CBSE Class 10 Maths HOTs Arithmetic Progressions Set G

Very Short Answer Type Questions

Question. Write the \(n^{th}\) term of the A.P. \(\frac{1}{m}, \frac{1+m}{m}, \frac{1+2m}{m}, \dots\dots\)
Answer: Let the first term be \(a\), common difference be \(d\) and \(n^{th}\) term be \(a_n\).
We have \(a = \frac{1}{m}\)
\(d = \frac{1+m}{m} - \frac{1}{m} = 1\)
\(a_n = \frac{1}{m} + (n - 1)1\)
Hence, \(a_n = \frac{1}{m} + n - 1\)

Question. What is the common difference of an A.P. which \(a_{21} - a_7 = 84\).
Answer: Let the first term be \(a\), common difference be \(d\) and \(n^{th}\) term be \(a_n\).
We have \(a_{21} - a_7 = 84\)
\(a + 20d - a - 6d = 84\)
\(14d = 84\)
\(d = \frac{84}{14} = 6\)
Hence common difference is 6.

Question. How many terms of the A.P. \(18, 16, 14, \dots\dots\) be taken so that their sum is zero?
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\)
Here \(a = 18, d = -2, S_n = 0\)
\(S_n = \frac{n}{2}[2a + (n - 1)d]\)
Let sum of \(n\) term be zero, then we have
\(\frac{n}{2}[36 + (n - 1)(-2)] = 0\)
\(n(38 - 2n) = 0\)
\(n = 19\)

Question. How many terms of the A.P. \(27, 24, 21, \dots\dots\) should be taken so that their sum is zero?
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\)
Here \(a = 27, d = -3, S_n = 0\)
\(S_n = \frac{n}{2}[2a + (n - 1)d]\)
Let sum of \(n\) term be zero, then we have
\(\frac{n}{2}[54 + (n - 1)(-3)] = 0\)
\(n(-3n + 57) = 0\)
\(n = 19\)

Question. Find the sum of sixteen terms of an A.P. \(-1, -5, -9, \dots\dots\)
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\)
Here, \(a_1 = -1, a_2 = -5\) and \(d = -4\)
Now \(S_n = \frac{n}{2}[2a + (n - 1)d]\)
\(S_{16} = \frac{16}{2}[2 \times (-1) + (16 - 1)(-4)]\)
\(= 8[-2 - 60] = 8(-62)\)
\(= -496\)

Question. If the \(1^{st}\) term of a series is 7 and \(13^{th}\) term is 35. Find the sum of 13 terms of the sequence.
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
Here \(a = 7, a_{13} = 35\)
\(a_n = a + (n - 1)d\)
\(a_{13} = a + 12d\)
\(35 = 7 + 12d \Rightarrow d = \frac{7}{3}\)
Now \(S_n = \frac{n}{2}[2a + (n - 1)d]\)
\(S_{13} = \frac{13}{2} \left[2 \times 7 + 12 \times \left(\frac{7}{3}\right)\right]\)
\(= \frac{13}{2} [14 + 28]\)
\(= \frac{13}{2} \times 42 = 273\)

Question. If the \(n^{th}\) term of a sequence is \(3 - 2n\). Find the sum of fifteen terms.
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\)
Here, \(a_n = 3 - 2n\)
Taking \(n = 1\), \(a_1 = 3 - 2 = 1\)
15th term, \(a_{15} = 3 - 2 \times 15 = 3 - 30 = -27\)
Now \(S_n = \frac{n}{2}(a + l)\)
\(S_{15} = \frac{15}{2} [1 + (-27)]\)
\(= \frac{15}{2} [-26]\)
\(= 15 \times (-13) = -195\)

Question. If \(S_n\) denotes the sum of \(n\) terms of an A.P. whose common difference is \(d\) and first term is \(a\), find \(S_n - 2S_{n-1} + S_{n-2}\).
Answer: We have \(a_n = S_n - S_{n-1}\)
\(a_{n-1} = S_{n-1} - S_{n-2}\)
\(S_n - 2S_{n-1} + S_{n-2} = S_n - S_{n-1} - S_{n-1} + S_{n-2}\)
\(= (S_n - S_{n-1}) - (S_{n-1} - S_{n-2})\)
\(= a_n - a_{n-1} = d\)

