Now, calculate the difference $a_{n+1} - a_n$:

From equation (i), we simplify:

$a_{n+1} - a_n = -1 - 4n - 3 + 4n$

$a_{n+1} - a_n = (-1 - 3) + (-4n + 4n)$

$a_{n+1} - a_n = -4 + 0$

$a_{n+1} - a_n = -4$

Since the difference between any term and its preceding term is the constant value -4, the sequence $a_1, a_2, a_3, \dots$ forms an AP.

The common difference is $d = -4$.

Now, we need to find the sum of the first 20 terms ($S_{20}$). The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}(2a + (n-1)d)$, where $a$ is the first term and $d$ is the common difference.

The first term is $a_1$. Using the given formula $a_n = 3 - 4n$ for $n=1$:

$a_1 = 3 - 4(1) = 3 - 4 = -1$.

So, the first term is $a = -1$. The common difference is $d = -4$, and we need to find the sum of the first $n=20$ terms.

Substitute the values into the sum formula:

From equation (ii), we calculate the sum:

$S_{20} = 10(2(-1) + (19)(-4))$

$S_{20} = 10(-2 - 76)$

$S_{20} = 10(-78)$

$S_{20} = -780$

The sum of the first 20 terms is -780.

The sequence $a_1, a_2, a_3, \dots$ forms an AP with first term -1 and common difference -4. The sum of the first 20 terms is -780.

The final answer is The sequence forms an AP. $S_{20} = -780$.

Given:

The sum of the first $n$ terms of an AP is given by the formula $S_n = n(4n + 1)$.

To Find:

The Arithmetic Progression (AP).

Solution:

The first term of an AP, denoted by $a_1$, is equal to the sum of the first term, $S_1$.

Substitute $n=1$ into the given formula for $S_n$:

From equation (i):

$S_1 = 1(4 + 1) = 1(5) = 5$

So, the first term is $a_1 = S_1 = 5$.

The sum of the first two terms of the AP is $S_2 = a_1 + a_2$. We can find $S_2$ by substituting $n=2$ into the formula for $S_n$:

From equation (ii):

$S_2 = 2(8 + 1) = 2(9) = 18$

Now we can find the second term $a_2$ using the relation $S_2 = a_1 + a_2$:

$a_2 = S_2 - a_1$

$a_2 = 18 - 5$

$a_2 = 13$

The common difference of the AP, denoted by $d$, is the difference between any term and its preceding term. Using the first two terms:

$d = 13 - 5$

$d = 8$

The AP is determined by its first term ($a_1$) and its common difference ($d$).

The first term is 5 and the common difference is 8. The terms of the AP are $a_1, a_1+d, a_1+2d, \dots$

$a_1 = 5$

$a_2 = 5 + 8 = 13$

$a_3 = 13 + 8 = 21$

and so on.

The AP is 5, 13, 21, 29, ...

The final answer is The AP is 5, 13, 21, 29, ....

Given:

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

The $k$-th term of the AP is $a_k = 164$.

To Find:

The value of $k$.

Solution:

The $n$-th term of an AP can be found using the formula $a_n = S_n - S_{n-1}$ for $n > 1$, and $a_1 = S_1$.

First, find the first term $a_1$ by setting $n=1$ in the formula for $S_n$:

From equation (i):

$S_1 = 1(3 \times 1 + 5) = 1(3 + 5) = 1(8) = 8$

So, the first term is $a_1 = 8$.

Next, find the formula for the $(n-1)$-th sum, $S_{n-1}$, by replacing $n$ with $(n-1)$ in the formula for $S_n$. This is valid for $n > 1$.

From equation (ii):

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

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

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

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

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

$S_{n-1} = 3n^3 - 4n^2 - n + 2$

A simpler way to calculate $S_{n-1}$ from $S_n = 3n^2 + 5n$ is:

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

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

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

$S_{n-1} = 3n^2 - n - 2$ for $n > 1$.

Now, find the $n$-th term, $a_n$, for $n > 1$ using $a_n = S_n - S_{n-1}$:

From equation (iii):

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

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

$a_n = 6n + 2$ for $n > 1$.

Let's verify if this formula for $a_n$ also holds for $n=1$:

$a_1 = 6(1) + 2 = 6 + 2 = 8$.

This matches the value of $a_1 = 8$ that we calculated directly from $S_1$. Therefore, the formula for the $n$-th term of the AP is $a_n = 6n + 2$, valid for all $n \geq 1$.

We are given that the $k$-th term of the AP is $a_k = 164$.

Using the formula for the $n$-th term with $n=k$, we get the expression for $a_k$:

Now, we set this expression equal to the given value of $a_k$:

From equation (v), we solve the linear equation for $k$:

$6k = 164 - 2$

$6k = 162$

Divide both sides by 6:

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

$k = 27$

The value of $k$ is 27.

The final answer is $k = 27$.

Given:

$S_n$ denotes the sum of the first $n$ terms of an Arithmetic Progression (AP).

To Prove:

$S_{12} = 3(S_8 – S_4)$

Proof:

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

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

Calculate the Left Hand Side (LHS), $S_{12}$:

$S_{12} = \frac{12}{2}(2a + (12-1)d)$

$S_{12} = 6(2a + 11d)$

Calculate the terms needed for the Right Hand Side (RHS), $S_8$ and $S_4$:

$S_8 = \frac{8}{2}(2a + (8-1)d)$

$S_8 = 4(2a + 7d)$

$S_4 = \frac{4}{2}(2a + (4-1)d)$

$S_4 = 2(2a + 3d)$

Now calculate the expression $(S_8 - S_4)$ using equations (ii) and (iii):

$S_8 - S_4 = (8a + 28d) - (4a + 6d)$

$S_8 - S_4 = 8a + 28d - 4a - 6d$}

$S_8 - S_4 = (8a - 4a) + (28d - 6d)$

Now calculate the RHS, $3(S_8 - S_4)$, using equation (iv):

$3(S_8 - S_4) = 3(4a + 22d)$

Compare the expression for the LHS from equation (i) and the expression for the RHS from equation (v):

LHS = $12a + 66d$

RHS = $12a + 66d$

Since LHS = RHS, the equality is proven.

Hence, $S_{12} = 3(S_8 – S_4)$.

The proof is complete.

Given:

The 4th term of the AP is $a_4 = -15$.

The 9th term of the AP is $a_9 = -30$.

To Find:

The sum of the first 17 terms of the AP, $S_{17}$.

Solution:

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

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

Using the given information, we can write two equations:

For the 4th term ($n=4$):

Substituting the given value $a_4 = -15$ into equation (i):

For the 9th term ($n=9$):

Substituting the given value $a_9 = -30$ into equation (ii):

Now we have a system of two linear equations (1) and (2) with two variables, $a$ and $d$.

Subtract equation (1) from equation (2):

From equation (iii):

$a + 8d - a - 3d = -30 + 15$

$5d = -15$

Divide both sides by 5:

$d = \frac{-15}{5}$

$d = -3$

Substitute the value of $d = -3$ into equation (1):

From equation (iv):

$a - 9 = -15$

Add 9 to both sides:

$a = -15 + 9$

$a = -6$

So, the first term of the AP is $a = -6$ and the common difference is $d = -3$.

