CBSE Class 10 Maths HOTs Arithmetic Progressions Set K

Very Short Answer Type Questions

Question. Find the sum of the first 100 natural numbers.
Answer: The list of first 100 natural numbers: 1, 2, 3, ......, 100, which forms an AP with \( a = 1, d = 1 \)
So, \( S_{100} = \frac{100}{2} [2 \times 1 + (100 - 1)(1)] \)
\( = 50 [101] \)
\( = 5050 \)

Question. If the sum of the first ‘p’ terms of an A.P. is ‘q’ and sum of the first ‘q’ terms is ‘p’; then show that the sum of the first (p + q) terms is -(p + q).
Answer: Consider an A.P., whose first term be \( 'a' \) and common difference be \( 'd' \).

Question. If the mean of the first \( n \) natural number is 15, then find \( n \).
Answer: The natural numbers are, 1, 2, 3 ..... \( n \)
Their mean = \( \frac{S_n}{n} \) and \( S_n = \frac{n}{2}[2a + (n - 1)d] \)
The mean of first \( n \) natural numbers is \( \frac{n(n+1)}{2n} = \frac{n+1}{2} \)
\( \therefore \frac{n+1}{2} = 15 \) (given)
\( \Rightarrow n = 29 \)

Question. If in an A.P., \( a = 15, d = -3 \) and \( a_n = 0 \), then find the value of \( n \).
Answer: Given, for an A.P., \( a = 15, d = -3 \) and \( a_n = 0 \)
Now, \( a_n = a + (n - 1)d \)
\( \Rightarrow 0 = 15 + (n - 1) \times (-3) \)
\( \Rightarrow (n - 1) = 5 \)
\( \Rightarrow n = 6 \)
Hence, the value of \( n \) is 6.

Question. In an AP, if the common difference (\( d \)) = – 4, and the seventh term (\( a_7 \)) is 4, then find the first term.
Answer: Given: common difference (\( d \)) of an A.P. = – 4
Seventh term (\( a_7 \)) of A.P. = 4
Let, ‘\( a \)’ be the first term of the A.P.
Then, \( a_7 = a + (n - 1)d \)
\( \Rightarrow 4 = a + (7 - 1)(-4) \)
\( \Rightarrow 4 = a - 24 \)
\( \Rightarrow a = 28 \)
Hence, the first term of the A.P. is 28.

Question. Write the \( n \)th term of the A.P. \( \frac{1}{m}, \frac{1+m}{m}, \frac{1+2m}{m}, ......... \)
Answer: Given, A.P. is \( \frac{1}{m}, \frac{1+m}{m}, \frac{1+2m}{m}, ...... \)
Here, first term \( a = \frac{1}{m} \)
Common difference, \( d = \frac{1+m}{m} - \frac{1}{m} = \frac{m}{m} = 1 \)
\( \therefore n \)th term, \( a_n = a + (n - 1)d \)
\( = \frac{1}{m} + (n - 1) \times 1 \)
\( = \frac{1}{m} + n - 1 \)
\( = \frac{1 + mn - m}{m} \)
Hence, the \( n \)th term of given A.P. is \( \frac{mn - m + 1}{m} \).

Question. Find the 9th term from the end (towards the first term) of the A.P. 5, 9, 13, ...., 185.
Answer: Given A.P. is 5, 9, 13, ...., 185, Since we have to find the 9th term from the end, We should reverse the A.P. Then, it will become easier to find the 9th term from the starting:
185, 181, ......, 13, 9, 5
Now, first term, \( a = 185 \)
Common difference, \( d = 181 - 185 = -4 \)
Then, 9th term of A.P.,
\( a_9 = a + (9 - 1) \times d \)
\( = 185 + 8 \times (-4) \)
\( = 185 - 32 = 153 \)
Hence, the 9th term from the end is 153.

Question. If the first three terms of an A.P. are \( b, c \) and \( 2b \), then find the ratio of \( b \) and \( c \).
Answer: \( b, c \) and \( 2b \) are in A.P.
\( \Rightarrow c = \frac{b + 2b}{2} \)
\( \Rightarrow 2c = 3b \)
\( \therefore b : c = 2 : 3 \)

Question. Find the 16th term of the AP: 2, 7, 12, 17, .....
Answer: Here, first term, \( a = 2 \) and common difference, \( d = 5 \).
Using formula, \( n \)th term,
\( a_n = a + (n - 1)d \)
So, \( a_{16} = a + 15d \)
\( = 2 + 15 \times 5 \)
\( = 2 + 75 = 77 \)

Short Answer Type Questions

Question. If the sum of the first 9 terms of an AP is equal to the sum of its first 11 terms, then find the sum of its first 20 terms.
Answer: Let, the first term of an A.P. be ‘\( a \)’ and its common difference be ‘\( d \)’.
Given, \( S_9 = S_{11} \)
\( \Rightarrow \frac{9}{2} [2a + (9 - 1)d] = \frac{11}{2} [2a + (11 - 1)d] \)
\( \Rightarrow 9[2a + 8d] = 11[2a + 10d] \)
\( \Rightarrow 18a + 72d = 22a + 110d \)
\( \Rightarrow 4a = - 38d \)
\( \Rightarrow 2a = - 19d \) ...(i)
Now, sum of first 20 terms:
\( S_{20} = \frac{20}{2} [2a + (20 - 1)d] \)
\( = 10[2a + 19d] \)
\( = 10[-19d + 19d] \) [from (i)]
\( = 10 \times 0 = 0 \)
Hence, the sum of first 20 terms is 0.

