CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set F

Question. How many two digit numbers are divisible by 7?
Answer: Two digit numbers which are divisible by 7 are: 14, 21, 28, ......, 98.
It forms an A.P.
Here, \( a = 14, d = 7 \) and \( a_n = 98 \)
Since, \( a_n = a + (n - 1)d \)
\( 98 = 14 + (n - 1)7 \)
\( 98 - 14 = 7n - 7 \)
\( 84 + 7 = 7n \)
or, \( 7n = 91 \)
or, \( n = 13 \)

Question. In a certain A.P. \( 32^{th} \) term is twice the \( 12^{th} \) term. Prove that \( 70^{th} \) term is twice the \( 31^{st} \) term.
Answer: Let the \( 1^{st} \) term be \( a \) and common difference be \( d \).
According to the question, \( a_{32} = 2a_{12} \)
\( \therefore a + 31d = 2(a + 11d) \)
\( a + 31d = 2a + 22d \)
\( a = 9d \)
Again, \( a_{70} = a + 69d \)
\( = 9d + 69d = 78d \)
\( \because a_{31} = a + 30d \)
\( = 9d + 30d = 39d \)
Hence, \( a_{70} = 2a_{31} \). Hence Proved.

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: Given, \( a_8 = 0 \) or, \( a + 7d = 0 \) or, \( a = -7d \)
or, \( a_{38} = a + 37d \)
or, \( a_{38} = -7d + 37d = 30d \)
And, \( a_{18} = a + 17d \)
\( = -7d + 17d = 10d \)
or, \( a_{38} = 30d = 3 \times 10d = 3 \times a_{18} \)
\( \therefore a_{38} = 3a_{18} \). 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 \( d \).
Then, \( a + 4d = 20 \) ...(i)
and \( a + 6d + a + 10d = 64 \)
\( a + 8d = 32 \) ...(ii)
Solving equations (i) and (ii), we get
\( d = 3 \)
Hence, common difference, \( d = 3 \).

Question. Find the middle term of the A.P. 213, 205, 197, ..... 37.
Answer: Here, \( a = 213, d = 205 - 213 = -8 \) and \( l = 37 \)
Let the number of terms be \( n \).
\( \because l = a + (n - 1)d \)
\( \therefore 37 = 213 + (n - 1)(-8) \)
or, \( 37 - 213 = -8(n - 1) \)
or, \( n - 1 = \frac{-176}{-8} = 22 \)
or, \( n = 22 + 1 = 23 \)
The middle term will be \( \frac{23 + 1}{2} = 12^{th} \)
\( \therefore a_{12} = a + (n - 1)d \)
\( = 213 + (12 - 1)(-8) \)
\( = 213 - 88 = 125 \)
Thus, the middle term will be 125.

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

Short Answer Type Questions-II

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, \( d = \frac{77}{4} - 20 = -\frac{3}{4} \)
Let \( t_n < 0 \)
\( \because t_n = a + (n - 1)d \)
\( \therefore 20 + (n - 1) \left( -\frac{3}{4} \right) < 0 \)
\( \Rightarrow 80 - 3n + 3 < 0 \)
\( \Rightarrow 83 - 3n < 0 \)
\( \Rightarrow n > \frac{83}{3} \)
\( \Rightarrow n > 27.6 \)
\( \Rightarrow n = 28 \)
Hence, \( 28^{th} \) term will be the first negative term.

Question. Find the middle term of the A.P. 7, 13, 19, ...., 247.
Answer: In this A.P., \( a = 7, d = 13 - 7 = 6 \) and \( t_n = 247 \)
\( \because t_n = a + (n - 1)d \)
\( \therefore 247 = 7 + (n - 1)6 \)
\( \Rightarrow 6(n - 1) = 240 \)
\( \Rightarrow n - 1 = 40 \)
\( \Rightarrow n = 41 \)
Hence, the middle term \( = \frac{n + 1}{2} = \frac{41 + 1}{2} = \frac{42}{2} = 21 \).
Hence, \( 21^{st} \) term will be the middle term.
\( \therefore t_{21} = a + 20d = 7 + 20 \times 6 = 7 + 120 = 127 \)

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: Given, first term, \( A = a \)
and second term \( = b \)
\( \Rightarrow \) common difference, \( d = b - a \)
Last term, \( l = c \)
\( \Rightarrow A + (n - 1)d = c \)
\( \Rightarrow a + (n - 1)(b - a) = c \)
\( \Rightarrow (b - a)(n - 1) = c - a \)
\( \Rightarrow n - 1 = \frac{c - a}{b - a} \)
\( \Rightarrow n = \frac{c - a}{b - a} + 1 = \frac{c - a + b - a}{b - a} = \frac{b + c - 2a}{b - a} \)
Now \( \text{sum} = \frac{n}{2} [A + l] \)
\( = \frac{\frac{b + c - 2a}{b - a}}{2} [a + c] \)
\( = \frac{(a + c)(b + c - 2a)}{2(b - a)} \). Hence Proved.