Now, we need to find the sum of the first 17 terms, $S_{17}$. The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}(2a + (n-1)d)$.

Substitute $n=17$, $a=-6$, and $d=-3$ into the sum formula:

From equation (v), we calculate the sum:

$S_{17} = \frac{17}{2}(2(-6) + 16(-3))$

$S_{17} = \frac{17}{2}(-12 - 48)$

$S_{17} = \frac{17}{2}(-60)$

$S_{17} = 17 \times \frac{-60}{2}$

$S_{17} = 17 \times (-30)$

$S_{17} = -510$

The sum of the first 17 terms of the AP is -510.

The final answer is $S_{17} = -510$.

Given:

Sum of the first 6 terms of an AP, $S_6 = 36$.

Sum of the first 16 terms of an AP, $S_{16} = 256$.

To Find:

The sum of the first 10 terms of the AP, $S_{10}$.

Solution:

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

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

Using the given information for $S_6 = 36$ (with $n=6$):

From equation (i):

$36 = 3(2a + 5d)$

Divide both sides by 3:

Using the given information for $S_{16} = 256$ (with $n=16$):

From equation (ii):

$256 = 8(2a + 15d)$

Divide both sides by 8:

Now we have a system of two linear equations (1) and (2) with two variables, $a$ and $d$.

Equation (1): $2a + 5d = 12$

Equation (2): $2a + 15d = 32$}

Subtract equation (1) from equation (2) to eliminate $a$:

From equation (iii):

$2a + 15d - 2a - 5d = 20$

$10d = 20$

Divide both sides by 10 to find $d$:

$d = \frac{20}{10}$

$d = 2$

Substitute the value of $d = 2$ into equation (1) to find $a$:

From equation (iv):

$2a + 10 = 12$

$2a = 12 - 10$

$2a = 2$

Divide both sides by 2 to find $a$:

$a = \frac{2}{2}$

$a = 1$

So, the first term of the AP is $a = 1$ and the common difference is $d = 2$.

Now, we need to find the sum of the first 10 terms, $S_{10}$. Use the formula for the sum of the first $n$ terms with $n=10$, $a=1$, and $d=2$:

From equation (v):

$S_{10} = 5(2(1) + 9(2))$

$S_{10} = 5(2 + 18)$

$S_{10} = 5(20)$

$S_{10} = 100$}

The sum of the first 10 terms of the AP is 100.

The final answer is $S_{10} = 100$.

Given:

The number of terms in the AP is $n = 11$.

The middle most term of the AP is 30.

To Find:

The sum of the first 11 terms of the AP, $S_{11}$.

Solution:

Since the number of terms is odd ($n=11$), the middle most term is the $(\frac{n+1}{2})$-th term.

The position of the middle term is $\frac{11+1}{2} = \frac{12}{2} = 6$-th term.

So, the 6th term of the AP is $a_6 = 30$.

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

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

Using this formula for the 6th term ($n=6$), we have:

We are given that $a_6 = 30$. Substituting this into equation (i):

Now, we need to find the sum of the first 11 terms, $S_{11}$. The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}(2a + (n-1)d)$.

Substitute $n=11$ into the sum formula:

From equation (ii), we simplify:

$S_{11} = \frac{11}{2}(2a + 10d)$

$S_{11} = \frac{11}{2} \times 2(a + 5d)$

$S_{11} = 11(a + 5d)$

From equation (1), we know that $a + 5d = 30$. Substitute this value into the expression for $S_{11}$:

From equation (iii):

$S_{11} = 330$

The sum of the first 11 terms of the AP is 330.

The final answer is $S_{11} = 330$.

Given:

The Arithmetic Progression (AP) is 8, 10, 12, ..., 126.

To Find:

The sum of the last ten terms of the given AP.

Solution:

The first term of the AP is $a = 8$.

The common difference $d = 10 - 8 = 2$.

The last term of the AP is $a_N = 126$, where $N$ is the total number of terms.

First, find the total number of terms ($N$) in the AP using the formula $a_N = a + (N-1)d$:

From equation (i), we solve for $N$:

$126 - 8 = (N-1)2$

$118 = 2(N-1)$

Divide both sides by 2:

$\frac{118}{2} = N-1$

$59 = N-1$

$N = 59 + 1$

$N = 60$

So, there are 60 terms in the AP.

We need to find the sum of the last ten terms. These terms are $a_{60-10+1}, a_{60-10+2}, \dots, a_{60}$.

The terms are $a_{51}, a_{52}, \dots, a_{60}$.

This is an AP consisting of 10 terms, where the first term is $a_{51}$ and the last term is $a_{60}$.

Find the 51st term ($a_{51}$) using the formula $a_n = a + (n-1)d$ with $n=51$:

From equation (ii), substitute $a=8$ and $d=2$:

$a_{51} = 8 + 50(2)$

$a_{51} = 8 + 100$}

$a_{51} = 108$

The sequence of the last ten terms starts with $a_{51} = 108$ and ends with $a_{60} = 126$. The number of terms in this sequence is 10.

The sum of these last ten terms can be found using the formula for the sum of an AP: $S = \frac{\text{Number of terms}}{2}(\text{First term} + \text{Last term})$.

Sum of last 10 terms $= \frac{10}{2}(a_{51} + a_{60})$

Sum of last 10 terms $= 5(108 + 126)$

Sum of last 10 terms $= 5(234)$

Calculating the product:

$5 \times 234 = 1170$

The sum of the last ten terms of the AP is 1170.

The final answer is 1170.

Given:

We need to find the sum of the first seven numbers which are multiples of both 2 and 9.

To Find:

The sum of the first 7 numbers that are multiples of both 2 and 9.

Solution:

A number that is a multiple of both 2 and 9 must be a multiple of their Least Common Multiple (LCM).

We find the LCM of 2 and 9:

LCM(2, 9) = $2 \times 9 = 18$

So, the numbers which are multiples of both 2 and 9 are the multiples of 18.

These numbers form an Arithmetic Progression (AP) where the terms are multiples of 18. The first such number is 18, the second is 36, and so on.

The sequence of these numbers is 18, 36, 54, ...

This AP has:

First term, $a = 18$.

Common difference, $d = 36 - 18 = 18$.

We need to find the sum of the first 7 terms, so $n = 7$.

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

Substitute the values $n=7$, $a=18$, and $d=18$ into the formula:

From equation (i):

$S_7 = \frac{7}{2}(36 + (6)18)$

$S_7 = \frac{7}{2}(36 + 108)$

$S_7 = \frac{7}{2}(144)$

$S_7 = 7 \times \frac{144}{2}$

$S_7 = 7 \times 72$}

$S_7 = 504$}

The sum of the first seven numbers which are multiples of 2 as well as of 9 is 504.

The final answer is 504.

Given:

The Arithmetic Progression (AP) is –15, –13, –11,...

The sum of the first $n$ terms is $S_n = -55$.

To Find:

The number of terms ($n$) needed to make the sum –55.

The reason for the double answer (if any).

Solution:

The first term of the AP is $a = -15$.

