CBSE Class 10 Arithmetic Progressions Sure Shot Questions Set H

Question. Fill in the blanks in the following table, given that ‘\( a \)’ is the first term, ‘\( d \)’ the common difference and \( a_n \) the \( n \)th term of the A.P.:
(i) \( a = 7, d = 3, n = 8, a_n = \dots \)
(ii) \( a = - 18, d = \dots, n = 10, a_n = 0 \)
(iii) \( a = \dots, d = - 3, n = 18, a_n = - 5 \)
(iv) \( a = - 18.9, d = 2.5, n = \dots, a_n = 3.6 \)
(v) \( a = 3.5, d = 0, n = 105, a_n = \dots \)
Answer: (i) \( a_n = a + (n - 1) d \)
\( \Rightarrow a_8 = 7 + (8 - 1) 3 \)
\( = 7 + 7 \times 3 \)
\( = 7 + 21 \)
\( \Rightarrow a_8 = 28 \)
(ii) \( a_n = a + (n - 1) d \)
\( \Rightarrow a_{10} = - 18 + (10 - 1) d \)
\( \Rightarrow 0 = - 18 + 9d \)
\( \Rightarrow 9d = 18 \Rightarrow d = \frac{18}{9} = 2 \)
\( \therefore d = 2 \)
(iii) \( a_n = a + (n - 1) d \)
\( \Rightarrow - 5 = a + (18 - 1) \times (- 3) \)
\( \Rightarrow - 5 = a + 17 \times (- 3) \)
\( \Rightarrow - 5 = a - 51 \)
\( \Rightarrow a = - 5 + 51 = 46 \)
Thus, \( a = 46 \)
(iv) \( a_n = a + (n - 1) d \)
\( \Rightarrow 3.6 = - 18.9 + (n - 1) \times 2.5 \)
\( \Rightarrow (n - 1) \times 2.5 = 3.6 + 18.9 \)
\( \Rightarrow (n - 1) \times 2.5 = 22.5 \)
\( \Rightarrow n - 1 = \frac{22.5}{2.5} = 9 \)
\( \Rightarrow n = 9 + 1 = 10 \)
Thus, \( n = 10 \)
(v) \( a_n = a + (n - 1) d \)
\( \Rightarrow a_n = 3.5 + (105 - 1) \times 0 \)
\( \Rightarrow a_n = 3.5 + 104 \times 0 \)
\( \Rightarrow a_n = 3.5 + 0 = 3.5 \)
Thus, \( a_n = 3.5 \)

Question. Choose the correct choice in the following and justify: 30th term of the A.P.: 10, 7, 4, ...., is
(a) 97
(b) 77
(c) \( - 77 \)
(d) \( - 87 \)
Answer: (c)
Justification: Here, \( a = 10, n = 30 \)
\( \because T_n = a + (n - 1) d \) and \( d = 7 - 10 = - 3 \)
\( \therefore T_{30} = 10 + (30 - 1) \times (- 3) \)
\( \Rightarrow T_{30} = 10 + 29 \times (- 3) \)
\( \Rightarrow T_{30} = 10 - 87 = - 77 \)
Thus, the correct choice is (C) \( - 77 \).

Question. Choose the correct choice in the following and justify: 11th term of the A.P.: \( - 3, - \frac{1}{2}, 2, \dots, \) is
(a) 28
(b) 22
(c) \( - 38 \)
(d) \( - 48 \frac{1}{2} \)
Answer: (b)
Justification: Here, \( a = - 3, n = 11 \) and \( d = - \frac{1}{2} - (- 3) = - \frac{1}{2} + 3 = \frac{5}{2} \)
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow T_{11} = - 3 + (11 - 1) \times \frac{5}{2} \)
\( \Rightarrow T_{11} = - 3 + 25 = 22 \)
Thus, the correct choice is (B) 22.

Question. In the following A.Ps., find the missing terms in the boxes:
(i) 2, [ ], 26
(ii) [ ], 13, [ ], 3
(iii) 5, [ ], [ ], \( 9 \frac{1}{2} \)
(iv) \( - 4, \text{[ ], [ ], [ ], [ ], } 6 \)
(v) [ ], 38, [ ], [ ], [ ], \( - 22 \)
Answer: (i) Here, \( a = 2, T_3 = 26 \)
Let common difference \( = d \)
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow T_3 = 2 + (3 - 1) d \)
\( \Rightarrow 26 = 2 + 2d \)
\( \Rightarrow 2d = 26 - 2 = 24 \)
\( \Rightarrow d = \frac{24}{2} = 12 \)
\( \therefore \) The missing term \( = a + d = 2 + 12 = 14 \)

(ii) Let the first term \( = a \) and common difference \( = d \)
Here, \( T_2 = 13 \) and \( T_4 = 3 \)
\( T_2 = a + d = 13 \)
\( T_4 = a + 3d = 3 \)
\( \therefore T_4 - T_2 = (a + 3d) - (a + d) = 3 - 13 \)
\( \Rightarrow 2d = - 10 \Rightarrow d = \frac{-10}{2} = - 5 \)
Now, \( a + d = 13 \Rightarrow a + (- 5) = 13 \Rightarrow a = 13 + 5 = 18 \)
Thus, missing terms are \( a \) and \( a + 2d \) or 18 and \( 18 + (- 10) = 8 \)
i.e., \( T_1 = 18 \) and \( T_3 = 8 \)