Question. Solve the equation: \( 1 + 4 + 7 + 10 + ... + x = 287 \).
Answer: Given, \( a = 1 \) and \( d = 4 - 1 = 3 \)
Let number of terms is the series be \( n \), then
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( \Rightarrow \frac{n}{2} [2 \times 1 + (n - 1)3] = 287 \)
\( \Rightarrow \frac{n}{2} [2 + 3n - 3] = 287 \)
\( \Rightarrow 3n^2 - n - 574 = 0 \)
\( \Rightarrow 3n^2 - 42n + 41n - 574 = 0 \)
\( \Rightarrow 3n(n - 14) + 41(n - 14) = 0 \)
\( \Rightarrow (n - 14)(3n + 41) = 0 \)
Either \( n = 14 \) or \( n = -\frac{41}{3} \), it is not possible.
Thus \( 14^{th} \) term is \( x \).
\( a_{14} = a + (14 - 1)d = 1 + 13 \times 3 = 40 \).

Question. If in an A.P., the sum of first \( m \) terms is \( n \) and the sum of its first \( n \) terms is \( m \), then prove that the sum of its first \( (m + n) \) terms is \( -(m + n) \).
Answer: Let \( 1^{st} \) term of series be \( a \) and common difference be \( d \), then
\( S_m = n \)
\( \Rightarrow \frac{m}{2} [2a + (m - 1)d] = n \)
\( \Rightarrow m[2a + (m - 1)d] = 2n \) ...(i)
and \( S_n = m \)
\( \Rightarrow \frac{n}{2} [2a + (n - 1)d] = m \)
\( \Rightarrow n[2a + (n - 1)d] = 2m \) ...(ii)
On subtracting eq. (ii) from eq. (i)
\( 2(n - m) = 2a(m - n) + d[m^2 - n^2 - (m - n)] \)
\( \Rightarrow 2(n - m) = (m - n)[2a + d(m + n - 1)] \)
\( \Rightarrow -2 = 2a + d(m + n - 1) \) ...(iii)
Now, \( S_{m + n} = \frac{m + n}{2} [2a + (m + n - 1)d] \)
\( = \frac{m + n}{2} (-2) \) [from (iii)]
\( = -(m + n) \). Hence Proved.

Question. Find the sum of all 11 terms of an A.P. whose middle term is 30.
Answer: In an A.P. with 11 terms,
middle term \( = \frac{11 + 1}{2} \text{ term} = 6^{th} \text{ term} \)
Now, sixth term i.e., \( a_6 = a + (6 - 1)d \)
i.e., \( a + 5d = 30 \) ...(i) [Since middle term i.e., \( a_6 = 30 \) (given)]
Now, the sum of 11 terms,
\( S_{11} = \frac{11}{2} [2a + (11 - 1)d] \)
\( = \frac{11}{2} [2a + 10d] \)
\( = \frac{11}{2} \times 2[a + 5d] \)
\( = 11 \times 30 = 330 \) [from (i)]

Question. If the sum of first \( m \) terms of an A.P. is the same as the sum of its first \( n \) terms, show that the sum of its first \( (m + n) \) terms is zero.
Answer: Sum of first \( m \) terms = Sum of first \( n \) terms
\( \Rightarrow S_m = S_n \)
\( \frac{m}{2} [2a + (m - 1)d] = \frac{n}{2} [2a + (n - 1)d] \)
\( m[2a + (m - 1)d] = n[2a + (n - 1)d] \)
\( 2a(m - n) + [m(m - 1) - n(n - 1)]d = 0 \)
\( 2a(m - n) + [m^2 - m - n^2 + n]d = 0 \)
\( 2a(m - n) + [(m - n)(m + n) - (m - n)]d = 0 \)
\( (m - n)[2a + (m + n - 1)d] = 0 \)
Here, \( (m - n) \) is not equal to zero.
So, \( [2a + (m + n - 1)d] = 0 \)
Hence, \( S_{m + n} = \frac{m + n}{2} [2a + (m + n - 1)d] = 0 \).