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

$d = -13 + 15 = 2$.}

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

Substitute the given values $S_n = -55$, $a = -15$, and $d = 2$ into the formula:

From equation (i), we solve for $n$:

$-55 = \frac{n}{2}(-30 + 2n - 2)$

$-55 = \frac{n}{2}(2n - 32)$

Multiply both sides by 2:

$-110 = n(2n - 32)$

$-110 = 2n^2 - 32n$

Rearrange the equation into a standard quadratic form ($Ax^2 + Bx + C = 0$):

$2n^2 - 32n + 110 = 0$

Divide the entire equation by 2 to simplify:

Solve the quadratic equation (ii) for $n$. We can factor the equation. We need two numbers that multiply to 55 and add up to -16. These numbers are -5 and -11.

$(n - 5)(n - 11) = 0$

This gives two possible solutions for $n$:

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

$n - 11 = 0 \implies n = 11$

Since $n$ must be a positive integer representing the number of terms, both $n=5$ and $n=11$ are valid solutions.

So, the sum of the first 5 terms is -55, and the sum of the first 11 terms is also -55.

Reason for the double answer:

The common difference of the AP is positive ($d=2$). The terms of the AP are increasing: –15, –13, –11, –9, –7, –5, –3, –1, 1, 3, 5, ...

The sum of the first 5 terms is $S_5 = -55$.

The sum of the terms from the 6th term to the 11th term is:

$a_6 = a + 5d = -15 + 5(2) = -15 + 10 = -5$

$a_7 = a + 6d = -15 + 6(2) = -15 + 12 = -3$

$a_8 = a + 7d = -15 + 7(2) = -15 + 14 = -1$

$a_9 = a + 8d = -15 + 8(2) = -15 + 16 = 1$

$a_{10} = a + 9d = -15 + 9(2) = -15 + 18 = 3$}

$a_{11} = a + 10d = -15 + 10(2) = -15 + 20 = 5$}

The sum of the terms from the 6th to the 11th term is $a_6 + a_7 + a_8 + a_9 + a_{10} + a_{11} = (-5) + (-3) + (-1) + 1 + 3 + 5$.

Sum $= (-5 + 5) + (-3 + 3) + (-1 + 1) = 0 + 0 + 0 = 0$.

The sum of the first 11 terms is the sum of the first 5 terms plus the sum of the terms from the 6th to the 11th term:

$S_{11} = S_5 + (a_6 + a_7 + a_8 + a_9 + a_{10} + a_{11})$

$S_{11} = -55 + 0$}

$S_{11} = -55$}

Since the sum of the terms from the 6th to the 11th term is zero, adding these terms to the sum of the first 5 terms results in the same sum (-55). This occurs because the common difference is positive, causing the terms to transition from negative values to positive values, and the positive and negative values cancel each other out over a certain range of terms whose sum is zero.

The final answer is 5 terms or 11 terms. The reason for the double answer is that the sum of the terms from the 6th to the 11th term is zero.

Given:

For the first AP: First term $a_1 = 8$, common difference $d_1 = 20$. The sum of the first $n$ terms is $S_{n,1}$.

For the second AP: First term $a_2 = -30$, common difference $d_2 = 8$. The sum of the first $2n$ terms is $S_{2n,2}$.

It is given that $S_{n,1} = S_{2n,2}$.

To Find:

The value of $n$.

Solution:

The formula for the sum of the first $k$ terms of an AP with first term $a$ and common difference $d$ is $S_k = \frac{k}{2}(2a + (k-1)d)$.

For the first AP, we have $a=a_1=8$, $d=d_1=20$, and number of terms $k=n$. The sum is $S_{n,1}$:

From equation (i):

$S_{n,1} = \frac{n}{2}(16 + 20n - 20)$

$S_{n,1} = \frac{n}{2}(20n - 4)$

For the second AP, we have $a=a_2=-30$, $d=d_2=8$, and number of terms $k=2n$. The sum is $S_{2n,2}$:

From equation (ii):

$S_{2n,2} = n(-60 + 16n - 8)$

We are given that $S_{n,1} = S_{2n,2}$. Equating the expressions from (1) and (2):

Rearrange equation (iii) to form a quadratic equation:

$0 = 16n^2 - 10n^2 - 68n + 2n$

$0 = 6n^2 - 66n$

Factor out $6n$ from the equation:

From equation (iv), the possible values for $n$ are when each factor is equal to zero:

$6n = 0 \implies n = 0$

or

$n - 11 = 0 \implies n = 11$

Since $n$ represents the number of terms in an AP, it must be a positive integer. A number of terms cannot be zero.

Therefore, the value of $n$ must be 11.

The value of $n$ is 11.

The final answer is $n = 11$.

Given:

Amount saved on Day 1 = $\textsf{₹}1$.

Amount saved on Day 2 = $\textsf{₹}2$.

Amount saved on Day 3 = $\textsf{₹}3$, and so on.

Period of saving: From Jan 1st, 2008, till the end of the month (January 2008).

Amount spent = $\textsf{₹}204$.

Amount remaining at the end of the month = $\textsf{₹}100$.

To Find:

The total pocket money for the month.

Solution:

First, determine the number of days Kanika saved money. The saving was from January 1st, 2008, till the end of January 2008. The year 2008 is a leap year (divisible by 4), but this only affects February. January has 31 days.

So, saving occurred for $n = 31$ days.

The amounts saved daily are 1, 2, 3, ..., 31. This sequence forms an Arithmetic Progression (AP) with:

First term, $a = 1$.

Common difference, $d = 2 - 1 = 1$.

Number of terms, $n = 31$.

The total amount saved is the sum of this AP, $S_n$.

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

Substitute the values $n=31$, $a=1$, and $d=1$ into the formula:

From equation (i):

$S_{31} = \frac{31}{2}(2 + 30)$

$S_{31} = \frac{31}{2}(32)$

$S_{31} = 31 \times 16$}

$S_{31} = 496$}

So, the total amount saved in the piggy bank is $\textsf{₹}496$.

Let $P$ be the total pocket money Kanika received at the beginning of the month.

The total pocket money is used for saving and spending, and the remaining amount is what is left.

The relationship between pocket money, saved amount, spent amount, and remaining amount is:

Total Pocket Money - Total Saved - Total Spent = Total Remaining

Or, equivalently:

From equation (ii), substitute the known values:

$P = \textsf{₹}496 + \textsf{₹}204 + \textsf{₹}100$

$P = \textsf{₹}(496 + 204 + 100)$

$P = \textsf{₹}(700 + 100)$

$P = \textsf{₹}800$

Kanika's pocket money for the month was $\textsf{₹}800$.

The final answer is $\textsf{₹}800$.

Given:

Savings in the first month ($a_1$) = $\textsf{₹}32$.

Savings in the second month ($a_2$) = $\textsf{₹}36$.

Savings in the third month ($a_3$) = $\textsf{₹}40$.

Total target savings ($S_n$) = $\textsf{₹}2000$.

To Find:

The number of months ($n$) in which Yasmeen will save $\textsf{₹}2000$.

Solution:

The monthly savings form a sequence: 32, 36, 40, ...

Let's check the difference between consecutive terms:

$36 - 32 = 4$

$40 - 36 = 4$

Since the difference between consecutive terms is constant, the sequence of monthly savings is an Arithmetic Progression (AP).