Question. If the \( p^{th} \) term of an A.P. is \( \frac{1}{q} \) and \( q^{th} \) term is \( \frac{1}{p} \). Prove that the sum of first \( pq \) terms of the A.P. is \( \frac{pq + 1}{2} \).
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) terms be \( S_n \)
\( a_p = a + (p - 1)d = \frac{1}{q} \) ...(1)
and \( a_q = a + (q - 1)d = \frac{1}{p} \) ...(2)
Solving (1) and (2) we get
\( a = \frac{1}{pq} \) and \( d = \frac{1}{pq} \)
\( S_{pq} = \frac{pq}{2} \left[ 2 \times \frac{1}{pq} + (pq - 1) \frac{1}{pq} \right] \)
\( = \frac{pq + 1}{2} \)

Question. The \( 8^{th} \) term of an A.P. is zero. Prove that its \( 38^{th} \) term is triple of its \( 18^{th} \) term.
Answer: Let the first term of an A.P. be \( a \), common difference be \( d \) and \( n^{th} \) term be \( a_n \).
We have, \( a_8 = 0 \) or, \( a + 7d = 0 \) or, \( a = -7d \)
Now \( a_{38} = a + 37d \)
\( a_{38} = -7d + 37d = 30d \)
\( a_{18} = a + 17d \)
\( = -7d + 17d = 10d \)
\( a_{38} = 30d = 3 \times 10d = 3 \times a_{18} \)
\( a_{38} = 3a_{18} \) Hence Proved

Question. If five times the fifth term of an A.P. is equal to eight times its eighth term, show that its \( 13^{th} \) term is zero.
Answer: Let the first term of an A.P. be \( a \), common difference be \( d \) and \( n^{th} \) term be \( a_n \).
Now \( 5a_5 = 8a_8 \)
\( 5(a + 4d) = 8(a + 7d) \)
\( 5a + 20d = 8a + 56d \)
\( 3a + 36d = 0 \)
\( 3(a + 12d) = 0 \)
\( a + 12d = 0 \)
\( a_{13} = 0 \) Hence Proved

Question. The fifth term of an A.P. is 20 and the sum of its seventh and eleventh terms is 64. Find the common difference.
Answer: Let the first term be \( a \) and common difference be \( d \).
\( a + 4d = 20 \dots(1) \)
\( a + 6d + a + 10d = 64 \)
\( a + 8d = 32 \dots(2) \)
Solving equations (1) and (2), we have
\( d = 3 \)

Short Answer Type Questions

Question. Which term of the progression \(20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4} \dots\dots\) is the first negative.
Answer: Let the first term be \(a\), common difference be \(d\) and \(n^{th}\) term be \(a_n\).
We have \(a = 20\) and \(d = -\frac{3}{4}\)
Let the \(n^{th}\) term be first negative term, then
\(a + (n - 1)d < 0\)
\(20 + (n - 1)\left(-\frac{3}{4}\right) < 0\)
\(20 - \frac{3}{4}n + \frac{3}{4} < 0\)
\(3n > 83\)
\(n > \frac{83}{3} = 27\frac{2}{3}\)
Hence \(28^{th}\) term is first negative.

Question. How many terms of the A.P. \(65, 60, 55, \dots\dots\) be taken so that their sum is zero?
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\)
We have \(a = 65, d = -5, S_n = 0\)
Now \(S_n = \frac{n}{2}[2a + (n - 1)d]\)
Let sum of \(n\) term be zero, then we have
\(\frac{n}{2}[130 + (n - 1)(-5)] = 0\)
\(\frac{n}{2}[130 - 5n + 5] = 0\)
\(135n - 5n^2 = 0\)
\(n(135 - 5n) = 0\)
\(5n = 135\)
\(n = 27\)