Question. In an AP, if \( S_5 + S_7 = 167 \) and \( S_{10} = 235 \), then find the AP, where \( S_n \) denotes the sum of its first \( n \) terms.
Answer: \( S_5 + S_7 = 167 \)
\( \Rightarrow \frac{5}{2}[2a + 4d] + \frac{7}{2}[2a + 6d] = 167 \)
\( 24a + 62d = 334 \)
or \( 12a + 31d = 167 \) ...(i)
\( S_{10} = 235 \)
\( \Rightarrow \frac{10}{2}[2a + 9d] = 235 \)
\( \Rightarrow 5[2a + 9d] = 235 \)
or \( 2a + 9d = 47 \) ...(ii)
Solving (i) and (ii) to get \( a = 1, d = 5 \).
Hence AP is 1, 6, 11, .....

Question. For what value of \( n \), are the \( n \)th terms of two A.Ps 63, 65, 67,... and 3, 10, 17,... equal?
Answer: Let \( a, d \) and \( A, D \) be the 1st term and common difference of the 2 A.Ps respectively.
For 1st A.P: \( a = 63, d = 2 \)
For 2nd A.P: \( A = 3, D = 7 \)
Given \( a_n = A_n \)
\( \Rightarrow a + (n - 1)d = A + (n - 1)D \)
\( 63 + (n - 1)2 = 3 + (n - 1)7 \)
\( 63 + 2n - 2 = 3 + 7n - 7 \)
\( 61 + 2n = 7n - 4 \)
\( 65 = 5n \)
\( 13 = n \)
When \( n \) is 13, the \( n \)th terms are equal.

Question. For an A.P., it is given that the first term (a) = 5, common difference (d) = 3, and the \( n^{th} \) term (\( a_n \)) = 50. Find \( n \) and the sum of first \( n \) terms (\( S_n \)) of the A.P.
Answer: Here, \( a = 5 \), \( d = 3 \) and \( a_n = a + (n – 1)d = 50 \)
\( \Rightarrow 5 + 3 (n – 1) = 50 \)
\( \Rightarrow 3(n – 1) = 45 \)
\( \Rightarrow n – 1 = 15 \)
or \( n = 16 \)
\( \therefore \) Number of terms, \( n = 16 \)
Now, \( S_{16} = \frac{16}{2} [2 \times 5 + (16-1)(3)] \)
[\( \because S_n = \frac{n}{2} (2a + (n – 1)d) \)]
\( = 8[10 + 45] \)
\( = 440 \).

Question. Find the sum of first 15 multiples of 8.
Answer: Multiples of 8 are :
8, 16, 24 ...,
Since, the difference between the numbers is constant, so it forms an A.P.
We need to find the sum of first 15 multiples.
\( S_n = \frac{n}{2} (2a + (n – 1)d) \)
Here, \( n = 15 \), \( a = 8 \) and \( d = 8 \)
\( \therefore S_{15} = \frac{15}{2} [2 \times 8 + (15 – 1) \times 8] \)
\( = \frac{15}{2} (16 + 14 \times 8) \)
\( = \frac{15}{2} [16 + 112] \)
\( = \frac{15}{2} \times 128 = 960 \)
Hence, the sum of first 15 multiples of 8 is 960.

Question. The sum of the first 30 terms of an A.P. is 1920. If the fourth term is 18, find its \( 11^{th} \) term.
Answer: Let \( a \) be the first term and \( d \), the common difference of the AP
Here, number of terms, \( n = 30 \)
Then, \( S_{30} = \frac{30}{2} [2a + 29d] = 1920 \)
or \( 2a + 29d = 128 \) ...(i)
Also, \( a_4 = a + 3d = 18 \) ...(ii)
On solving equations (i) and (ii), we get:
\( d = 4 \) and \( a = 6 \)
Thus, \( a_{11} = a + 10d = 6 + 40 = 46 \).
Hence, \( 11^{th} \) term of the AP is 46.

Question. Which term of the A.P. 20, \( 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, \dots \) is the first negative term.
Answer: Here, first term \( a = 20 \) and common difference \( d = -\frac{3}{4} \)
Let \( a_n \) be the first negative term, i.e.,
\( a + (n – 1)d < 0 \)
\( \Rightarrow 20 - \frac{3}{4}(n - 1) < 0 \)
or \( 20 - \frac{3}{4}n + \frac{3}{4} < 0 \)
or \( -\frac{3}{4}n < -\frac{83}{4} \)
or \( n > \frac{83}{3} \)
So, the nth term is 28.