The first term of the AP is $a = 32$.

The common difference of the AP is $d = 4$.

Let $n$ be the number of months required for the total savings to be $\textsf{₹}2000$. The total savings after $n$ months is the sum of the first $n$ terms of this AP, denoted by $S_n$.

We are given that $S_n = 2000$.

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

Substitute the given values of $S_n$, $a$, and $d$ into the formula:

From equation (i), simplify the expression inside the parenthesis:

$2000 = \frac{n}{2}(64 + 4n - 4)$

$2000 = \frac{n}{2}(4n + 60)$

Distribute the $\frac{n}{2}$ term:

$2000 = \frac{n}{2}(4n) + \frac{n}{2}(60)$

$2000 = 2n^2 + 30n$

Rearrange the equation into a standard quadratic form ($An^2 + Bn + C = 0$):

$2n^2 + 30n - 2000 = 0$

Divide the entire equation by 2 to simplify the coefficients:

Solve the quadratic equation (ii) for $n$. We can use the quadratic formula, $n = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$.

Here, $A=1$, $B=15$, and $C=-1000$.

From equation (iii), calculate the value under the square root (the discriminant):

Discriminant $= 15^2 - 4(1)(-1000) = 225 + 4000 = 4225$.

Now, calculate the square root of the discriminant:

$\sqrt{4225} = 65$

Substitute this value back into the quadratic formula for $n$:

$n = \frac{-15 \pm 65}{2}$

This gives two possible values for $n$:

$n_1 = \frac{-15 + 65}{2} = \frac{50}{2} = 25$

$n_2 = \frac{-15 - 65}{2} = \frac{-80}{2} = -40$

Since $n$ represents the number of months, it must be a positive integer. A negative number of months does not make sense in this context.

Therefore, the valid value for $n$ is 25.

Yasmeen will save $\textsf{₹}2000$ in 25 months.

The final answer is 25 months.

Given:

The sum of four consecutive numbers in an AP is 32.

The ratio of the product of the first and the last terms to the product of the two middle terms is 7 : 15.

To Find:

The four consecutive numbers in the AP.

Solution:

Let the four consecutive numbers in the Arithmetic Progression be $a - 3d$, $a - d$, $a + d$, and $a + 3d$. In this representation, the common difference of the AP is $(a-d) - (a-3d) = 2d$.

According to the first condition, the sum of these four numbers is 32:

$(a - 3d) + (a - d) + (a + d) + (a + 3d) = 32$

$a + a + a + a - 3d - d + d + 3d = 32$

$4a = 32$

Divide by 4:

According to the second condition, the ratio of the product of the first and the last terms to the product of the two middle terms is 7 : 15.

First term = $a - 3d$

Last term = $a + 3d$

Middle terms = $a - d$ and $a + d$

The ratio is:

Using the difference of squares formula, $(x-y)(x+y) = x^2 - y^2$, we simplify equation (2):

$\frac{a^2 - (3d)^2}{a^2 - d^2} = \frac{7}{15}$

Now, substitute the value of $a=8$ from equation (1) into equation (3):

From equation (4):

Cross-multiply equation (5):

$15(64 - 9d^2) = 7(64 - d^2)$

$15 \times 64 - 15 \times 9d^2 = 7 \times 64 - 7d^2$

$960 - 135d^2 = 448 - 7d^2$

Rearrange the terms to solve for $d^2$:

$960 - 448 = 135d^2 - 7d^2$

$512 = 128d^2$

$d^2 = \frac{512}{128}$

$d^2 = 4$

Taking the square root of both sides:

Now we find the four numbers using $a=8$ and the two values of $d$ from equation (6).

Case 1: $d = 2$

The numbers are:

$a - 3d = 8 - 3(2) = 8 - 6 = 2$

$a - d = 8 - 2 = 6$

$a + d = 8 + 2 = 10$

$a + 3d = 8 + 3(2) = 8 + 6 = 14$

The numbers are 2, 6, 10, 14. The common difference of this AP is $2d = 2(2) = 4$.

Case 2: $d = -2$

The numbers are:

$a - 3d = 8 - 3(-2) = 8 + 6 = 14$

$a - d = 8 - (-2) = 8 + 2 = 10$

$a + d = 8 + (-2) = 8 - 2 = 6$

$a + 3d = 8 + 3(-2) = 8 - 6 = 2$

The numbers are 14, 10, 6, 2. The common difference of this AP is $2d = 2(-2) = -4$.

Both sets of numbers satisfy the given conditions.

The four consecutive numbers in the AP are 2, 6, 10, and 14 (or in reverse order).

Given:

The given equation is the sum of a series: $1 + 4 + 7 + 10 + \dots + x = 287$.

To Find:

The value of $x$.

Solution:

The given series $1 + 4 + 7 + 10 + \dots + x$ is an Arithmetic Progression (AP).

The first term is $a = 1$.

The common difference is $d = 4 - 1 = 3$. We can verify this with other terms: $7 - 4 = 3$, $10 - 7 = 3$.

Let the number of terms in the AP be $n$. The last term is $x$, which is the $n$-th term, $a_n$.

The sum of the first $n$ terms is given as $S_n = 287$.

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

Substitute the given values $S_n = 287$, $a = 1$, and $d = 3$ into the formula:

From equation (i), we simplify the expression:

$287 = \frac{n}{2}(2 + 3n - 3)$

$287 = \frac{n}{2}(3n - 1)$

Multiply both sides by 2:

$2 \times 287 = n(3n - 1)$

$574 = 3n^2 - n$

Rearrange the equation into a standard quadratic form ($An^2 + Bn + C = 0$):

We solve the quadratic equation (ii) for $n$ using the quadratic formula, $n = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$.

Here, $A=3$, $B=-1$, and $C=-574$.

From equation (iii), calculate the value under the square root (the discriminant):

Discriminant $= (-1)^2 - 4(3)(-574) = 1 - 12(-574) = 1 + 6888 = 6889$.

The square root of the discriminant is $\sqrt{6889} = 83$.

Substitute this value back into the quadratic formula:

$n = \frac{1 \pm 83}{6}$

This gives two possible values for $n$:

$n_1 = \frac{1 + 83}{6} = \frac{84}{6} = 14$

$n_2 = \frac{1 - 83}{6} = \frac{-82}{6} = -\frac{41}{3}$

Since $n$ represents the number of terms in the AP, it must be a positive integer. Therefore, $n = 14$ is the valid solution.

The last term of the series is $x$, which is the 14th term ($a_{14}$).

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

Substitute $a=1$, $d=3$, and $n=14$ to find $x$:

From equation (iv):

$x = 1 + (13)3$

$x = 1 + 39$}

$x = 40$}

The value of $x$ is 40.

The final answer is $x = 40$.

Given:

Sum of the first 5 terms and the first 7 terms of an AP is 167, i.e., $S_5 + S_7 = 167$.

Sum of the first 10 terms of the same AP is 235, i.e., $S_{10} = 235$.

To Find:

The sum of the first 20 terms of the AP, $S_{20}$.

