CBSE Class 10 Maths HOTs Arithmetic Progressions Set H

Very Short Answer Type Questions

Question. The ninth term of an A.P. is -32 and the sum of its eleventh and thirteenth term is -94. Find the common difference of the A.P.
Answer: Let the first term be \( a \) and common difference be \( d \).
Now \( a + 8d = a_9 \)
\( a + 8d = -32 \dots(1) \)
and \( a_{11} + a_{13} = -94 \)
\( a + 10d + a + 12d = -94 \)
\( a + 11d = -47 \dots(2) \)
Solving equation (1) and (2), we have
\( d = -5 \)

Question. The seventeenth term of an A.P. exceeds its \( 10^{th} \) term by 7. Find the common difference.
Answer: Let the first term be \( a \) and common difference be \( d \).
Now \( a_{17} = a_{10} + 7 \)
\( a + 16d = a + 9d + 7 \)
\( 16d - 9d = 7 \)
\( 7d = 7 \)
\( d = 1 \)
Thus common difference is 1.

Question. If the number \( x + 3, 2x + 1 \) and \( x - 7 \) are in A.P. find the value of \( x \).
Answer: If \( x, y \) and \( z \) are three consecutive terms of an A.P. then we have
\( y - x = z - y \)
\( (2x + 1) - (x + 3) = (x - 7) - (2x + 1) \)
\( 2x + 1 - x - 3 = x - 7 - 2x - 1 \)
\( x - 2 = -x - 8 \)
\( 2x = -6 \)
\( x = -3 \)

Question. The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{th} \) term be \( a_n \).
We have, \( a_4 = 0 \)
\( a + 3d = 0 \) [\( a + (n - 1)d = a_n \)]
\( 3d = -a \)
\( -3d = a \) ...(1)
Now, \( a_{25} = a + 24d = -3d + 24d = 21d \) ...(2)
\( a_{11} = a + 10d = -3d + 10d = 7d \) ...(3)
From eqn (2) and eq (3) we have
\( a_{25} = 3a_{11} \) Hence Proved.

Question. The fourth term of an A.P. is 11. The sum of the fifth and seventh terms of the A.P. is 34. Find the common difference.
Answer: Let the first term be \( a \) and common difference be \( d \).
Now \( a_4 = 11 \)
\( a + 3d = 11 \dots(1) \)
and \( a_5 + a_7 = 34 \)
\( a + 4d + a + 6d = 34 \)
\( 2a + 10d = 34 \)
\( a + 5d = 17 \dots(2) \)
Solving equations (1) and (2) we have
\( d = 3 \)

Question. Find the middle term of the A.P. 213, 205, 197, .... 37.
Answer: Let the first term of an A.P. be \( a \), common difference be \( d \) and number of terms be \( m \).
Here, \( a = 213, d = 205 - 213 = -8, a_m = 37 \)
\( a_m = a + (m - 1)d \)
\( 37 = 213 + (m - 1)(-8) \)
\( 37 - 213 = -8(m - 1) \)
\( m - 1 = \frac{-176}{-8} = 22 \)

Question. Find the sum of first 15 multiples of 8.
Answer: Let the first term be \( a = 8 \), common difference be \( d = 8 \), \( n^{th} \) term be \( a_n \) and sum of \( n \) terms be \( S_n \).
First term of given A.P. be 8 and common difference be 8. Then
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( S_{15} = \frac{15}{2}[2 \times 8 + (15 - 1)8] \)
\( = \frac{15}{2}[16 + 112] \)
\( = \frac{15}{2} \times 128 = 960 \)
Hence, the sum of 15 terms is 960.

Question. Which of the term of A.P. \(5, 2, -1, \dots\) is \(-49\)?
Answer: Let the first term of an A.P. be \( a \) and common difference \( d \).
We have \( a = 5, d = -3 \)
Now \( a_n = a + (n - 1)d \)
Substituting all values we have
\( -49 = 5 + (n - 1)(-3) \)
\( -49 = 5 - 3n + 3 \)
\( 3n = 49 + 5 + 3 \)
\( n = \frac{57}{3} = 19^{th} \) term.