Question. In an A.P., if \(S_3 + S_7 = 167\) and \(S_{10} = 235\), then find the A.P., where \(S_n\) denotes the sum of first \(n\) terms.
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\)
\(S_n = \frac{n}{2}[2a + (n - 1)d]\)
\(S_3 + S_7 = 167\)
\(\frac{5}{2}(2a + 4d) + \frac{7}{2}(2a + 6d) = 167\)
\(24a + 62d = 334\)
\(12a + 31d = 167\) ...(1)
\(S_{10} = 235\)
\(5(2a + 9d) = 235\)
\(2a + 9d = 47\) ...(2)
Solving (1) and (2), we get
\(a = 1, d = 5\)
Thus AP is 1, 6, 11....

Question. If the \(n^{th}\) term of an A.P. is \(7 - 3n\), find the sum of twenty five terms.
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\)
Here \(n = 25, a_n = 7 - 3n\)
Taking \(n = 1, 2, 3, \dots\dots\) we have
\(a_1 = 7 - 3 \times 1 = 4\)
\(a_2 = 7 - 3 \times 2 = 1\)
\(a_3 = 7 - 3 \times 3 = -2\)
Thus required AP is 4, 1, -2, .....
Here, \(a = 4, d = 1 - 4 = -3\)
Now, \(S_n = \frac{n}{2}[2a + (n - 1)d]\)
\(S_{25} = \frac{25}{2}[2 \times 4 + (25 - 1)(-3)]\)
\(= \frac{25}{2}[8 + 24(-3)]\)
\(= \frac{25}{2}(8 - 72) = -800\)

Question. The sum of first \(n\) terms of an A.P. is \(5n - n^2\). Find the \(n^{th}\) term of the A.P.
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
We have, \(S_n = 5n - n^2\)
Now, \(n^{th}\) term of A.P.
\(a_n = S_n - S_{n-1}\)
\(= (5n - n^2) - [5(n - 1) - (n - 1)^2]\)
\( = 5n - n^2 - [5n - 5 - (n^2 + 1 - 2n)] \)
\( = 5n - n^2 - (5n - 5 - n^2 - 1 + 2n) \)
\( = 5n - n^2 - n + 6 + n^2 \)
\( = -2n + 6 \)
\( a_n = -2(n - 3) \)
Thus \( n^{\text{th}} \) term is \( -2(n - 3) \)

Question. The first and last term of an A.P. are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{\text{th}} \) term be \( a_n \) and sum of \( n \) term be \( S_n \).
We have \( a = 5, a_n = 45 \)
Now \( 45 = 5 + (n - 1)d \)
\( (n - 1)d = 40 \) ...(1)
Given, \( S_n = 400 \)
Now \( S_n = \frac{n}{2}(a + l) \)
\( 400 = \frac{n}{2}(5 + 45) \)
\( 800 = 50n \)
\( n = 16 \)
Substituting this value of \( n \) in (1) we have
\( (16 - 1)d = 40 \)
\( 15d = 40 \)
\( d = \frac{40}{15} = \frac{8}{3} \)

Question. If the sum of first \( k \) terms of an A.P. is \( 3k^2 - k \) and its common difference is 6. What is the first term?
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Let the sum of \( k \) terms of A.P. is \( S_k = 3k^2 - k \)
We have \( S_k = 3k^2 - k \)
Now \( k^{th} \) term of A.P.
\( a_k = S_k - S_{k-1} \)
\( a_k = (3k^2 - k) - [3(k - 1)^2 - (k - 1)] \)
\( = 3k^2 - k - [3(k^2 - 2k + 1) - k + 1] \)
\( = 3k^2 - k - [3k^2 - 6k + 3 - k + 1] \)
\( = 3k^2 - k - [3k^2 - 7k + 4] \)
\( = 6k - 4 \)
First term \( a = 6 \times 1 - 4 = 2 \)