Solution:

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

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

Using the given information $S_5 + S_7 = 167$, we can write:

$S_5 = \frac{5}{2}(2a + (5-1)d) = \frac{5}{2}(2a + 4d) = 5(a + 2d) = 5a + 10d$

$S_7 = \frac{7}{2}(2a + (7-1)d) = \frac{7}{2}(2a + 6d) = 7(a + 3d) = 7a + 21d$

So, $S_5 + S_7 = (5a + 10d) + (7a + 21d) = 12a + 31d$.

We are given $S_5 + S_7 = 167$. Therefore,

Using the given information $S_{10} = 235$, we can write:

$S_{10} = \frac{10}{2}(2a + (10-1)d) = 5(2a + 9d) = 10a + 45d$

We are given $S_{10} = 235$. Therefore,

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

(1) $12a + 31d = 167$

(2) $10a + 45d = 235$

Divide equation (2) by 5 to simplify it:

From equation (iii), we get:

From equation (3), express $2a$ in terms of $d$:

Rewrite equation (1) as $6(2a) + 31d = 167$. Substitute the expression for $2a$ from equation (4) into this rewritten equation (1):

From equation (v):

$282 - 54d + 31d = 167$

$282 - 23d = 167$

Subtract 167 from both sides and add $23d$ to both sides:

$282 - 167 = 23d$

$115 = 23d$}

Divide both sides by 23 to find $d$:

$d = \frac{115}{23}$

$d = 5$

Substitute the value of $d=5$ back into equation (3) to find $a$:

From equation (vi):

$2a + 45 = 47$

$2a = 47 - 45$}

$2a = 2$

Divide both sides by 2 to find $a$:

$a = \frac{2}{2}$

$a = 1$}

So, the first term of the AP is $a = 1$ and the common difference is $d = 5$.

Now, we need to find the sum of the first 20 terms, $S_{20}$. Use the formula for the sum of the first $n$ terms with $n=20$, $a=1$, and $d=5$:

From equation (vii):

$S_{20} = 10(2(1) + 19(5))$

$S_{20} = 10(2 + 95)$

$S_{20} = 10(97)$

$S_{20} = 970$

The sum of the first twenty terms of the AP is 970.

The final answer is $S_{20} = 970$.

Solution (i):

We need to find the sum of integers between 1 and 500 that are multiples of both 2 and 5.

Integers that are multiples of both 2 and 5 are multiples of their Least Common Multiple (LCM).

LCM(2, 5) = 10.

We are looking for multiples of 10 strictly between 1 and 500. This means the numbers must be greater than 1 and less than 500.

The first multiple of 10 greater than 1 is 10.

The last multiple of 10 less than 500 is 490.

The sequence of these numbers is an Arithmetic Progression (AP): 10, 20, 30, ..., 490.

The first term is $a = 10$.

The common difference is $d = 10$.

The last term is $a_n = 490$.

Let $n$ be the number of terms in this AP. The formula for the $n$-th term is $a_n = a + (n-1)d$.

From equation (i.1):

$490 - 10 = (n-1)10$}

$480 = 10(n-1)$

Divide both sides by 10:

$\frac{480}{10} = n-1$

$48 = n-1$

$n = 48 + 1 = 49$

There are 49 terms in this AP.

Now, find the sum of these 49 terms using the formula for the sum of an AP when the first and last terms are known: $S_n = \frac{n}{2}(a + a_n)$.

From equation (i.2):

$S_{49} = \frac{49}{2}(500)$

$S_{49} = 49 \times \frac{500}{2}$

$S_{49} = 49 \times 250$

$S_{49} = 12250$

The sum of the integers between 1 and 500 which are multiples of 2 as well as of 5 is 12250.

The final answer for part (i) is 12250.

Solution (ii):

We need to find the sum of integers from 1 to 500 that are multiples of both 2 and 5.

These are multiples of LCM(2, 5) = 10.

We are looking for multiples of 10 between 1 and 500, inclusive. This means $1 \leq \text{number} \leq 500$.

The first multiple of 10 greater than or equal to 1 is 10.

The last multiple of 10 less than or equal to 500 is 500.

The sequence of these numbers is an AP: 10, 20, 30, ..., 500.

The first term is $a = 10$.

The common difference is $d = 10$.

The last term is $a_n = 500$.

Let $n$ be the number of terms in this AP. The formula for the $n$-th term is $a_n = a + (n-1)d$.

From equation (ii.1):

$500 - 10 = (n-1)10$}

$490 = 10(n-1)$

Divide both sides by 10:

$\frac{490}{10} = n-1$

$49 = n-1$

$n = 49 + 1 = 50$

There are 50 terms in this AP.

Now, find the sum of these 50 terms using the formula $S_n = \frac{n}{2}(a + a_n)$.

From equation (ii.2):

$S_{50} = 25(510)$

$S_{50} = 12750$

The sum of the integers from 1 to 500 which are multiples of 2 as well as of 5 is 12750.

The final answer for part (ii) is 12750.

Solution (iii):

We need to find the sum of integers from 1 to 500 that are multiples of 2 or 5.

Using the Principle of Inclusion-Exclusion, the sum of integers which are multiples of 2 or 5 is given by:

$S_{\text{multiples of 2 or 5}} = S_{\text{multiples of 2}} + S_{\text{multiples of 5}} - S_{\text{multiples of both 2 and 5}}$.

First, find the sum of multiples of 2 from 1 to 500.

The sequence is 2, 4, 6, ..., 500.

This is an AP with first term $a = 2$, common difference $d = 2$, and last term $a_n = 500$.

Find the number of terms, $n_2$:

From equation (iii.1):

$498 = (n_2-1)2$

Divide both sides by 2:

$249 = n_2-1$

$n_2 = 249 + 1 = 250$

Find the sum, $S_{n_2}$, using $S_n = \frac{n}{2}(a + a_n)$: $S_{250} = \frac{250}{2}(2 + 500) = 125(502)$.

$125 \times 502 = 62750$.

So, the sum of multiples of 2 from 1 to 500 is 62750.

Next, find the sum of multiples of 5 from 1 to 500.

The sequence is 5, 10, 15, ..., 500.

This is an AP with first term $a = 5$, common difference $d = 5$, and last term $a_n = 500$.

Find the number of terms, $n_5$:

From equation (iii.2):

$495 = (n_5-1)5$

Divide both sides by 5:

$99 = n_5-1$

$n_5 = 99 + 1 = 100$

Find the sum, $S_{n_5}$, using $S_n = \frac{n}{2}(a + a_n)$: $S_{100} = \frac{100}{2}(5 + 500) = 50(505)$.

$50 \times 505 = 25250$.

So, the sum of multiples of 5 from 1 to 500 is 25250.

Finally, find the sum of multiples of both 2 and 5 (i.e., multiples of 10) from 1 to 500.

This was calculated in part (ii).

$S_{\text{multiples of 10}} = 12750$.

Now, apply the Principle of Inclusion-Exclusion:

$S_{\text{multiples of 2 or 5}} = S_{\text{multiples of 2}} + S_{\text{multiples of 5}} - S_{\text{multiples of 10}}$

From equation (iii.3):

$S_{\text{multiples of 2 or 5}} = 88000 - 12750$

$S_{\text{multiples of 2 or 5}} = 75250$

The sum of the integers from 1 to 500 which are multiples of 2 or 5 is 75250.