Question. If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first \( n \) terms.
Answer: \( S_4 = 40 \Rightarrow 2(2a + 3d) = 40 \Rightarrow 2a + 3d = 20 \) ...(i)
\( S_{14} = 280 \Rightarrow 7(2a + 13d) = 280 \Rightarrow 2a + 13d = 40 \) ...(ii)
Solving eq. (i) and (ii), we get
\( a = 7 \) and \( d = 2 \)
\( \therefore S_n = \frac{n}{2} [2 \times 7 + (n - 1)2] \)
\( = \frac{n}{2} [14 + 2n - 2] \)
\( = \frac{n}{2} (12 + 2n) \)
\( = 6n + n^2 \)
Hence, Sum of \( n \) terms \( = 6n + n^2 \)

Question. For what value of \( n \), are the \( n^{th} \) terms of two A.P.s 63, 65, 67,.... and 3, 10, 17,.... equal ?
Answer: Let \( a, d \) and \( A, D \) be the \( 1^{st} \) term and common different of the 2 A.P.s respectively.
Here, \( a = 63, d = 2 \) and \( A = 3, D = 7 \)
Given, \( a_n = A_n \)
\( \Rightarrow a + (n - 1)d = A + (n - 1)D \)
\( \Rightarrow 63 + (n - 1)2 = 3 + (n - 1)7 \)
\( \Rightarrow 63 + 2n - 2 = 3 + 7n - 7 \)
\( \Rightarrow 61 + 2n = 7n - 4 \)
\( \Rightarrow 5n = 65 \)
\( \Rightarrow n = 13 \)
\( \therefore \) When \( n \) is 13, the \( n^{th} \) terms are equal i.e., \( a_{13} = A_{13} \)

Question. If the \( 10^{th} \) term of an A.P. is 52 and the \( 17^{th} \) term is 20 more than the \( 13^{th} \) term, find A.P.
Answer: \( a_{10} = 52 \Rightarrow a + 9d = 52 \) ...(i)
Also \( a_{17} - a_{13} = 20 \)
\( a + 16d - (a + 12d) = 20 \)
\( 4d = 20 \Rightarrow d = 5 \)
Substituting, the value of \( d \) in (i), we get \( a = 7 \)
Hence, A.P. \( = 7, 12, 17, 22 \dots \)

Question. How many terms of an A.P. 9, 17, 25, .... must be taken to give a sum of 636 ?
Answer: \( a = 9, d = 8, S_n = 636 \)
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( 636 = \frac{n}{2} [18 + (n - 1)8] \)
\( 636 = n[9 + 4n - 4] \)
\( 4n^2 + 5n - 636 = 0 \)
\( 4n^2 - 48n + 53n - 636 = 0 \)
\( 4n(n - 12) + 53(n - 12) = 0 \)
\( n = 12 \) or \( n = -\frac{53}{4} \). Since \( n \) is a natural number, \( n = 12 \).
\( \therefore \) 12 terms are required to give sum 636.

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: \( S_n = \left[4 - \frac{1}{n}\right] + \left[4 - \frac{2}{n}\right] + \left[4 - \frac{3}{n}\right] + \dots \) up to \( n \) terms
\( = (4 + 4 + 4 + \dots \text{ up to } n \text{ terms}) - \frac{1}{n} (1 + 2 + 3 + \dots \text{ up to } n \text{ terms}) \)
\( = 4n - \frac{1}{n} \times \frac{n(n + 1)}{2} \)
\( = 4n - \frac{n + 1}{2} = \frac{8n - n - 1}{2} = \frac{7n - 1}{2} \).
Hence, sum of \( n \) terms \( = \frac{7n - 1}{2} \).

Question. If the sum of the first 14 terms of an A.P. is 1050 and its first term is 10, find its \( 20^{th} \) term.
Answer: Given, \( a = 10 \), and \( S_{14} = 1050 \). Let common difference be \( d \).
Since, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( \therefore S_{14} = \frac{14}{2} [2 \times 10 + (14 - 1)d] = 1050 \)
\( 20 + 13d = \frac{1050}{7} = 150 \)
\( 13d = 130 \Rightarrow d = 10 \)
\( a_{20} = a + (20 - 1)d = 10 + 19 \times 10 = 200 \).
Hence, \( a_{20} = 200 \).