Question. If the sum of the first 7 terms of an A.P. is 49 and that of the first 17 terms is 289, find the sum of its first \( n \) terms.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{\text{th}} \) term be \( a_n \) and sum of \( n \) term be \( S_n \).
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
Now \( S_7 = \frac{7}{2}(2a + 6d) = 49 \)
\( a + 3d = 7 \) ...(1)
and \( S_{17} = \frac{17}{2}(2a + 16d) = 289 \)
\( a + 8d = 17 \)
Subtracting (1) from (2), we get
\( 5d = 10 \Rightarrow d = 2 \)
Substituting this value of \( d \) in (1) we have
\( a = 1 \)
Now \( S_n = \frac{n}{2}[2 \times 1 + (n - 1)2] \)
\( = \frac{n}{2}[2 + 2n - 2] = n^2 \)
Hence, sum of \( n \) terms is \( n^2 \).

Question. How many terms of the A.P. \( -6, -\frac{11}{2}, -5, -\frac{9}{2}, \dots \) are needed to give their sum zero.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{\text{th}} \) term be \( a_n \) and sum of \( n \) term be \( S_n \).
We have \( a = -6, d = -\frac{11}{2} - (-6) = \frac{1}{2} \)
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
Let sum of \( n \) term be zero, then we have
\( \frac{n}{2}[2 \times (-6) + (n - 1)\frac{1}{2}] = 0 \)
\( \frac{n}{2}[-12 + \frac{n - 1}{2}] = 0 \)
\( \frac{n}{2}[\frac{-24 + n - 1}{2}] = 0 \)
\( n^2 - 25n = 0 \)
\( n(n - 25) = 0 \)
\( n = 25 \)
Hence 25 terms are needed.

Question. Which term of the A.P. \( 3, 12, 21, 30, \dots \) will be 90 more than its \( 50^{\text{th}} \) term.
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{\text{th}} \) term be \( a_n \).
We have \( a = 3, d = 9 \)
Now \( a_n = a + (n - 1)d \)
\( a_{50} = 3 + 49 \times 9 = 444 \)
Now, \( a_n - a_{50} = 90 \)
\( 3 + (n - 1)9 - 444 = 90 \)
\( (n - 1)9 = 90 + 441 \)
\( (n - 1) = \frac{531}{9} = 59 \)
\( n = 49 + 1 = 60 \)

Question. The \( 10^{\text{th}} \) term of an A.P. is \(-4\) and its \( 22^{\text{nd}} \) term is \(-16\). Find its \( 38^{\text{th}} \) term.
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{\text{th}} \) term be \( a_n \).
\( a_{10} = a + 9d = -4 \) ...(1)
and \( a_{22} = a + 21d = -16 \) ...(2)
Subtracting (2) from (1) we have
\( -12d = 12 \Rightarrow d = -1 \)
Substituting this value of \( d \) in (1) we get
\( a = 5 \)
Thus \( a_{38} = 5 + 37 \times -1 = -32 \)
Hence, \( a_{38} = -32 \)

Question. Find how many integers between 200 and 500 are divisible by 8.
Answer: Number divisible by 8 are 208, 216, 224, .... 496.
Which is an A.P.
Let the first term be \( a \), common difference be \( d \) and \( n^{\text{th}} \) term be \( a_n \).
We have \( a = 208, d = 8 \) and \( a_n = 496 \)
Now \( a + (n - 1)d = a_n \)
\( 208 + (n - 1)8 = 496 \)
\( (n - 1)8 = 496 - 208 \)
\( n - 1 = \frac{288}{8} = 36 \)
\( n = 36 + 1 = 37 \)
Hence, required numbers divisible by 8 is 37.

Question. The fifth term of an A.P. is 26 and its \( 10^{\text{th}} \) term is 51. Find the A.P.
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{\text{th}} \) term be \( a_n \).
\( a_5 = a + 4d = 26 \) ...(1)
\( a_{10} = a + 9d = 51 \) ...(2)
Subtracting (1) from (2) we have
\( 5d = 25 \)
\( d = 5 \)
Substituting this value of \( d \) in (1) we get
\( a = 6 \)
Hence, the AP is 6, 11, 16, 21, ....