The final answer for part (iii) is 75250.

Given:

The eighth term of an AP is half its second term: $a_8 = \frac{1}{2}a_2$.

The eleventh term exceeds one third of its fourth term by 1: $a_{11} = \frac{1}{3}a_4 + 1$.

To Find:

The 15th term of the AP, $a_{15}$.

Solution:

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

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

From the first condition, $a_8 = \frac{1}{2}a_2$:

$a + (8-1)d = \frac{1}{2}(a + (2-1)d)$

$a + 7d = \frac{1}{2}(a + d)$

Multiply both sides by 2:

$2(a + 7d) = a + d$}

$2a + 14d = a + d$}

Rearrange the terms:

$2a - a + 14d - d = 0$

From the second condition, $a_{11} = \frac{1}{3}a_4 + 1$:

$a + (11-1)d = \frac{1}{3}(a + (4-1)d) + 1$

$a + 10d = \frac{1}{3}(a + 3d) + 1$

$a + 10d = \frac{a}{3} + \frac{3d}{3} + 1$

$a + 10d = \frac{a}{3} + d + 1$

Multiply both sides by 3 to clear the denominator:

$3(a + 10d) = 3\left(\frac{a}{3} + d + 1\right)$

$3a + 30d = a + 3d + 3$}

Rearrange the terms:

$3a - a + 30d - 3d = 3$

Now we have a system of two linear equations (1) and (2) with two variables, $a$ and $d$:

(1) $a + 13d = 0$}

(2) $2a + 27d = 3$}

From equation (1), express $a$ in terms of $d$:

Substitute the expression for $a$ from equation (3) into equation (2):

From equation (iv):

$-26d + 27d = 3$

$d = 3$

Substitute the value of $d=3$ back into equation (3) to find $a$:

From equation (v):

$a = -39$

So, the first term of the AP is $a = -39$ and the common difference is $d = 3$.

Now, we need to find the 15th term of the AP, $a_{15}$. Use the formula $a_n = a + (n-1)d$ with $n=15$, $a=-39$, and $d=3$:

From equation (vi):

$a_{15} = -39 + (14)3$}

$a_{15} = -39 + 42$}

$a_{15} = 3$

The 15th term of the AP is 3.

The final answer is $a_{15} = 3$.

Given:

Number of terms in the AP, $n = 37$.

The sum of the three middle most terms is 225.

The sum of the last three terms is 429.

To Find:

The Arithmetic Progression (AP).

Solution:

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

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

Since the number of terms is 37 (which is odd), the middle most term is the $(\frac{n+1}{2})$-th term.

Position of the middle term = $\frac{37+1}{2} = \frac{38}{2} = 19$-th term.

The three middle most terms are the term before the 19th, the 19th term, and the term after the 19th.

These are the 18th, 19th, and 20th terms.

The three middle most terms are $a_{18}$, $a_{19}$, and $a_{20}$.

$a_{18} = a + (18-1)d = a + 17d$

$a_{19} = a + (19-1)d = a + 18d$

$a_{20} = a + (20-1)d = a + 19d$

The sum of the three middle most terms is 225:

$a_{18} + a_{19} + a_{20} = 225$

$(a + 17d) + (a + 18d) + (a + 19d) = 225$

$3a + (17 + 18 + 19)d = 225$

$3a + 54d = 225$

Divide the entire equation by 3:

The last three terms are the 35th, 36th, and 37th terms ($a_{35}$, $a_{36}$, and $a_{37}$).

$a_{35} = a + (35-1)d = a + 34d$

$a_{36} = a + (36-1)d = a + 35d$

$a_{37} = a + (37-1)d = a + 36d$

The sum of the last three terms is 429:

$a_{35} + a_{36} + a_{37} = 429$

$(a + 34d) + (a + 35d) + (a + 36d) = 429$

$3a + (34 + 35 + 36)d = 429$

$3a + 105d = 429$

Divide the entire equation by 3:

Now we have a system of two linear equations (1) and (2) with two variables, $a$ and $d$:

(1) $a + 18d = 75$}

(2) $a + 35d = 143$

Subtract equation (1) from equation (2) to eliminate $a$:

From equation (iii):

$a + 35d - a - 18d = 68$

$17d = 68$

Divide both sides by 17 to find $d$:

$d = \frac{68}{17}$

$d = 4$

Substitute the value of $d = 4$ into equation (1) to find $a$:

From equation (iv):

$a + 72 = 75$

$a = 75 - 72$}

$a = 3$

The first term is $a = 3$ and the common difference is $d = 4$.

The AP is given by $a, a+d, a+2d, a+3d, \dots$

The terms are:

1st term: $a = 3$

2nd term: $a+d = 3 + 4 = 7$

3rd term: $a+2d = 3 + 2(4) = 3 + 8 = 11$

and so on.

The Arithmetic Progression is 3, 7, 11, 15, ...

The final answer is The AP is 3, 7, 11, 15, ....

Solution (i):

We need to find the sum of integers between 100 and 200 that are divisible by 9.

The integers are strictly greater than 100 and strictly less than 200.

The first multiple of 9 greater than 100 is found by dividing 100 by 9: $100 \div 9 \approx 11.11$. The next integer is 12. So, the first multiple of 9 is $12 \times 9 = 108$.

The last multiple of 9 less than 200 is found by dividing 200 by 9: $200 \div 9 \approx 22.22$. The previous integer is 22. So, the last multiple of 9 is $22 \times 9 = 198$.

The sequence of these numbers is an Arithmetic Progression (AP): 108, 117, 126, ..., 198.

The first term is $a = 108$.

The common difference is $d = 9$.

The last term is $a_n = 198$.

Let $n$ be the number of terms in this AP. The formula for the $n$-th term is $a_n = a + (n-1)d$.

From equation (i.1):

$198 - 108 = (n-1)9$}

$90 = 9(n-1)$

Divide both sides by 9:

$\frac{90}{9} = n-1$

$10 = n-1$

$n = 10 + 1 = 11$

There are 11 terms in this AP.

Now, find the sum of these 11 terms using the formula for the sum of an AP when the first and last terms are known: $S_n = \frac{n}{2}(a + a_n)$.

From equation (i.2):

$S_{11} = \frac{11}{2}(306)$

$S_{11} = 11 \times \frac{306}{2}$}

$S_{11} = 11 \times 153$}

$S_{11} = 1683$}

The sum of the integers between 100 and 200 that are divisible by 9 is 1683.

The final answer for part (i) is 1683.

Solution (ii):

We need to find the sum of the integers between 100 and 200 that are not divisible by 9.

The integers are strictly between 100 and 200, which means they are from 101 to 199, inclusive.

The total number of integers between 100 and 200 is $199 - 101 + 1 = 99$.

These integers form an AP: 101, 102, ..., 199.

The first term is $A = 101$.

The common difference is $D = 1$.

The last term is $A_{total} = 199$.

The number of terms is $N = 99$.

The sum of all integers between 100 and 200 (i.e., from 101 to 199) is $S_{total} = \frac{N}{2}(A + A_{total})$.