Question. Find the middle term of the A.P. 7, 13, 19, ...., 247.
Answer: Here, first term, \( a = 7 \), common difference \( d = 6 \)
Let AP contains ‘n’ terms. Then, \( a_n = 247 \)
\( \Rightarrow a + (n – 1)d = 247 \)
\( \Rightarrow 7 + 6(n – 1) = 247 \)
\( \Rightarrow n – 1 = 40 \)
or \( n = 41 \)
So, the middle term is 21st term
\( a_{21} = a + (n – 1)d \)
\( = 7 + 20 \times 6 \)
\( = 7 + 120, i.e., 127 \)

Question. Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.
Answer: Let \( (a – d), a \) and \( (a + d) \) be three parts of 207 such that these are in AP.
It is given that sum of these numbers = 207
\( \Rightarrow (a – d) + a + (a + d) = 207 \)
\( \Rightarrow 3a = 207 \Rightarrow a = 69 \)
It is also given that product of two smaller parts = 4623
\( \Rightarrow a(a – d) = 4623 \)
\( \Rightarrow 69(69 – d) = 4623 \)
\( \Rightarrow 69 – d = \frac{4623}{69} \)
\( \Rightarrow 69 – d = 67 \)
\( \Rightarrow d = 69 – 67 = 2 \)
So, first part \( = a – d = 69 – 2 = 67 \),
second part \( = a = 69 \)
and third part \( = a + d = 69 + 2 = 71 \)
Hence, the three parts are 67, 69 and 71.

Question. How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?
Answer: Numbers that lie between 10 and 300 are 11, 12, ... 299.
Numbers between 10 and 300 which when divided by 4 leave a remainder 3 and are 11, 15, 19, ...299
Here, first number, \( a = 11 \)
and common difference, \( d = 4 \)
number of terms, \( n = ? \)
We know that \( a_n = a + (n – 1)d \)
\( a + (n – 1)d = 299 \)
\( \Rightarrow 11 + (n – 1)(4) = 299 \)
\( \Rightarrow 11 + 4n – 4 = 299 \)
\( \Rightarrow 4n = 292 \Rightarrow n = 73 \)

Question. Show that the sum of all terms of an A.P. whose first term is \( a \), the second term is \( b \) and the last term is \( c \) is equal to \( \frac{(a + c)(b + c - 2a)}{2(b - a)} \)
Answer: Here, first term is \( a \), and, common difference = \( b – a = d \) and last term is \( c \).
Let A.P. contains ‘n’ terms. Then, \( a_n = c \), i.e., \( a + (n – 1)d = c \)
\( \Rightarrow (n - 1)(b - a) = c - a \)
\( \Rightarrow n - 1 = \frac{c - a}{b - a} \)
\( \Rightarrow n = \frac{c - a}{b - a} + 1 = \frac{c - a + b - a}{b - a} = \frac{b + c - 2a}{b - a} \)
Now \( S_n = \frac{n}{2}[a + c] \) [Since c is last term]
\( = \frac{b + c - 2a}{2(b - a)} [a + c] \)
\( = \frac{(a + c)(b + c - 2a)}{2(b - a)} \) Hence, proved.

Question. Find the sum of n terms of the series \( \left(4 - \frac{1}{n}\right) + \left(4 - \frac{2}{n}\right) + \left(4 - \frac{3}{n}\right) + \dots \)
Answer: Given series is:
\( \left(4 - \frac{1}{n}\right) + \left(4 - \frac{2}{n}\right) + \left(4 - \frac{3}{n}\right) + \dots n \) terms.
\( = (4 + 4 + 4 + \dots n \text{ terms}) - \left(\frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \dots + n \text{ terms}\right) \)
\( = 4(1 + 1 + 1 + \dots \text{upto } n \text{ terms}) – \frac{1}{n}(1 + 2 + 3 + \dots + \text{upto } n \text{ terms}) \)
\( = 4n - \frac{1}{n} \times \frac{n(n + 1)}{2} \)
\( = 4n - \frac{(n + 1)}{2} \)
\( = \frac{8n - n - 1}{2} = \frac{7n - 1}{2} \)
Hence, the sum of the series is \( \frac{7n - 1}{2} \).

Question. Find the sum of the first 40 positive integers which give a remainder 1 when divided by 6.
Answer: The numbers which when divided by 6 gives 1 as remainder are :
7, 13, 19, 25, 31, 37, ...
They form an A.P., as their common difference is the same, \( d = 6 \)
Here: first term, \( a = 7 \), common difference, \( d = 6 \)
Sum of first 40 positive integers:
\( \therefore S_{40} = \frac{n}{2} [2a + (n – 1)d] \)
\( = \frac{40}{2} [2 \times 7 + (40 – 1) \times 6] \)
\( = 20[14 + 39 \times 6] \)
\( = 20 \times 248 \)
\( = 4,960 \)
Hence, the sum of the first 40 positive integers is 4,960.