Question. Find the A.P. whose third term is 5 and seventh term is 9.
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{\text{th}} \) term be \( a_n \).
Now \( a_3 = a + 2d = 5 \) ...(1)
and \( a_7 = a + 6d = 9 \) ...(2)
Subtracting (1) from (2) we have
\( 4d = 4 \Rightarrow d = 1 \)
Substituting this value of \( d \) in (1) we get
\( a = 3 \)
Hence AP is 3, 4, 5, 6, ......

Question. Find whether -150 is a term of the A.P. 11, 8, 5, 2, ....
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{\text{th}} \) term be \( a_n \).
Let the \( n^{\text{th}} \) term of given A.P. 11, 8, 5, 2, .... be -150
Hence \( a = 11, d = 8 - 11 = -3 \) and \( a_n = -150 \)
\( a + (n - 1)d = a_n \)
\( 11 + (n - 1)(-3) = -150 \)
\( (n - 1)(-3) = -161 \)
\( n - 1 = \frac{-161}{-3} = 53\frac{2}{3} \)
which is not a whole number. Hence -150 is not a term of given A.P.

Question. If seven times the \( 7^{\text{th}} \) term of an A.P. is equal to eleven times the \( 11^{\text{th}} \) term, then what will be its \( 18^{\text{th}} \) term.
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{\text{th}} \) term be \( a_n \).
\( 7a_7 = 11a_{11} \)
Now \( 7(a + 6d) = 11(a + 10d) \)
\( 7a + 42d = 11a + 110d \)
\( 11a - 7a = 42d - 110d \)
\( 4a = -68d \)
\( 4a + 68d = 0 \)
\( 4(a + 17d) = 0 \)
\( a + 17d = 0 \)
Hence, \( a_{18} = 0 \)

Question. In an A.P. of 50 terms, the sum of the first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the A.P.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{\text{th}} \) term be \( a_n \) and sum of \( n \) term be \( S_n \).
\( S_{10} = 210 \)
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( \frac{10}{2}(2a + 9d) = 210 \)
\( 10a + 45d = 210 \Rightarrow 2a + 9d = 42 \) ...(1)
Total terms = 50
Sum of last 15 terms = \( S_{50} - S_{35} \)
\( a_{36} = a + 35d \)
\( a_{50} = a + 49d \)
Sum of last 15 terms \( = \frac{n}{2}(a_{36} + a_{50}) \)
\( 2565 = \frac{15}{2}(a + 35d + a + 49d) \)
\( 171 = \frac{1}{2}(2a + 84d) \)
\( a + 42d = 171 \) ...(2)
Solving (1) and (2) we get
\( a = 3 \) and \( d = 4 \)
Hence, AP is 3, 7, 11, .....

Question. The sum of first \(n\) terms of three arithmetic progressions are \(S_1, S_2\) and \(S_3\) respectively. The first term of each A.P. is 1 and common differences are 1, 2 and 3 respectively. Prove that \(S_1 + S_3 = 2S_2\).
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
We have \(S_1 = 1 + 2 + 3 + \dots + n\)
\(S_2 = 1 + 3 + 5 + \dots\) up to \(n\) terms
\(S_3 = 1 + 4 + 7 + \dots\) upto \(n\) terms
Now \(S_1 = \frac{n(n + 1)}{2}\)
\(S_2 = \frac{n}{2}[2 \times 1 + (n - 1)2] = \frac{n}{2}[2n] = n^2\)
and \(S_3 = \frac{n}{2}[2 \times 1 + (n - 1)3] = \frac{n(3n - 1)}{2}\)
Now, \(S_1 + S_3 = \frac{n(n + 1)}{2} + \frac{n(3n - 1)}{2}\)
\(= \frac{n[n + 1 + 3n - 1]}{2}\)
\(= \frac{n[4n]}{2}\)
\(= 2n^2 = 2S_2\) Hence Proved

HOTS QUESTIONS

Question. Find the sum of the two digits numbers divisible by 6.
Answer: Series of two digits numbers divisible by 6 is:
12, 18, 24, ...........96. It forms and AP.
Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Here \( a = 12, d = 18 - 12 = 6, a_n = 96 \)
\( a_n = a + (n - 1)d \)
\( 96 = 12 + (n - 1) \times 6 \)
\( 84 = 6(n - 1) \)
\( n = 14 + 1 = 15 \)
\( S_n = \frac{n}{2}[a + a_n] \)
\( = \frac{15}{2}[12 + 96] \)
\( = \frac{15 \times 108}{2} \)
\( = 15 \times 54 = 810 \)
Hence the sum of given AP is 810.