Question. Find the sum of all odd numbers between 0 and 50.
Answer: Given, \( 1 + 3 + 5 + 7 + \dots + 49 \)
Let total odd number of terms be \( n \).
\( a_n = 1 + (n - 1) \times 2 = 49 \)
\( (n - 1) \times 2 = 48 \Rightarrow n - 1 = 24 \Rightarrow n = 25 \)
\( S_{25} = \frac{25}{2} (1 + 49) = 25 \times 25 = 625 \).
Hence, sum of odd numbers between 0 and 50 is 625.

Question. If \( m^{th} \) term of A.P. is \( \frac{1}{n} \) and \( n^{th} \) term is \( \frac{1}{m} \), find the sum of first \( mn \) terms.
Answer: Let first term be \( a \) and common difference be \( d \).
\( a_m = a + (m - 1)d = \frac{1}{n} \) ...(i)
\( a_n = a + (n - 1)d = \frac{1}{m} \) ...(ii)
Subtracting (ii) from (i), we get \( (m - n)d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn} \Rightarrow d = \frac{1}{mn} \)
From (i), \( a = \frac{1}{n} - (m - 1)\frac{1}{mn} = \frac{1}{mn} \)
Now, \( S_{mn} = \frac{mn}{2} [2a + (mn - 1)d] = \frac{mn}{2} [\frac{2}{mn} + \frac{mn - 1}{mn}] \)
\( = \frac{mn}{2} \left[ \frac{2 + mn - 1}{mn} \right] = \frac{mn + 1}{2} \).
Hence, the sum of first \( mn \) terms \( = \frac{1}{2}(mn + 1) \).

Question. Find the sum of all two digit natural numbers which are divisible by 4.
Answer: First two digit multiple of 4 is 12 and last is 96.
So, \( a = 12, d = 4 \) and \( l = 96 \)
Let \( n^{th} \) term be last term \( = 96 \)
\( a_n = a + (n - 1)d = l \Rightarrow 12 + (n - 1)4 = 96 \Rightarrow n - 1 = 21 \Rightarrow n = 22 \)
Now, \( S_{22} = \frac{22}{2} [12 + 96] = 11 \times 108 = 1188 \).

Question. Find the sum of the following series: \( 5 + (– 41) + 9 + (– 39) + 13 + (– 37) + 17 + \dots + (– 5) + 81 + (– 3) \).
Answer: The series can be written as \( (5 + 9 + 13 + \dots + 81) + [(-41) + (-39) + (-37) + \dots + (-5) + (-3)] \)
For the series \( (5 + 9 + 13 + \dots + 81) \): \( a = 5, d = 4, a_n = 81 \).
\( 81 = 5 + (n - 1)4 \Rightarrow (n - 1)4 = 76 \Rightarrow n = 20 \).
\( S_{20} = \frac{20}{2} (5 + 81) = 860 \).
For series \( (-41) + (-39) + \dots + (-3) \): \( a = -41, d = 2, a_n = -3 \).
\( -3 = -41 + (n - 1)2 \Rightarrow (n - 1)2 = 38 \Rightarrow n = 20 \).
\( S_{20} = \frac{20}{2} [-41 + (-3)] = -440 \).
Hence, the sum of the series \( = 860 - 440 = 420 \).

Question. The ninth term of an A.P. is equal to seven times the second term and twelfth term exceeds five times the third term by 2. Find the first term and the common difference.
Answer: Let the first term be \( a \) and common difference be \( d \).
Given, \( a_9 = 7a_2 \Rightarrow a + 8d = 7(a + d) \Rightarrow 6a - d = 0 \) ...(i)
and \( a_{12} = 5a_3 + 2 \Rightarrow a + 11d = 5(a + 2d) + 2 \Rightarrow 4a + d = -2 \) ...(ii)
Solving (i) and (ii), \( 10a = -2 \). Wait, check math: \( a + 8d = 7a + 7d \implies -6a + d = 0 \).
From (ii), \( a + 11d = 5a + 10d + 2 \implies -4a + d = 2 \).
Subtracting equations: \( -2a = -2 \Rightarrow a = 1 \).
From \( -6(1) + d = 0 \Rightarrow d = 6 \).
Hence, first term = 1 and common difference = 6.