Question. Divide 56 in four parts in A.P such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5 : 6.
Answer: Let the four parts of the A.P. are \( (a – 3d), (a – d), (a + d), (a + 3d) \)
Now, \( a – 3d + a – d + a + d + a + 3d = 56 \)
\( \Rightarrow 4a = 56 \)
\( \Rightarrow a = 14 \)
According to question,
\( \frac{(a-3d)(a+3d)}{(a-d)(a+d)} = \frac{5}{6} \)
\( \Rightarrow \frac{(14-3d)(14+3d)}{(14-d)(14+d)} = \frac{5}{6} \) [Putting a = 14]
\( \Rightarrow \frac{196 - 9d^2}{196 - d^2} = \frac{5}{6} \)
\( \Rightarrow 1176 – 54d^2 = 980 – 5d^2 \)
\( \Rightarrow 49d^2 = 196 \)

Question. Among the natural numbers 1 to 49, find a number x, such that the sum of numbers preceding it is equal to sum of numbers succeeding it.
Answer: Let, the number be \( x \).
\( 1, 2, 3, 4, ..., x - 1, x, x + 1, ... 49 \)
\( 1 + 2 + 3 + 4 + ... + x - 1 = x + 1 + ... + 49 \)
\( S_{x - 1} = S_{50} - S_x \)
\( \Rightarrow \frac{(x-1)x}{2} = \frac{49 \times 50}{2} - \frac{(x+1)x}{2} \)
\( \Rightarrow x^2 - x = 2450 - x^2 - x \)
\( \Rightarrow x^2 = 1225 \)
\( \Rightarrow x = \pm 35 \) (\(-35\) is not between 1 to 49 therefore rejected)
\( x = 35 \).

Question. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
Answer: Let the required numbers in A.P. are \( (a - d), a, (a + d) \) respectively.
Now, \( a - d + a + a + d = 15 \)
\( 3a = 15 \Rightarrow a = 5 \)
According to question, number is
\( 100(a - d) + 10a + a + d = 111a - 99d \)
Number on reversing the digits is
\( 100(a + d) + 10a + a - d = 111a + 99d \)
Now, as per given condition in question,
\( (111a - 99d) - (111a + 99d) = 594 \)
\( \Rightarrow -198d = 594 \Rightarrow d = -3 \)
So, digits of number are \( [5 - (-3), 5, (5 + (-3)] \)
\( = 8, 5, 2 \)
Required number is \( 111 \times (5) - 99 \times (-3) \)
\( = 555 + 297 = 852 \)
The number is 852.

Question. Find the sum of first 24 terms of an A.P. whose \( n^{th} \) term is given by \( a_n = 3 + 2n \).
Answer: Given, \( n^{th} \) term of an A.P., \( a_n = 3 + 2n \)
First term of A.P., \( a_1 = 3 + 2 \times 1 = 5 \)
Second term of A.P., \( a_2 = 3 + 2 \times 2 = 7 \)
third term of A.P., \( a_3 = 3 + 2 \times 3 = 9 \)
and \( 24^{th} \) term of A.P., \( a_{24} = 3 + 2 \times 24 = 51 \)
common difference of A.P., \( d = a_2 - a_1 = a_3 - a_2 \)
\( = 7 - 5 = 9 - 7 = 2 \)
Sum of 24 terms, \( S_{24} = \frac{n}{2} [2a + (n - 1)d] \)
Here, \( a = 5, n = 24, d = 2 \)
\( \therefore S_{24} = \frac{24}{2} [2 \times 5 + (24 - 1) \times 2] \)
\( = 12 [10 + 46] \)
\( = 12 \times 56 = 672 \)
Hence the sum of first 24 terms of an A.P. is 676.

Question. Determine the A.P. whose third term is 16 and the \( 7^{th} \) term exceeds the \( 5^{th} \) term by 12.
Answer: Let, the first term of A.P. be \( 'a' \) and its common difference be \( 'd' \).
Given, \( a_3 = 16 \)
i.e. \( a + (3 - 1)d = 16 \quad [\because a_n = a + (n - 1)d] \)
\( \Rightarrow a + 2d = 19 \quad ...(i) \)
And \( a_7 = a_5 + 12 \)
\( \Rightarrow a + (7 - 1)d = a + (5 - 1)d + 12 \)
\( \Rightarrow a + 6d = a + 4d + 12 \)
\( \Rightarrow 2d = 12 \)
\( \Rightarrow d = 2 \)
If we put the value of \( 'd' \) in equation (i), we get:
\( a + 2 \times 2 = 16 \)
\( \Rightarrow a = 16 - 4 = 12 \)
First term of AP = 12
Second term of AP = \( 12 + 2 = 14 \)
Third term of AP = \( 14 + 2 = 16 \) and so on.
Hence, the required A.P is 12, 14, 16, 18, ....