Question. Find the first four terms of an A.P. Whose first term is \(-2\) and common difference is \(-2\).
Answer: We have \( a_1 = -2 \),
\( a_2 = a_1 + d = -2 + (-2) = -4 \)
\( a_3 = a_2 + d = -4 + (-2) = -6 \)
\( a_4 = a_3 + d = -6 + (-2) = -8 \)
Hence first four terms are \(-2, -4, -6, -8\).

Question. Find the tenth term of the sequence \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \dots \)
Answer: Let the first term of an A.P. be \( a \) and common difference be \( d \).
Given AP is \( \sqrt{2}, \sqrt{8}, \sqrt{18} \) or \( \sqrt{2}, 2\sqrt{2}, 3\sqrt{2} \dots \)
where, \( a = \sqrt{2}, d = \sqrt{2}, n = 10 \)
Now \( a_n = a + (n - 1)d \)
\( a_{10} = \sqrt{2} + (10 - 1)\sqrt{2} \)
\( = \sqrt{2} + 9\sqrt{2} \)
\( = 10\sqrt{2} \)
Therefore tenth term of the given sequence \( \sqrt{200} \).

Question. Find the next term of the series \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32} \dots \)
Answer: Let the first term of an A.P. be \( a \) and common difference \( d \).
Here, \( a = \sqrt{2} \), \( a + d = \sqrt{8} = 2\sqrt{2} \)
\( d = 2\sqrt{2} - \sqrt{2} = \sqrt{2} \)
Next term \( = \sqrt{32} + \sqrt{2} \)
\( = 4\sqrt{2} + \sqrt{2} \)
\( = 5\sqrt{2} \)
\( = \sqrt{50} \)

Question. Is series \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots \) an A.P.? Give reason.
Answer: Let common difference be \( d \) then we have
\( d = a_2 - a_1 = \sqrt{6} - \sqrt{3} = \sqrt{3}(\sqrt{2} - 1) \)
\( d = a_3 - a_2 = \sqrt{9} - \sqrt{6} = 3 - \sqrt{6} \)
\( d = a_4 - a_3 = \sqrt{12} - \sqrt{9} = 2\sqrt{3} - 3 \)
As common difference are not equal, the given series is not in A.P.

Question. What is the next term of an A.P. \( \sqrt{7}, \sqrt{28}, \sqrt{63}, \dots \)?
Answer: Let the first term of an A.P. be \( a \) and common difference be \( d \).
Here, \( a = \sqrt{7}, a + d = \sqrt{28} \)
\( d = \sqrt{28} - \sqrt{7} = 2\sqrt{7} - \sqrt{7} = \sqrt{7} \)
Next term \( = \sqrt{63} + \sqrt{7} \)
\( = 3\sqrt{7} + \sqrt{7} = 4\sqrt{7} \)
\( = \sqrt{7 \times 16} \)
\( = \sqrt{112} \)

Question. If the common difference of an A.P. is \(-6\), find \( a_{16} - a_{12} \).
Answer: Let the first term of an A.P. be \( a \) and common difference be \( d \).
Now \( d = -6 \)
\( a_{16} = a + (16 - 1)(-6) = a - 90 \)
\( a_{12} = a + (12 - 1)(-6) = a - 66 \)
\( a_{16} - a_{12} = (a - 90) - (a - 66) = a - 90 - a + 66 \)
\( = -24 \)

Question. For what value of \( k \) will the consecutive terms \( 2k + 1, 3k + 3 \) and \( 5k - 1 \) from an A.P.?
Answer: If \( x, y, \) and \( z \) are in AP we have
\( y - x = z - y \)
Thus if \( 2k + 1, 3k + 3, 5k - 1 \) are in A.P. then
\( (3k + 3) - (2k + 1) = (5k - 1) - (3k + 3) \)
\( 3k + 3 - 2k - 1 = 5k - 1 - 3k - 3 \)
\( k + 2 = 2k - 4 \)
\( 2k - k = 4 + 2 \)
\( k = 6 \)