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 three digits be \( a - d, a, a + d \).
Sum \( = 15 \Rightarrow 3a = 15 \Rightarrow a = 5 \).
Original number \( = 100(a - d) + 10a + (a + d) = 111a - 99d \).
Reversed number \( = 100(a + d) + 10a + (a - d) = 111a + 99d \).
Given, \( (111a - 99d) - (111a + 99d) = 594 \)
\( -198d = 594 \Rightarrow d = -3 \).
The digits are \( 5 - (-3) = 8, 5, 5 + (-3) = 2 \).
Hence the number is 852.

Question. Divide 56 in four parts in A.P. such that the ratio of the product of their extremes \( (1^{st} \text{ and } 4^{th}) \) to the product of middle \( (2^{nd} \text{ and } 3^{rd}) \) is 5 : 6.
Answer: Let the four parts be \( a - 3d, a - d, a + d, a + 3d \).
Sum \( = 56 \Rightarrow 4a = 56 \Rightarrow a = 14 \).
Ratio: \( \frac{(14 - 3d)(14 + 3d)}{(14 - d)(14 + d)} = \frac{5}{6} \)
\( \Rightarrow \frac{196 - 9d^2}{196 - d^2} = \frac{5}{6} \)
\( 1176 - 54d^2 = 980 - 5d^2 \)
\( 49d^2 = 196 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2 \).
The parts are 8, 12, 16, 20 or 20, 16, 12, 8.

Question. The \( p^{th}, q^{th} \text{ and } r^{th} \) terms of an A.P. are \( a, b \text{ and } c \) respectively. Show that \( a(q – r) + b(r – p) + c(p – q) = 0 \).
Answer: Let first term be \( a' \) and common difference be \( d \).
\( a = a' + (p - 1)d, b = a' + (q - 1)d, c = a' + (r - 1)d \).
\( a(q - r) + b(r - p) + c(p - q) = [a' + (p - 1)d](q - r) + [a' + (q - 1)d](r - p) + [a' + (r - 1)d](p - q) \)
\( = a'[q - r + r - p + p - q] + d[(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)] \)
\( = a' \times 0 + d[pq - pr - q + r + qr - qp - r + p + rp - rq - p + q] \)
\( = 0 + d \times 0 = 0 \). Hence Proved.

Question. The sum of first \( n \) terms of three arithmetic progressions are \( S_1, S_2 \text{ and } S_3 \) 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: \( S_1 = \frac{n}{2} [2(1) + (n - 1)1] = \frac{n(n + 1)}{2} \)
\( S_2 = \frac{n}{2} [2(1) + (n - 1)2] = n^2 \)
\( S_3 = \frac{n}{2} [2(1) + (n - 1)3] = \frac{n(3n - 1)}{2} \)
\( S_1 + S_3 = \frac{n(n + 1) + n(3n - 1)}{2} = \frac{n^2 + n + 3n^2 - n}{2} = \frac{4n^2}{2} = 2n^2 = 2S_2 \). Hence Proved.

Question. If the sum of the first \( n \) terms of an A.P. is \( \frac{1}{2} [3n^2 + 7n] \), then find its \( n^{th} \) term. Hence write its \( 20^{th} \) term.
Answer: \( S_n = \frac{3n^2 + 7n}{2} \)
\( a_1 = S_1 = \frac{3 + 7}{2} = 5 \)
\( S_2 = \frac{12 + 14}{2} = 13 \)
\( a_2 = S_2 - S_1 = 13 - 5 = 8 \)
\( d = a_2 - a_1 = 8 - 5 = 3 \).
\( a_n = a_1 + (n - 1)d = 5 + (n - 1)3 = 3n + 2 \).
\( a_{20} = 3(20) + 2 = 62 \).

Question. Prove that the \( n^{th} \) term of an A.P. can not be \( n^2 + 1 \). Justify your answer.
Answer: Let \( a_n = n^2 + 1 \).
\( a_1 = 1^2 + 1 = 2 \)
\( a_2 = 2^2 + 1 = 5 \)
\( a_3 = 3^2 + 1 = 10 \)
Common differences: \( a_2 - a_1 = 3 \), \( a_3 - a_2 = 5 \).
Since the common differences are not equal, the sequence is not an A.P. Hence, \( n^2 + 1 \) cannot be the \( n^{th} \) term of an A.P.