Long Short Answer Type Questions

Question. The first term of an AP is –5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
Answer: Let \( a \) be the first term, \( d \) be the common difference and \( n \) be the number of terms.
It is given that
first term, \( a = –5 \)
Last term, \( L = 45 \)
We know that, if the last term of an AP is known, then the sum of n terms of an AP is
\( S_n = \frac{n}{2}(a + l) \)
\( 120 = \frac{n}{2}(-5 + 45) \)
\( 120 \times 2 = 40 \times n \)
\( \Rightarrow n = 6 \)
\( L = a + (n – 1)d \)
\( \Rightarrow 45 = –5 + (6 – 1)d \)
\( \Rightarrow 50 = 5d \Rightarrow d = 10 \)
Hence, number of terms = 6
and common difference = 10

Question. If \( S_n \) denotes the sum of first \( n \) terms of an AP, prove that \( S_{12} = 3(S_8 – S_4) \)
Answer: We know that sum of n terms of an AP,
\( S_n = \frac{n}{2} [2a + ( n – 1) d] \)
\( S_4 = \frac{4}{2} [2a + (4 – 1)d] \)
\( = 2 [2a +3d] = 4a + 6d \)
\( S_8 = \frac{8}{2} [2a + (8 – 1)d] \)
\( = 4 [2a +7d] = 8a + 28d \)
\( (S_8 – S_4) = (8a + 28d) – (4a +6d) \)
\( = 8a + 28d – 4a – 6d \)
\( = 4a + 22d \) ...(i)
\( S_{12} = \frac{12}{2} [2a + (12 – 1)d] \)
\( = 6 [2a +11d] = 12a + 66d \)
\( = 3 [4a + 22d] \)
\( S_{12} = 3 [S_8 – S_4] \) [using equation (i)]
Hence proved.

Question. If the sum of the first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of the first 10 terms.
Answer: It is given that
\( S_6 = 36 \) and \( S_{16} = 256 \)
Let \( a \) be the first term and \( d \) be the common difference of an AP.
We know that sum of n terms of an AP,
\( S_n = \frac{n}{2} [2a + (n – 1)d] \)
Now \( S_6 = 36 \)(Given)
\( \Rightarrow \frac{6}{2} [2a + (6 – 1)d] = 36 \)
\( \Rightarrow 3 [2a + (6 – 1)d] = 36 \)
\( \Rightarrow 2a + 5d = 12 \) ...(i)
Also, \( S_{16} = 256 \)
\( \Rightarrow \frac{16}{2} [2a + (16 – 1)d] = 256 \)
\( \Rightarrow 8[2a + 15d] = 256 \)
\( \Rightarrow 2a + 15d = 32 \) ...(ii)
Subtracting equation (i) from equation (ii), we get
\( (2a + 15d) – (2a + 5d) = 32 – 12 \)
\( \Rightarrow 2a + 15d – 2a – 5d = 20 \)
\( \Rightarrow 10d = 20 \)
\( \Rightarrow d = 2 \)
Putting the value of d in equation (i), we get
\( 2a + 5d = 12 \)
\( 2a + 5(2) = 12 \)
\( \Rightarrow 2a + 10 = 12 \)
\( \Rightarrow 2a = 2 \Rightarrow a = 1 \)
We have to find sum of first 10 terms, \( S_{10} = ? \)
\( S_{10} = \frac{10}{2} [2a + (10 – 1)d] \)
\( = 5 [2(1) + 9(2)] \)
\( = 5 [2 + 18] \)
\( = 5 \times 20 \)
\( S_{10} = 100 \)
Hence, the required sum of the first 10 terms is 100.

Question. The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is –30 and the common difference is 8. Find n.
Answer: It is given that
first term of first AP, \( a = 8 \)
and common difference of first AP, \( d = 20 \)
Let \( n \) be the number of terms in first AP.
We know that sum of first \( n \) terms of an AP,
\( S_n = \frac{n}{2} [2a + (n – 1)d] \)
\( = \frac{n}{2} [2 \times 8 + (n – 1)20] \)
\( = \frac{n}{2} [16 + 20n – 20] \)
\( = \frac{n}{2} [20n – 4] = n[10n – 2] \)
\( \Rightarrow S_n = n[10n – 2] \) ...(i)
Now, first term of second AP (\( a’ \)) = –30
Common difference of secod (\( d’ \)) = 8
\( \therefore \) Sum of first 2n terms of second AP
\( S_{2n} = \frac{2n}{2} [2a’ + (2n – 1)d’] \)
\( = n[2(–30) + (2n –1)8] \)
\( = n[–60 + 16n – 8] \)
\( S_{2n} = n[16n – 68] \) ...(ii)
By given condition,
Sum of first n terms of first AP
= sum of first 2n terms of second AP
\( \Rightarrow S_n = S_{2n} \)
Using equation (i) and equation (ii), we get
\( \Rightarrow n(10n – 2) = n(16n – 68) \)
\( \Rightarrow 10n – 2 – 16n + 68= 0 \)
\( \Rightarrow –6n + 66 = 0 \)
\( \Rightarrow –6(n – 11) = 0 \)
\( \Rightarrow n = 11 \)
Hence, the required value of n is 11.