From equation (ii.1):

$S_{total} = \frac{99}{2}(300)$

$S_{total} = 99 \times 150$}

$S_{total} = 14850$}

The sum of the integers between 100 and 200 that are not divisible by 9 is equal to the sum of all integers between 100 and 200 minus the sum of integers between 100 and 200 that are divisible by 9.

Sum of integers not divisible by 9 = $S_{total} - S_{\text{divisible by 9}}$

From part (i), $S_{\text{divisible by 9}} = 1683$.

From equation (ii.2):

Sum of integers not divisible by 9 = $13167$

The sum of the integers between 100 and 200 that are not divisible by 9 is 13167.

The final answer for part (ii) is 13167.

Given:

The ratio of the 11th term to the 18th term of an AP is 2 : 3.

$\frac{a_{11}}{a_{18}} = \frac{2}{3}$

To Find:

1. The ratio of the 5th term to the 21st term ($\frac{a_5}{a_{21}}$).

2. The ratio of the sum of the first five terms to the sum of the first 21 terms ($\frac{S_5}{S_{21}}$).

Solution:

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

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

Using the given ratio $\frac{a_{11}}{a_{18}} = \frac{2}{3}$:

$a_{11} = a + (11-1)d = a + 10d$

$a_{18} = a + (18-1)d = a + 17d$

So, we have:

Cross-multiply equation (i):

$3(a + 10d) = 2(a + 17d)$

$3a + 30d = 2a + 34d$}

Rearrange the terms to find a relationship between $a$ and $d$:

$3a - 2a = 34d - 30d$

Part 1: Ratio of the 5th term to the 21st term ($\frac{a_5}{a_{21}}$)

The 5th term is $a_5 = a + (5-1)d = a + 4d$.

The 21st term is $a_{21} = a + (21-1)d = a + 20d$.

The ratio is $\frac{a_5}{a_{21}} = \frac{a + 4d}{a + 20d}$.

Substitute the relationship $a = 4d$ from equation (1) into this ratio:

From equation (ii):

$\frac{a_5}{a_{21}} = \frac{8d}{24d}$

Assuming $d \neq 0$ (if $d=0$, all terms are equal, and the initial ratio would be 1:1, not 2:3), we can cancel $d$:

The ratio of the 5th term to the 21st term is 1 : 3.

Part 2: Ratio of the sum of the first five terms to the sum of the first 21 terms ($\frac{S_5}{S_{21}}$)

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

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

The sum of the first 21 terms is $S_{21} = \frac{21}{2}(2a + (21-1)d) = \frac{21}{2}(2a + 20d)$.

The ratio is $\frac{S_5}{S_{21}} = \frac{\frac{5}{2}(2a + 4d)}{\frac{21}{2}(2a + 20d)}$.

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

Substitute the relationship $a = 4d$ from equation (1) into equation (iv):

From equation (v):

$\frac{S_5}{S_{21}} = \frac{5(8d + 4d)}{21(8d + 20d)}$

$\frac{S_5}{S_{21}} = \frac{5(12d)}{21(28d)}$

$\frac{S_5}{S_{21}} = \frac{60d}{588d}$

Assuming $d \neq 0$, cancel $d$ and simplify the fraction:

The ratio of the sum of the first five terms to the sum of the first 21 terms is 5 : 49.

The ratio of the 5th term to the 21st term is 1 : 3.

The ratio of the sum of the first five terms to the sum of the first 21 terms is 5 : 49.

The final answer is The ratio of the 5th term to the 21st term is 1 : 3. The ratio of the sum of the first five terms to the sum of the first 21 terms is 5 : 49.

Given:

First term of the AP, $a_1 = a$.

Second term of the AP, $a_2 = b$.

Last term of the AP, $a_n = c$, where $n$ is the number of terms.

To Show:

The sum of the AP, $S_n$, is equal to $\frac{(a \;+\; c) (b \;+\; c \;-\; 2a)}{2 (b \;-\; a)}$.

Proof:

Let the first term of the AP be $a_1 = a$ and the common difference be $d$.

The common difference $d$ is the difference between the second term and the first term:

We assume $b \neq a$, otherwise the common difference is 0 and the AP is constant ($a, a, a, \dots$), which would imply $c=a=b$ and the denominator of the expression we need to show would be zero.

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

We are given the last term is $c$, so $a_n = c$. Substitute the values of $a_1$ and $d$:

From equation (2), we solve for the number of terms, $n$:

$c - a = (n-1)(b - a)$

Divide both sides by $(b - a)$ (which is non-zero):

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

Add 1 to both sides:

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

$n = \frac{c - a + (b - a)}{b - a}$

The formula for the sum of the first $n$ terms of an AP when the first term ($a_1$) and the last term ($a_n$) are known is $S_n = \frac{n}{2}(a_1 + a_n)$.

Substitute the expressions for $n$, $a_1$, and $a_n$ from equations (3), (given), and (given) into the sum formula:

From equation (4), we simplify the expression:

$S_n = \frac{1}{2} \times \frac{b + c - 2a}{b - a} \times (a + c)$

$S_n = \frac{(b + c - 2a)(a + c)}{2(b - a)}$

Rearrange the terms in the numerator to match the required format:

We have successfully shown that the sum of the AP is equal to the given expression.

Hence, the sum of the AP is $\frac{(a \;+\; c) (b \;+\; c \;-\; 2a)}{2 (b \;-\; a)}$.

The proof is complete.

Given:

The given equation is the sum of a series: $-4 + (-1) + 2 + \dots + x = 437$.

To Find:

The value of $x$.

Solution:

The given series $-4 + (-1) + 2 + \dots + x$ is an Arithmetic Progression (AP).

The first term is $a = -4$.

The common difference is $d = (-1) - (-4) = -1 + 4 = 3$. We can verify this with the next term: $2 - (-1) = 2 + 1 = 3$.

Let the number of terms in the AP be $n$. The last term is $x$, which is the $n$-th term, $a_n$.

The sum of the first $n$ terms is given as $S_n = 437$.

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

Substitute the given values $S_n = 437$, $a = -4$, and $d = 3$ into the formula:

From equation (i), we simplify the expression:

$437 = \frac{n}{2}(-8 + 3n - 3)$

$437 = \frac{n}{2}(3n - 11)$

Multiply both sides by 2:

$2 \times 437 = n(3n - 11)$

$874 = 3n^2 - 11n$

Rearrange the equation into a standard quadratic form ($An^2 + Bn + C = 0$):

We solve the quadratic equation (ii) for $n$ using the quadratic formula, $n = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$.

Here, $A=3$, $B=-11$, and $C=-874$.

From equation (iii), calculate the value under the square root (the discriminant):

Discriminant $= (-11)^2 - 4(3)(-874) = 121 - 12(-874) = 121 + 10488 = 10609$.

The square root of the discriminant is $\sqrt{10609} = 103$.

Substitute this value back into the quadratic formula:

$n = \frac{11 \pm 103}{6}$

This gives two possible values for $n$:

$n_1 = \frac{11 + 103}{6} = \frac{114}{6} = 19$

$n_2 = \frac{11 - 103}{6} = \frac{-92}{6} = -\frac{46}{3}$

Since $n$ represents the number of terms in the AP, it must be a positive integer. Therefore, $n = 19$ is the valid solution.