Question. Find the number of terms of the A.P. -12, -9, -6, ..., 21. If 1 is added to each term of this A.P., then find the sum of all the terms of the A.P. thus obtained.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) terms be \( S_n \).
We have \( a = -12, d = -9 - (-12) = 3 \)
\( a_n = a + (n - 1)d \)
\( 21 = -12 + (n - 1) \times 3 \)
\( 21 + 12 = (n - 1) \times 3 \)
\( 33 = (n - 1) \times 3 \)
\( n - 1 = 11 \)
\( n = 11 + 1 = 12 \)
Now, if 1 is added to each term we have a New A.P. with
\( -12 + 1, -9 + 1, -6 + 1, ......, 21 + 1 \)
Now we have \( a = -11, d = 3, \) and \( a_n = 22 \) and \( n = 12 \)
Sum of this obtained A.P.
\( S_{12} = \frac{12}{2} [-11 + 22] \)
\( = 6 \times 11 = 66 \)
Hence the sum of new AP is 66.

Question. How many terms of the A.P. -6, \( -\frac{11}{2} \), -5, ..... are needed to give the sum -25? Explain the double answer.
Answer: A.P. is -6, \( -\frac{11}{2} \), -5 .......
Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Here we have \( a = -6 \)
\( d = -\frac{11}{2} + 6 = \frac{1}{2} \)
\( S_n = -25 \)
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( -25 = \frac{n}{2} [-12 + (n - 1) \times \frac{1}{2}] \)
\( -50 = n \left[ \frac{-24 + (n - 1)}{2} \right] \)
\( -100 = n [n - 25] \)
\( n^2 - 25n + 100 = 0 \)
\( (n - 20)(n - 5) = 0 \)
\( n = 20, 5 \),
or, \( S_{20} = S_5 \)
Here we have got two answers because the sum of certain terms of the AP is zero. Since \( a \) is negative and \( d \) is positive; the sum of the terms from \( 6^{th} \) to \( 20^{th} \) is zero.

Question. If \( S_1, S_2, S_3 \) be the sum of \( n, 2n, 3n \) terms respectively of an A.P. Prove that \( S_3 = 3(S_2 - S_1) \).
Answer: Let the first term be \( a \), and common difference be \( d \).
Now \( S_1 = \frac{n}{2} [2a + (n - 1)d] \)
\( S_2 = \frac{2n}{2} [2a + (2n - 1)d] \)
\( S_3 = \frac{3n}{2} [2a + (3n - 1)d] \)
\( 3(S_2 - S_1) \)
\( = 3 \left[ \frac{2n}{2} [2a + (2n - 1)d] - \frac{n}{2} [2a + (n - 1)d] \right] \)
\( = 3 \left[ \frac{n}{2} [4a + 2(2n - 1)d - \{2a + (n - 1)d\}] \right] \)
\( = 3 \left[ \frac{n}{2} (4a + 4nd - 2d - 2a - nd + d) \right] \)
\( = 3 \left[ \frac{n}{2} (2a + 3nd - d) \right] \)
\( = \frac{3n}{2} [2a + (3n - 1)d] = S_3 \)