Question. If \( m^{th} \) term of an A.P. is \( \frac{1}{n} \) and \( n^{th} \) term is \( \frac{1}{m} \), find the sum of its first \( mn \) terms.
Answer: Let, the first term of an A. P. be ‘a’ and its common difference be ‘d’.
Now, \( a_m = a + (m – 1)d \)
\( \Rightarrow \frac{1}{n} = a + (m – 1)d \) ...(i) (given)
and \( a_n = a + (n – 1)d \)
\( \Rightarrow \frac{1}{m} = a + (n – 1)d \) ...(ii) (given)
On subtracting equation (ii) from (i), we get:
\( \frac{1}{n} - \frac{1}{m} = [(m – 1) – (n – 1)]d \)
\( \Rightarrow \frac{m - n}{mn} = (m – n)d \)
\( \Rightarrow d = \frac{1}{mn} \) ...(iii)
If we put the value of d in equation (i), we get
\( \frac{1}{n} = a + (m – 1) \times \frac{1}{mn} \)
\( \Rightarrow a = \frac{1}{n} - \left(\frac{m - 1}{mn}\right) = \frac{m - m + 1}{mn} = \frac{1}{mn} \) ...(iv)
Sum of \( mn \) terms,
\( S_{mn} = \frac{mn}{2} \left[2 \times \frac{1}{mn} + (mn – 1) \times \frac{1}{mn}\right] \)
[\( \because S_n = \frac{n}{2} (2a + (n – 1)d] \)
\( S_{mn} = \frac{1}{2} [2 + mn - 1] = \frac{mn + 1}{2} \)
Hence, the sum of (mn) terms is \( (mn + 1) / 2 \).

Question. Two AP's have the same common difference. The difference between their \( 100^{th} \) terms is 100, what is the difference between \( 1000^{th} \) terms.
Answer: Let \( a_1 \) and \( a_2 \) are two AP’s and their common difference be \( d \).
According to question,
\( [a_1 + (100 – 1)d] – [a_2 + (100 – 1)d] = 100 \)
[\( \because a_n = a + (n – 1)d \)]
\( \Rightarrow [a_1 + 99d] – [a_2 + 99d] = 100 \)
\( \Rightarrow a_1 – a_2 =100 \) ...(i)
Now, to find the difference between their \( 1000^{th} \) terms
\( [a_1 + (1000 – 1)d] – [a_2 + (1000 – 1)d] \)
\( = a_1 + 999d – a_2 – 999d \)
\( = a_1 – a_2 \)
By equation (i), we get
\( [a_1 + (1000 – 1)d] – [a_2 + (1000 – 1)d] = a_1 – a_2 = 100 \)
Therefore, difference between their \( 1000^{th} \) terms would be equal to 100.

Question. Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why?
Answer: Let \( d \) be the common difference of two APs
First term of first AP is 2. First AP will be 2, 2 + \( d \), 2 + 2\( d \)...
First term of second AP is 7. Second AP will be 7, 7 + \( d \), 7 + 2\( d \)...
We know that \( a_n = a + (n - 1)d \)
10th term of first AP = 2 + 9\( d \)
10th term of second AP = 7 + 9\( d \)
Difference of their 10th term = (7 + 9\( d \)) - (2 + 9\( d \)) = 5
Again, 21st term of first AP = 2 + 20\( d \)
21st term of second AP = 7 + 20\( d \)
Difference of their 21st term = (7 + 20\( d \)) - (2 + 20\( d \)) = 5
Thus, we can say that if \( a_n \) and \( b_n \) are \( n \)th terms of first and second AP respectively, then
\( b_n - a_n = [7 + (n - 1)d] - [2 + (n - 1)d] \)
\( = 7 + (n - 1)d - 2 - (n - 1)d = 5 \)
Hence, the difference between any two corresponding terms of such AP’s is the same as the difference between their first terms.

Question. Justify whether it is true to say that the following are the \( n^{th} \) terms of an AP:
(i) \( 2n – 3 \)
(ii) \( 3n^2 + 5 \)
(iii) \( 1 + n + n^2 \)
Answer:
(i) Yes, \( (2n – 3) \) is the \( n^{th} \) term of an A.P.
It is given that \( a_n = 2n – 3 \)
Put \( n = 1, a_1 = 2(1) – 3 = 2 – 3 = –1 \)
\( n = 2, a_2 = 2(2) – 3 = 4 – 3 = 1 \)
\( n = 3, a_3 = 2(3) – 3 = 6 – 3 = 3 \)
\( n = 4, a_4 = 2(4) – 3 = 8 – 3 = 5 \)
List of numbers becomes –1, 1, 3, 5...
Here, \( a_2 – a_1 = 1 – (–1) = 2 \)
\( a_3 – a_2 = 3 – 1 = 2 \)
\( a_4 – a_3 = 5 – 3 = 2 \)
Clearly, \( a_2 – a_1 = a_3 – a_2 = a_4 – a_3 \)
Hence, \( 2n – 3 \) is the \( n^{th} \) term of an AP.
(ii) No, \( (3n^2 + 5) \) is not the \( n^{th} \) term of an AP.
It is given that \( a_n = 3n^2 + 5 \)
Put \( n = 1, a_1 = 3(1)^2 + 5 = 3 + 5 = 8 \)
\( n = 2, a_2 = 3(2)^2 + 5 = 12 + 5 = 17 \)
\( n = 3, a_3 = 3(3)^2 + 5 = 27 + 5 = 32 \)
List of number becomes 8, 17, 32 ...
Here, \( a_2 – a_1 = 17 – 8 = 9 \),
\( a_3 – a_2 = 32 – 17 = 15 \)
Clearly, \( a_2 – a_1 \neq a_3 – a_2 \)
Hence, \( (3n^2 + 5) \) is not the \( n^{th} \) term of an AP.
(iii) No, \( (1 + n + n^2) \) is not the \( n^{th} \) term of an AP.
It is given that \( a_n = 1 + n + n^2 \)
Put \( n = 1, a_1 = 1 + (1) + (1)^2 = 1 + 1 + 1 = 3 \)
\( n = 2, a_2 = 1 + (2) + (2)^2 = 1 + 2 + 4 = 7 \)
\( n = 3, a_3 = 1 + (3) + (3)^2 = 1 + 3 + 9 = 13 \)
List of number becomes 3, 7, 13 ...
Here, \( a_2 – a_1 = 7 – 3 = 4 \)
\( a_3 – a_2 = 13 – 7 = 6 \)
Clearly, \( a_2 – a_1 \neq a_3 – a_2 \)
Hence, \( (1 + n + n^2) \) is not the \( n^{th} \) term of an AP.