Question. If \( S_n \) denotes, the sum of the first \( n \) terms of an A.P. prove that \( S_{12} = 3(S_8 – S_4) \).
Answer: Let first term be \( a \) and common difference be \( d \).
\( S_{12} = \frac{12}{2} [2a + 11d] = 12a + 66d \) ...(i)
\( S_8 = \frac{8}{2} [2a + 7d] = 8a + 28d \)
\( S_4 = \frac{4}{2} [2a + 3d] = 4a + 6d \)
\( 3(S_8 - S_4) = 3(4a + 22d) = 12a + 66d \).
From (i), \( S_{12} = 3(S_8 - S_4) \). Hence Proved.

Question. The \( 14^{th} \) term of an A.P. is twice its \( 8^{th} \) term. If the \( 6^{th} \) term is – 8, then find the sum of its first 20 terms.
Answer: Given \( a_{14} = 2a_8 \Rightarrow a + 13d = 2(a + 7d) \Rightarrow a = -d \) ...(i)
Also \( a_6 = -8 \Rightarrow a + 5d = -8 \).
From (i), \( -d + 5d = -8 \Rightarrow 4d = -8 \Rightarrow d = -2 \).
\( a = -(-2) = 2 \).
\( S_{20} = \frac{20}{2} [2(2) + 19(-2)] = 10[4 - 38] = -340 \).

Long Answer Type Questions

Question. The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last term to the product of two middle terms is 7 : 15. Find the numbers.
Answer: Let the four consecutive terms of A.P. be \( (a - 3d), (a - d), (a + d) \) and \( (a + 3d) \).
Sum \( = 32 \Rightarrow 4a = 32 \Rightarrow a = 8 \).
Given \( \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15} \)
\( \frac{a^2 - 9d^2}{a^2 - d^2} = \frac{7}{15} \Rightarrow 15(64 - 9d^2) = 7(64 - d^2) \)
\( 960 - 135d^2 = 448 - 7d^2 \Rightarrow 128d^2 = 512 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2 \).
Numbers are 2, 6, 10, 14 or 14, 10, 6, 2.

Question. If \( m \) times the \( m^{th} \) term of an Arithmetic Progression is equal to \( n \) times its \( n^{th} \) term and \( m \neq n \), show that the \( (m + n)^{th} \) term of the A.P. is zero.
Answer: Let first term be \( a \) and common difference be \( d \).
Given \( m \cdot a_m = n \cdot a_n \)
\( m[a + (m - 1)d] = n[a + (n - 1)d] \)
\( am + m(m - 1)d = an + n(n - 1)d \)
\( a(m - n) + d[m^2 - m - n^2 + n] = 0 \)
\( a(m - n) + d[(m - n)(m + n) - (m - n)] = 0 \)
Dividing by \( (m - n) \) since \( m \neq n \):
\( a + d[m + n - 1] = 0 \)
This is the expression for \( a_{m+n} \). Hence \( a_{m+n} = 0 \). Hence Proved.

Question. The first term of an A.P. is 3, the last term is 83 and the sum of all its terms is 903. Find the number of terms and the common difference of the A.P.
Answer: \( a = 3, a_n = 83, S_n = 903 \).
\( S_n = \frac{n}{2} (a + a_n) \Rightarrow 903 = \frac{n}{2} (3 + 83) \Rightarrow 903 = 43n \Rightarrow n = 21 \).
Also \( a_n = a + (n - 1)d \Rightarrow 83 = 3 + 20d \Rightarrow 20d = 80 \Rightarrow d = 4 \).
Number of terms is 21 and common difference is 4.

Question. An A.P. consists of 50 terms of which \( 3^{rd} \) term is 12 and last term is 106. Find the \( 29^{th} \) term.
Answer: \( n = 50, a_3 = 12, a_{50} = 106 \).
\( a + 2d = 12 \) and \( a + 49d = 106 \).
Subtracting: \( 47d = 94 \Rightarrow d = 2 \).
\( a + 4 = 12 \Rightarrow a = 8 \).
\( a_{29} = a + 28d = 8 + 28 \times 2 = 64 \).