Question. Find the \( 25^{th} \) term of the A.P. \( -5, \frac{-5}{2}, 0, \frac{5}{2} \dots \)
Answer: Let the first term of an A.P. be \( a \) and common difference be \( d \).
Here, \( a = -5, d = \frac{-5}{2} - (-5) = \frac{5}{2} \)
\( a_n = a + (n - 1)d \)
\( a_{25} = -5 + (25 - 1) \times \left( \frac{5}{2} \right) \)
\( = -5 + 60 \)
\( = 55 \)

Question. The first three terms of an A.P. are \( 3y - 1, 3y + 5 \) and \( 5y + 1 \) respectively then find \( y \).
Answer: If \( x, y, \) and \( z \) are in AP then we have
\( y - x = z - y \)
Therefore if \( 3y - 1, 3y + 5 \) and \( 5y + 1 \) in A.P.
\( (3y + 5) - (3y - 1) = (5y + 1) - (3y + 5) \)
\( 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5 \)
\( 6 = 2y - 4 \)
\( 2y = 6 + 4 \)
\( y = \frac{10}{2} = 5 \)

Short Answer Type Questions

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 the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
\(S_n = \frac{n}{2}[2a + (n - 1)d]\)
\(S_{12} = 6[2a + 11d] = 12a + 66d\)
\(S_8 = 4[2a + 7d] = 8a + 28d\)
\(S_4 = 2[2a + 3d] = 4a + 6d\)
\(3(S_8 - S_4) = 3[(8a + 28d) - (4a + 6d)]\)
\(= 3[4a + 22d] = 12a + 66d\)
\(= 6[2a + 11d] = S_{12}\) 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: 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, \(a_{14} = 2a_8\) and \(a_6 = -8\)
Now \(a + 13d = 2(a + 7d)\)
\(a + 13d = 2a + 14d\)
\(a = -d\) ...(1)
and \(a_6 = -8\)
\(a + 5d = -8\) ...(2)
Solving (1) and (2), we get
\(a = 2, d = -2\)
Now \(S_{20} = \frac{20}{2}[2 \times 2 + (20 - 1)(-2)]\)
\(= 10[4 + 19 \times (-2)]\)
\(= 10(4 - 38)\)
\(= 10 \times (-34) = -340\)

Question. If the ratio of the sums of first \(n\) terms of two A.P.’s is \((7n + 1) : (4n + 27)\), find the ratio of their \(m^{th}\) terms.
Answer: Let \(a\), and \(A\) be the first term and \(d\) and \(D\) be the common difference of two AP’s, then we have
\(\frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a + (n - 1)d]}{\frac{n}{2}[2A + (n - 1)D]} = \frac{7n + 1}{4n + 27}\)
\(\frac{2a + (n - 1)d}{2A + (n - 1)D} = \frac{7n + 1}{4n + 27}\)
\(\frac{a + (\frac{n - 1}{2})d}{A + (\frac{n - 1}{2})D} = \frac{7n + 1}{4n + 27}\)
Putting \(\frac{n - 1}{2} = m - 1\) or \(n = 2m - 1\) we get
\(\frac{a + (m - 1)d}{A + (m - 1)D} = \frac{7(2m - 1) + 1}{4(2m - 1) + 27} = \frac{14m - 6}{8m + 23}\)
Hence, \(\frac{a_m}{A_m} = \frac{14m - 6}{8m + 23}\)