Question. Find \( a \), \( b \) and \( c \) such that the following numbers in AP: \( a, 7, b, 23, c \).
Answer: It is given that \( a, 7, b, 23, c \) are in AP
They have a common difference
i.e., \( 7 – a = b – 7 = 23 – b = c – 23 \)
Taking second and third terms, we get
\( b – 7 = 23 – b \Rightarrow 2b = 30 \Rightarrow b = 15 \)
Taking first and second terms, we get
\( 7 – a = b – 7 \)
\( \Rightarrow 7 – a = 15 – 7 \) [As \( b = 15 \)]
\( \Rightarrow 7 – a = 8 \Rightarrow a = –1 \)
Taking third and fourth terms, we get
\( 23 – b = c – 23 \)
\( \Rightarrow 23 – 15 = c – 23 \) [As \( b = 15 \)]
\( \Rightarrow 8 = c – 23 \Rightarrow c = 31 \)
Hence, \( a = –1 \), \( b = 15 \) and \( c = 31 \).

Question. Determine the AP whose \( 5^{th} \) term is 19 and the difference of the \( 8^{th} \) term from the \( 13^{th} \) term is 20.
Answer: Let the first term of an AP be \( a \) and common difference be \( d \).
It is given that \( a_5 = 19 \)
and \( a_{13} – a_8 = 20 \)
We know that \( a_n = a + (n – 1) d \)
\( a_5 = a + (5 – 1) d = a + 4d = 19 \) ...(i)
Also, \( a_{13} – a_8 = 20 \)
\( \Rightarrow (a + 12d) – (a + 7d) = 20 \)
\( \Rightarrow a + 12d – a – 7d = 20 \)
\( \Rightarrow 5d = 20 \Rightarrow d = 4 \)
Putting the value of \( d = 4 \) in equation (i), we get
\( a + 4d = 19 \)
\( \Rightarrow a + 4(4) = 19 \)
\( \Rightarrow a + 16 = 19 \)
\( \Rightarrow a = 3 \)
So, the required AP will be:
\( a, a + d, a + 2d, a + 3d ... \)
i.e., \( 3, 3 + 4, 3 + 2(4), 3 + 3(4) ... \)
i.e., \( 3, 7, 11, 15 ... \)

Question. The \( 26^{th} \), \( 11^{th} \) and the last term of an AP are 0, 3 and \( -\frac{1}{5} \) respectively. Find the common difference and the number of terms.
Answer: It is given that
\( 26^{th} \) term of AP, \( a_{26} = 0 \)
\( 11^{th} \) term of AP, \( a_{11} = 3 \)
Last term, \( L = -\frac{1}{5} \)
Let the AP contain \( n \) term, last term (L) is the \( n^{th} \) term.
Let first term be \( a \) and common difference be \( d \) of an AP.
We know that
\( a_n = a + (n - 1)d \)
Also, \( a_{26} = 0 \) [Given]
\( \Rightarrow a_{26} = a + (26 - 1)d \)
\( \Rightarrow 0 = a + 25d \)
\( \Rightarrow a + 25d = 0 \quad ...(i) \)
Also, \( a_{11} = 3 \)
\( \Rightarrow a + (11 - 1) d = 3 \)
\( \Rightarrow a + 10d = 3 \quad ...(ii) \)
Last term \( L = -\frac{1}{5} \)
\( a + (n - 1) d = -\frac{1}{5} \quad ...(iii) \)
Subtracting equation (ii) from equation (i), we get
\( \Rightarrow (a + 25d) - (a + 10d) = 0 - 3 \)
\( \Rightarrow a + 25d - a - 10d = 0 - 3 \)
\( \Rightarrow 15d = -3 \)
\( \Rightarrow d = \frac{-3}{15} = -\frac{1}{5} \)
Putting the value of \( d \) in equation (i)
\( \Rightarrow a + 25 \left(-\frac{1}{5}\right) = 0 \)
\( \Rightarrow a - 5 = 0 \)
\( \Rightarrow a = 5 \)
Putting the value of \( a \) and \( d \) in equation (iii)
\( \Rightarrow a + (n - 1) d = -\frac{1}{5} \)
\( \Rightarrow 5 + (n - 1)\left(-\frac{1}{5}\right) = -\frac{1}{5} \)
\( \Rightarrow 25 - (n - 1) = -1 \)
\( \Rightarrow 25 + 1 = (n - 1) \)
\( \Rightarrow n = 27 \)
Hence, the common difference = \( -\frac{1}{5} \)
and number of terms = 27.