Question. If the ratio of the sum of the first \( n \) terms of two A.Ps is \( (7n + 1) : (4n + 27) \), then find the ratio of their \( 9^{th} \) terms.
Answer: \( \frac{S_n}{S'_n} = \frac{\frac{n}{2} [2a + (n - 1)d]}{\frac{n}{2} [2a' + (n - 1)d']} = \frac{7n + 1}{4n + 27} \)
\( \frac{a + \frac{n - 1}{2} d}{a' + \frac{n - 1}{2} d'} = \frac{7n + 1}{4n + 27} \)
For ratio of \( 9^{th} \) terms, replace \( \frac{n - 1}{2} = 8 \Rightarrow n = 17 \).
\( \frac{a_9}{a'_9} = \frac{7(17) + 1}{4(17) + 27} = \frac{119 + 1}{68 + 27} = \frac{120}{95} = \frac{24}{19} \).
The ratio is 24 : 19.

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: \( \frac{S_m}{S_n} = \frac{m^2}{n^2} \Rightarrow \frac{\frac{m}{2} [2a + (m - 1)d]}{\frac{n}{2} [2a + (n - 1)d]} = \frac{m^2}{n^2} \)
\( \frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n} \)
\( 2an + mnd - nd = 2am + mnd - md \)
\( 2an - 2am = nd - md \Rightarrow 2a(n - m) = d(n - m) \Rightarrow d = 2a \).
\( \frac{a_m}{a_n} = \frac{a + (m - 1)2a}{a + (n - 1)2a} = \frac{2am - a}{2an - a} = \frac{2m - 1}{2n - 1} \). Hence Proved.

Question. If the \( p^{th} \) term of an A.P. is \( 1/q \) and \( q^{th} \) term is \( 1/p \). Prove that the sum of first \( pq \) term of the A.P. is \( [\frac{pq+1}{2}] \).
Answer: (As solved in Short Answer II Q15).
\( S_{pq} = \frac{pq}{2} [2a + (pq - 1)d] = \frac{pq}{2} [\frac{2}{pq} + \frac{pq - 1}{pq}] = \frac{pq + 1}{2} \). Hence Proved.

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 term to the sum of its first 10 terms.
Answer: \( \frac{a + 10d}{a + 17d} = \frac{2}{3} \Rightarrow 3a + 30d = 2a + 34d \Rightarrow a = 4d \).
\( \frac{S_5}{S_{10}} = \frac{\frac{5}{2} [2a + 4d]}{\frac{10}{2} [2a + 9d]} = \frac{5[8d + 4d]}{10[8d + 9d]} = \frac{60d}{170d} = \frac{6}{17} \).
The ratio is 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: Middle term \( = \frac{37 + 1}{2} = 19^{th} \). Middle most terms are \( a_{18}, a_{19}, a_{20} \).
\( (a + 17d) + (a + 18d) + (a + 19d) = 225 \Rightarrow 3a + 54d = 225 \Rightarrow a + 18d = 75 \) ...(i)
Last three terms are \( a_{35}, a_{36}, a_{37} \).
\( (a + 34d) + (a + 35d) + (a + 36d) = 429 \Rightarrow 3a + 105d = 429 \Rightarrow a + 35d = 143 \) ...(ii)
Subtracting (i) from (ii): \( 17d = 68 \Rightarrow d = 4 \).
\( a + 72 = 75 \Rightarrow a = 3 \).
The A.P. is 3, 7, 11, ..., 147.

Question. The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers.
Answer: Let numbers be \( a - d, a, a + d \).
Sum \( = 12 \Rightarrow 3a = 12 \Rightarrow a = 4 \).
Sum of cubes \( = (4 - d)^3 + 4^3 + (4 + d)^3 = 288 \).
\( 64 - 48d + 12d^2 - d^3 + 64 + 64 + 48d + 12d^2 + d^3 = 288 \)
\( 192 + 24d^2 = 288 \Rightarrow 24d^2 = 96 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2 \).
Numbers are 2, 4, 6 or 6, 4, 2.

Question. Find the value of \( a, b \) and \( c \) such that the numbers \( a, 7, b, 23 \) and \( c \) are in A.P.
Answer: Let common difference be \( d \).
\( a + d = 7 \) ...(i)
\( a + 3d = 23 \) ...(ii)
Subtracting: \( 2d = 16 \Rightarrow d = 8 \).
\( a + 8 = 7 \Rightarrow a = -1 \).
\( b = a + 2d = -1 + 16 = 15 \).
\( c = a + 4d = -1 + 32 = 31 \).
Values are \( a = -1, b = 15, c = 31 \).