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: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
Sum of \(n\) term \(S_n = \frac{1}{2}[3n^2 + 7n]\)
Sum of 1 term \(S_1 = \frac{1}{2}[3 \times (1)^2 + 7(1)] = \frac{1}{2}[3 + 7] = \frac{1}{2} \times 10 = 5\)
Sum of 2 term \(S_2 = \frac{1}{2}[3(2)^2 + 7 \times 2] = \frac{1}{2}[12 + 14] = \frac{1}{2} \times 26 = 13\)
Now \(a_1 = S_1 = 5\)
\(a_2 = S_2 - S_1 = 13 - 5 = 8\)
\(d = a_2 - a_1 = 8 - 5 = 3\)
Now, A.P. is \(5, 8, 11, \dots\)
\(n^{th}\) term, \(a_n = a + (n - 1)d = 5 + (n - 1)3 = 5 + 3n - 3 = 3n + 2\)
\(20^{th}\) term, \(a_{20} = 5 + (20 - 1)(3) = 62\)
Hence, \(a_{20} = 62\)

Question. In an A.P., if the \(12^{th}\) term is \(-13\) and the sum of its first four terms is 24, find the sum of its first ten terms.
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
\(a_{12} = a + 11d = -13\) ...(1)
\(S_n = \frac{n}{2}[2a + (n - 1)d]\)
Now \(S_4 = 2[2a + 3d] = 24\)
\(2a + 3d = 12\) ...(2)
Multiplying (1) by 2 and subtracting (2) from it we get
\((2a + 22d) - (2a + 3d) = -26 - 12\)
\(19d = -38\)
\(d = -2\)
Substituting the value of \(d\) in (1) we get
\(a + 11 \times -2 = -13\)
\(a = -13 + 22\)
\(a = 9\)
Now, \(S_n = \frac{n}{2}[2a + (n - 1)d]\)
\(S_{10} = \frac{10}{2}(2 \times 9 + 9 \times -2)\)
\(= 5 \times (18 - 18) = 0\)
Hence, \(S_{10} = 0\)

Question. The tenth term of an A.P., is \(-37\) and the sum of its first six terms is \(-27\). Find the sum of its first eight terms.
Answer: Let the first term be \(a\), common difference be \(d\), \(n^{th}\) term be \(a_n\) and sum of \(n\) term be \(S_n\).
\(a_n = a + (n - 1)d\)
\(S_n = \frac{n}{2}[2a + (n - 1)d]\)
\(a + 9d = -37\) ...(1)
\(3(2a + 5d) = -27\)
\(2a + 5d = -9\) ...(2)
Multiplying (1) by 2 and subtracting (2) from it, we get
\((2a + 18d) - (2a + 5d) = -74 + 9\)
\(13d = -65\)
\(d = -5\)
Substituting the value of \(d\) in (1) we get
\(a + 9 \times -5 = -37\)
\(a = -37 + 45\)
\(a = 8\)
Now \(S_n = \frac{n}{2}[2a + (n - 1)d]\)
\(S_8 = \frac{8}{2}[2 \times 8 + (8 - 1)(-5)]\)
\(= 4[16 - 35]\)
\(= 4 \times -19 = -76\)
Hence, \(S_8 = -76\)

Question. Find the sum of first seventeen terms of A.P. whose \(4^{th}\) and \(9^{th}\) terms are \(-15\) and \(-30\) respectively.
Answer: Let the first term be \(a\), common difference be \(d\) and \(n^{th}\) term be \(a_n\).
Now \(a_4 = a + 3d = -15\) ...(1)
\(a_9 = a + 8d = -30\) ...(2)
Subtracting eqn (1) from eqn (2), we obtain
\((a + 8d) - (a + 3d) = -30 - (-15)\)
\(5d = -15 \implies d = \frac{-15}{5} = -3\)
Substituting the value of \(d\) in (1) we get
\(a + 3(-3) = -15\)
\(a = -15 + 9 = -6\)
Now \(S_{17} = \frac{17}{2}[2 \times (-6) + (17 - 1)(-3)]\)
\(= \frac{17}{2}[-12 + 16 \times (-3)]\)
\(= \frac{17}{2}[-12 - 48]\)
\(= \frac{17}{2}[-60] = 17 \times (-30)\)
\(= -510\)
Thus \(S_{17} = -510\)