Question. A spiral is made up of successive semi-circles with centres alternately at A and B starting with A, of radii 1 cm, 2 cm, 3 cm, ...... as shown in the figure. What is the total length of spiral made up of eleven consecutive semi-circles? (Use \( \pi = 3.14 \).)
Answer: Let \( r_1, r_2, ............. \) be the radii of semi-circles and \( l_1, l_2, ............ \) be the lengths of circumferences of semi-circles, then
\( l_1 = \pi r_1 = \pi(1) = \pi \text{ cm} \)
\( l_2 = \pi r_2 = \pi(2) = 2\pi \text{ cm} \)
\( l_3 = 3\pi \text{ cm} \)
...............
\( l_{11} = 11\pi \text{ cm} \)
Total length of spiral
\( L = l_1 + l_2 + ........ + l_{11} \)
\( = \pi + 2\pi + 3\pi + ........ + 11\pi \)
\( = \pi (1 + 2 + 3 + ....... + 11) \)
\( = \pi \times \frac{11 \times 12}{2} \)
\( = 66 \times 3.14 \)
\( = 207.24 \text{ cm} \)

Question. The ratio of the sums of first \( m \) and first \( n \) terms of an A.P. is \( m^2 : n^2 \). Show that the ratio of its \( m^{th} \) and \( n^{th} \) terms is \( (2m - 1) : (2n - 1) \).
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) terms be \( S_n \)
\( \frac{S_m}{S_n} = \frac{m^2}{n^2} \)
\( \frac{\frac{m}{2} [2a + (m - 1)d]}{\frac{n}{2} [2a + (n - 1)d]} = \frac{m^2}{n^2} \)
\( \frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m^2}{n^2} \times \frac{n}{m} = \frac{m}{n} \)
\( n[2a + (m - 1)d] = m[2a + (n - 1)d] \)
\( 2ma + mnd - md = 2na + nmd - nd \)
\( 2ma - 2na = md - nd \)
\( d = 2a \)
Now, \( \frac{a_m}{a_n} = \frac{a + (m - 1)d}{a + (n - 1)d} \)
\( = \frac{a + (m - 1) \times 2a}{a + (n - 1) \times 2a} \)
\( = \frac{a + 2ma - 2a}{a + 2na - 2a} \)
\( = \frac{2ma - a}{2na - a} = \frac{a(2m - 1)}{a(2n - 1)} \)
\( = (2m - 1) : (2n - 1) \)

Question. If the ratio of the \( 11^{th} \) term of an A.P. to its \( 18^{th} \) term is 2 : 3, find the ratio of the sum of the first five terms to the sum of its first 10 terms.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) terms be \( S_n \)
Now \( \frac{a_{11}}{a_{18}} = \frac{a + 10d}{a + 17d} = \frac{2}{3} \)
\( 3(a + 10d) = 2(a + 17d) \)
\( 3a + 30d = 2a + 34d \)
\( a = 4d \) ...(1)
Now, \( \frac{S_5}{S_{10}} = \frac{\frac{5}{2}(2a + 4d)}{\frac{10}{2}(2a + 9d)} = \frac{(a + 2d)}{2a + 9d} \)
Substituting the value \( a = 4d \) we have
or, \( \frac{S_5}{S_{10}} = \frac{4d + 2d}{8d + 9d} = \frac{6d}{17d} = \frac{6}{17} \)
Hence \( S_5 : S_{10} = 6 : 17 \)

Question. An A.P. Consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the A.P.
Answer: Let the middle most terms of the A.P. be \( (x - d), x, (x + d) \)
We have \( x - d + x + x + d = 225 \)
\( 3x = 225 \)
or, \( x = 75 \)
and the middle term \( = \frac{37 + 1}{2} = 19^{th} \) term
Thus AP is
\( (x - 18d), ...., (x - 2d), (x - d), x, (x + d), (x + 2d), ......., (x + 18d) \)
Sum of last three terms,
\( (x + 18d) + (x + 17d) + (x + 16d) = 429 \)
\( 3x + 51d = 429 \)
\( 3(75) + 51d = 429 \)
\( 225 + 51d = 429 \) or, \( d = 4 \)
First term \( a_1 = x - 18d = 75 - 18 \times 4 = 3 \)
\( a_2 = 3 + 4 = 7 \)
Hence A.P. = 3, 7, 11, ......., 147.