The last term of the series is $x$, which is the 19th term ($a_{19}$).

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

Substitute $a=-4$, $d=3$, and $n=19$ to find $x$:

From equation (iv):

$x = -4 + (18)3$}

$x = -4 + 54$}

$x = 50$}

The value of $x$ is 50.

The final answer is $x = 50$.

Given:

Total loan amount = $\textsf{₹}118000$.

First instalment = $\textsf{₹}1000$.

Increase in instalment every month = $\textsf{₹}100$.

To Find:

1. The amount paid in the 30th instalment.

2. The amount of loan remaining after the 30th instalment.

Solution:

The monthly instalments form an Arithmetic Progression (AP). The first term is the first instalment, and the common difference is the monthly increase.

First term of the AP, $a = 1000$.

Common difference of the AP, $d = 100$.

Part 1: Amount paid in the 30th instalment

The amount paid in the 30th instalment is the 30th term of the AP, $a_{30}$.

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

Substitute $n=30$, $a=1000$, and $d=100$ into the formula:

From equation (i):

$a_{30} = 1000 + (29)100$}

$a_{30} = 1000 + 2900$}

$a_{30} = 3900$}

The amount paid in the 30th instalment is $\textsf{₹}3900$.

Part 2: Amount of loan remaining after the 30th instalment

First, find the total amount paid in the first 30 instalments. This is the sum of the first 30 terms of the AP, $S_{30}$.

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

Substitute $n=30$, $a=1000$, and $d=100$ into the formula:

From equation (ii):

$S_{30} = 15(2000 + (29)100)$

$S_{30} = 15(2000 + 2900)$

$S_{30} = 15(4900)$

$S_{30} = 73500$}

The total amount paid in the first 30 instalments is $\textsf{₹}73500$.

The amount of loan still to pay after the 30th instalment is the total loan amount minus the total amount paid in the first 30 instalments.

Remaining loan amount = Total loan amount - $S_{30}$.

Remaining loan amount = $\textsf{₹}118000 - \textsf{₹}73500$.

Remaining loan amount = $\textsf{₹}44500$.

The amount paid in the 30th instalment is $\textsf{₹}3900$. The amount of loan remaining after the 30th instalment is $\textsf{₹}44500$.

The final answer is Amount paid in 30th instalment = $\textsf{₹}3900$, Amount of loan remaining = $\textsf{₹}44500$.

Given:

Total number of flags = 27.

Interval between flags = 2 m.

Flags are stored at the position of the middle most flag.

Ruchi carries one flag at a time and returns to the storage position each time.

Ruchi ends by returning to the storage position (where her books are).

To Find:

1. Total distance covered by Ruchi in completing the job and returning to her books.

2. The maximum distance she travelled carrying a flag.

Solution:

The flags are placed at intervals of 2m along a straight passage. Let's denote the positions where the flags are fixed as $P_1, P_2, \dots, P_{27}$.

The number of flags is 27, which is an odd number. The middle most flag is the $(\frac{27+1}{2})$-th flag, which is the 14th flag ($P_{14}$).

The flags are stored at the position of the 14th flag, $P_{14}$. This is Ruchi's starting point and the location of her books.

The flags are placed symmetrically around the middle flag. There are $27 - 1 = 26$ intervals between the 27 flags.

There are $(27-1)/2 = 13$ flags on one side of the middle flag and 13 flags on the other side.

The flags are at positions relative to the storage point ($P_{14}$):

To one side: $P_{15}, P_{16}, \dots, P_{27}$ (13 flags).

To the other side: $P_{13}, P_{12}, \dots, P_1$ (13 flags).

The distance between adjacent flags is 2m.

The distance from the storage point ($P_{14}$) to $P_{15}$ is 2m.

The distance from the storage point ($P_{14}$) to $P_{13}$ is 2m.

The distance from the storage point ($P_{14}$) to $P_{16}$ is $2 \times 2 = 4$m.

The distance from the storage point ($P_{14}$) to $P_{12}$ is $2 \times 2 = 4$m.

In general, the distance from the storage point ($P_{14}$) to the $k$-th flag position away from the center is $k \times 2$ meters.

The flags are at distances of 2m, 4m, 6m, ..., $13 \times 2 = 26$m from the storage point, on each side.

Ruchi takes one flag from the storage, goes to a position, fixes the flag, and returns to the storage. She does this for each of the 26 flags that are not at the storage location.

Consider the flags on one side (say, $P_{15}$ to $P_{27}$). There are 13 such flags.

For the flag at $P_{15}$ (distance 2m from storage): Ruchi goes 2m, fixes flag, returns 2m. Total distance = $2 \times 2$ m.

For the flag at $P_{16}$ (distance 4m from storage): Ruchi goes 4m, fixes flag, returns 4m. Total distance = $2 \times 4$ m.

... For the flag at $P_{27}$ (distance 26m from storage): Ruchi goes 26m, fixes flag, returns 26m. Total distance = $2 \times 26$ m.

The total distance covered for the 13 flags on one side is the sum of the distances for each flag:

Distance for one side $= (2 \times 2) + (2 \times 4) + (2 \times 6) + \dots + (2 \times 26)$

Distance for one side $= 2(2 + 4 + 6 + \dots + 26)$

The series $2 + 4 + 6 + \dots + 26$ is an AP with first term $a=2$, common difference $d=2$, and number of terms $n=13$.

The sum of this AP can be calculated using $S_n = \frac{n}{2}(a + a_n)$:

Sum of distances to positions on one side $= \frac{13}{2}(2 + 26) = \frac{13}{2}(28) = 13 \times 14 = 182$ m.

The total distance covered for placing these 13 flags and returning each time is $2 \times (\text{Sum of distances to positions on one side}) = 2 \times 182 = 364$ m.

Since there are 13 flags on the other side ($P_1$ to $P_{13}$) at the same distances (2m, 4m, ..., 26m), the total distance covered for these 13 flags is also 364 m.

The flag at the storage position ($P_{14}$) does not require any travel to place it.

The total distance covered by Ruchi in completing this job (placing the 26 flags) and returning to collect her books (which are at the storage location) is the sum of the distances for each side.

Total distance = Distance for the 13 flags on one side + Distance for the 13 flags on the other side.

Total distance = 364 m + 364 m = 728 m.

Part 2: Maximum distance travelled carrying a flag

Ruchi carries a flag from the storage point ($P_{14}$) to a flag position. The distance travelled while carrying a flag is the distance from the storage point to the position where the flag is fixed.

The flags are fixed at distances of 2m, 4m, 6m, ..., 26m from the storage point.

The maximum distance she travelled carrying a flag is the distance to the farthest flag position from the storage point.

The farthest flag positions are $P_1$ and $P_{27}$.

The distance from $P_{14}$ to $P_1$ is $(14-1) \times 2 = 13 \times 2 = 26$ m.

The distance from $P_{14}$ to $P_{27}$ is $(27-14) \times 2 = 13 \times 2 = 26$ m.

The maximum distance she travelled carrying a flag is 26 m.

The total distance covered is 728 m. The maximum distance she travelled carrying a flag is 26 m.

The final answer is Total distance covered = 728 m, Maximum distance travelled carrying a flag = 26 m.