Question. The common difference of an A.P. is \(-2\). Find its sum, if first term is 100 and last term is \(-10\).
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 \(a = 100, d = -2, a_n = -10\)
Now \(a_n = a + (n - 1)d\)
\(-10 = 100 + (n - 1)(-2)\)
\(-10 = 100 - 2n + 2\)
\(2n = 112\)
\(n = 56\)
Thus \(56^{th}\) term is \(-10\) and number of terms in A.P. are 56.
Now \(S_n = \frac{n}{2}(a + l)\)
\(S_{56} = \frac{56}{2}(100 - 10)\)
\(= \frac{56}{2}(90) = 56 \times 45 = 2520\)
Thus \(S_n = 2520\)

Question. The \( 16^{th} \) term of an A.P. is five times its third term. If its \( 10^{th} \) term is 41, then find the sum of its first 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 \( S_n \).
We have, \( a_{16} = 5a_3 \)
\( a + 15d = 5(a + 2d) \)
\( 4a = 5d \) ...(1)
and \( a_{10} = 41 \)
\( a + 9d = 41 \) ...(2)
Solving (1) and (2), we get
\( a = 5, d = 4 \)
Now \( S_{15} = \frac{15}{2}[2 \times 5 + (15 - 1) \times 4] \)
\( = \frac{15}{2}[10 + 56] \)
\( = \frac{15}{2} \times 66 = 15 \times 33 = 495 \)
Thus \( S_{15} = 495 \)

Question. The \( 13^{th} \) term of an A.P. is four times its \( 3^{rd} \) term. If the fifth term is 16, then find the sum of its first ten 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 \( a_{13} = 4a_3 \)
\( a + 12d = 4(a + 2d) \)
\( 3a = 4d \) ...(1)
and \( a_5 = 16 \)
\( a + 4d = 16 \) ...(2)
Substituting the value of \( a = \frac{4}{3}d \) in (2)
\( \frac{4}{3}d + 4d = 16 \)
\( 16d = 48 \Rightarrow d = 3 \)
Thus \( a = 4 \) and \( d = 3 \)
Now \( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( S_{10} = \frac{10}{2}[2 \times 4 + (10 - 1)3] \)
\( = 5[8 + 27] = 5 \times 35 = 175 \)
Thus \( S_{10} = 175 \)

Question. The \( n^{th} \) term of an A.P. is given by \( (-4n + 15) \). Find the sum of first 20 terms of this 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 \( a_n = -4n + 15 \)
\( a_1 = -4 \times 1 + 15 = 11 \)
\( a_2 = -4 \times 2 + 15 = 7 \)
\( a_3 = -4 \times 3 + 15 = 3 \)
\( d = a_2 - a_1 = 7 - 11 = -4 \)
Now, we have \( a = 11, d = -4 \)
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( S_{20} = \frac{20}{2}[2 \times 11 + (20 - 1) \times (-4)] \)
\( = 10[22 - 76] \)
\( = 10 \times (-54) = -540 \)
Thus \( S_{20} = -540 \)

Question. The sum of first 7 terms of an A.P. is 63 and sum of its next 7 terms is 161. Find \( 28^{th} \) term of 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 \( S_n \).
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
Now, \( S_7 = 63 \)
\( \frac{7}{2}[2a + 6d] = 63 \)
\( 2a + 6d = 18 \) ...(1)
Also, sum of next 7 terms,
\( S_{14} = S_{first 7} + S_{next 7} = 63 + 161 \)
\( \frac{14}{2}[2a + 13d] = 224 \)
\( 2a + 13d = 32 \) ...(2)
Subtracting (1) from (2)
\( 7d = 14 \Rightarrow d = 2 \)
Substituting the value of \( d \) in (1) we get
\( a = 3 \)
Now \( a_n = a + (n - 1)d \)
\( a_{28} = 3 + 2 \times (27) \)
\( = 57 \)
Thus \( 28^{th} \) term is 57.