Question. Find the sum of the following series :
5 + (– 41) + 9 + (– 39) + 13 + (– 37) + 17 + .... + (– 5) + 81 + (– 3)
Answer: Given: series is
5 + (– 41) + 9 + (– 39) + 13 + (– 37) + 17 + (– 5) + 81 + (– 3)
= [5 + 9 + 13 + ... + 81) + [(– 41) + (– 39) + (– 37) + (– 5) + (– 3)]
Here are the two series:
1. \( 5 + 9 + 13 + ...... + 81 \)
2. \( (- 41) + (- 39) + (- 37) + .... + (- 3) \)
For the first series :
first term \( a = 5 \), common difference, \( d = 9 - 5 = 4 \)
last term, \( l(a_n) = 81 \)
Then, \( a_n = a + (n - 1)d \) [where ‘n’ is the number of terms]
\( 81 = 5 + (n - 1) \times 4 \)
\( \Rightarrow (n - 1) = \frac{76}{4} = 19 \)
\( \Rightarrow n = 20 \)
Sum of the first series, \( S_{20} \)
\( S_{20} = \frac{20}{2} [2 \times 5 + (20 - 1) \times 4] \)
\( = 10(10 + 19 \times 4) \)
\( = 10(10 + 76) \)
\( = 860 \)
For the second series :
first term, \( a' = - 41 \),
common difference, \( d' = (- 39) - (- 41) \)
\( = (- 37) - (- 39) \)
\( = 2 \)
last term, \( l'(a_n) = - 3 \)
Then, \( a_n' = a' + (n' - 1)d' \)
[where \( n' \) are the number of terms]
\( \Rightarrow - 3 = - 41 + (n' - 1) \times 2 \)
\( \Rightarrow (n' - 1) = 19 \)
\( \Rightarrow n' = 20 \)
Sum of the second series, \( S'_{20} = \frac{n'}{2} [2a' + (n' - 1)d'] \)
\( = \frac{20}{2} [2 \times (- 41) + (19) \times 2] \)
\( = 10[- 82 + 38] \)
\( = 10 \times (- 44) \)
\( = - 440 \)
\( \therefore \) Total sum = \( S_{20} + S'_{20} \)
\( = 860 - 440 = 420 \)
Hence, the total sum of series is 420.

Question. The sum of four consecutive number in AP is 32 and the ratio of the product of the first and last terms to the product of two middle term is 7 : 15. Find the numbers.
Answer: Let \( 'a' \) be the first term and \( 'd' \) be the common difference to the AP. Then,
\( a - 3d, a - d, a + d, a + 3d \),
are four consecutive terms of the AP.
As per the question,
\( (a - 3d) + (a - d) + (a + d) + (a + 3d) = 32 \)
\( \Rightarrow 4a = 32 \), or \( a = 8 \)
...(i)
and \( \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15} \)
\( \Rightarrow \frac{a^2 - 9d^2}{a^2 - d^2} = \frac{7}{15} \)
\( \Rightarrow 15a^2 - 135d^2 = 7a^2 - 7d^2 \)
\( \Rightarrow 8a^2 = 128 d^2 \)
Using (i), we have:
\( 8 \times 8^2 = 128 d^2 \)
\( \Rightarrow d^2 = 4 \), or \( d = \pm 2 \)
Thus, the four numbers are 2, 6, 10 and 14

Question. Solve : \( 1 + 4 + 7 + 10 + .... + x = 287 \).
Answer: In the given AP, \( a = 1 \) and \( d = 3 \).
Let AP contains \( 'n' \) terms. Then, \( a_n = x \)
\( \Rightarrow a + (n - 1)d = x \)
\( \Rightarrow 1 + 3(n - 1) = x \)
\( \Rightarrow n = \frac{x + 2}{3} \)
Further, \( S_n = \frac{n}{2} \) [first term + last term]
\( \Rightarrow 287 = \frac{x + 2}{3 \times 2} [1 + x] \)
\( \Rightarrow (x + 1) (x + 2) = 1722 \)
or \( x^2 + 3x - 1720 = 0 \)
\( \Rightarrow x^2 + 43x - 40x - 1720 = 0 \)
\( \Rightarrow x(x + 43) - 40(x + 43) = 0 \)
\( \Rightarrow (x + 43) (x - 40) = 0 \)
\( \Rightarrow x - 40 = 0 \quad (\because x + 43 \neq 0) \)
\( \Rightarrow x = 40 \).
Thus, \( x = 40 \).