Question. The sum of first \( n \) terms of an A.P. is given by \( S_n = 3n^2 - 4n \). Determine the A.P. and the \( 12^{th} \) term.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \).
\( S_n = 3n^2 - 4n \)
\( S_1 = 3(1)^2 - 4(1) = -1 \)
\( S_2 = 3(2)^2 - 4(2) = 4 \)
\( a_1 = S_1 = -1 \)
\( a_2 = S_2 - S_1 = 4 - (-1) = 5 \)
\( d = a_2 - a_1 = 5 - (-1) = 6 \)
Thus AP is -1, 5, 11, ....
Now \( a_{12} = a + 11d \)
\( = -1 + 11 \times 6 = 65 \)

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, \( a = 12, d = 4 \). Let \( n^{th} \) term be last term \( a_n = 96 \)
Now \( a + (n - 1)d = a_n \)
\( 12 + (n - 1)4 = 96 \)
\( (n - 1)4 = 96 - 12 = 84 \)
\( n - 1 = 21 \)
\( n = 21 + 1 = 22 \)
Now, \( S_{22} = \frac{22}{2}[12 + 96] \)
\( = 11 \times 108 \)
\( = 1188 \)

Question. Find the sum of the following series.
\( 5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + .... + (-5) + 81 + (-3) \)
Answer: The series can be written as
\( (5 + 9 + 13 + ... + 81) + [(-41) + (-39) + (-37) + (-35) + ... + (-5) + (-3)] \)
For the series \( (5 + 9 + 13 + ..... + 81) \)
\( a = 5 \)
\( d = 4 \)
and \( a_n = 81 \)
Now \( a_n = 5 + (n - 1)4 = 81 \)
\( 81 = 5 + (n - 1)4 \)
\( (n - 1)4 = 76 \)
\( n = 20 \)
\( S_n = \frac{20}{2}(5 + 81) = 860 \)
For series \( (-41) + (-39) + (-37) + ... + (-5) + (-3) \)
\( a_n = -3 \)
\( a = -41 \)
\( d = 2 \)
\( a_n = -41 + (n - 1)(2) \)
\( -3 = -41 + 2n - 2 \Rightarrow n = 20 \)
Now \( S_n = \frac{20}{2}[-41 + (-3)] = -440 \)
Sum of the series \( = 860 - 440 = 420 \)

Question. If \( m^{th} \) term of an AP is \( \frac{1}{n} \) and \( n^{th} \) term is \( \frac{1}{m} \) find the sum of first \( mn \) terms.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \).
Now \( a_m = a + (m - 1)d = \frac{1}{n} \) ...(1)
\( a_n = a + (n - 1)d = \frac{1}{m} \) ...(2)
Subtracting (2) from (1) we get
\( (m - n)d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn} \)
\( d = \frac{1}{mn} \)
Substituting this value of \( d \) in (1), we get
\( a = \frac{1}{mn} \)
Now, \( S_{mn} = \frac{mn}{2} \left[ \frac{2}{mn} + (mn - 1)\frac{1}{mn} \right] \)
\( = 1 + \frac{mn}{2} - \frac{1}{2} = \frac{1}{2} + \frac{mn}{2} \)
\( = \frac{1}{2}[mn + 1] \)
Hence, the sum on \( mn \) terms is \( \frac{1}{2}[mn + 1] \).

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

Question. The minimum age of children to be eligible to participate in a painting competition is 8 years. It is observed that the age of youngest boy was 8 years and the ages of rest of participants are having a common difference of 4 months. If the sum of ages of all the participants is 168 years, find the age of eldest participant in the painting competition. 
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 \( a = 8, d = 4 \text{ months} = \frac{1}{3} \text{ years}, S_n = 168 \)
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( 168 = \frac{n}{2} \left[ 2(8) + (n - 1)\frac{1}{3} \right] \)
\( n^2 + 47n - 1008 = 0 \)
\( n^2 + 63n - 16n - 1008 = 0 \)
\( (n - 16)(n + 63) = 0 \)
\( n = 16 \) or \( n = -63 \)
As \( n \) cannot be negative, we take \( n = 16 \)
Age of the eldest participant \( = a + 15d = 13 \text{ years} \)