(iii) Here, \( a = 5 \) and \( T_4 = 9 \frac{1}{2} \)
since, \( T_4 = a + 3d \)
\( \Rightarrow 9 \frac{1}{2} = 5 + 3d \)
\( \Rightarrow 3d = 9 \frac{1}{2} - 5 = 4 \frac{1}{2} \)
\( \Rightarrow d = 4 \frac{1}{2} \div 3 = \frac{9}{2} \times \frac{1}{3} = \frac{3}{2} \)
\( \therefore \) The missing terms are:
\( T_2 = a + d = 5 + \frac{3}{2} = 6 \frac{1}{2} \)
\( T_3 = a + 2d = 5 + 2 \left( \frac{3}{2} \right) = 8 \)

(iv) Here, \( a = - 4 \) and \( T_6 = 6 \)
\( \because T_n = a + (n - 1) d \)
\( \therefore T_6 = - 4 + (6 - 1) d \)
\( \Rightarrow 6 = - 4 + 5d \)
\( \Rightarrow 5d = 6 + 4 = 10 \)
\( \Rightarrow d = 10 \div 5 = 2 \)
\( \therefore T_2 = a + d = - 4 + 2 = - 2 \)
\( T_3 = a + 2d = - 4 + 2 (2) = 0 \)
\( T_4 = a + 3d = - 4 + 3 (2) = 2 \)
\( T_5 = a + 4d = - 4 + 4 (2) = 4 \)
\( \therefore \) The missing terms are \( - 2, 0, 2, 4 \)

(v) Here, \( T_2 = 38 \) and \( T_6 = - 22 \)
\( \therefore T_2 = a + d = 38 \)
\( T_6 = a + 5d = - 22 \)
\( \Rightarrow T_6 - T_2 = a + 5d - (a + d) = - 22 - 38 \)
\( \Rightarrow 4d = - 60 \Rightarrow d = \frac{-60}{4} = - 15 \)
\( \therefore a + d = 38 \Rightarrow a + (- 15) = 38 \Rightarrow a = 38 + 15 = 53 \)
Now, \( T_3 = a + 2d = 53 + 2 (- 15) = 53 - 30 = 23 \)
\( T_4 = a + 3d = 53 + 3 (- 15) = 53 - 45 = 8 \)
\( T_5 = a + 4d = 53 + 4 (- 15) = 53 - 60 = - 7 \)
Thus, the missing terms are 53, 23, 8, \( - 7 \).

Question. Which term of the A.P.: 3, 8, 13, 18, ..., is 78?
Answer: Let the \( n \)th term \( = 78 \)
Here, \( a = 3, \Rightarrow T_1 = 3 \) and \( T_2 = 8 \)
\( \therefore d = T_2 - T_1 = 8 - 3 = 5 \)
Now, \( T_n = a + (n - 1) d \)
\( \Rightarrow 78 = 3 + (n - 1) \times 5 \)
\( \Rightarrow 78 - 3 = (n - 1) \times 5 \)
\( \Rightarrow 75 = (n - 1) \times 5 \)
\( \Rightarrow (n - 1) = 75 \div 5 = 15 \)
\( \Rightarrow n = 15 + 1 = 16 \)
Thus, 78 is the 16th term of the given A.P.

Question. Find the number of terms in each of the following A.Ps.:
(i) 7, 13, 19, ..., 205
(ii) \( 18, 15 \frac{1}{2}, 13, \dots, - 47 \)
Answer: (i) Here, \( a = 7 \)
\( d = 13 - 7 = 6 \)
Let the number of terms be \( n \)
\( \therefore T_n = 205 \)
Now, \( T_n = a + (n - 1) \times d \)
\( \Rightarrow 7 + (n - 1) \times 6 = 205 \)
\( \Rightarrow (n - 1) \times 6 = 205 - 7 = 198 \)
\( \Rightarrow n - 1 = \frac{198}{6} = 33 \)
\( \therefore n = 33 + 1 = 34 \)
Thus, the required number of terms is 34.

(ii) Here, \( a = 18 \)
\( d = 15 \frac{1}{2} - 18 = - 2 \frac{1}{2} = - \frac{5}{2} \)
Let the \( n \)th term \( = - 47 \)
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow - 47 = 18 + (n - 1) \times \left( - \frac{5}{2} \right) \)
\( \Rightarrow - 47 - 18 = (n - 1) \times \left( - \frac{5}{2} \right) \)
\( \Rightarrow - 65 = (n - 1) \times \left( - \frac{5}{2} \right) \)
\( \Rightarrow n - 1 = - 65 \times \left( - \frac{2}{5} \right) \)
\( \Rightarrow n - 1 = (- 13) \times (- 2) = 26 \)
\( \Rightarrow n = 26 + 1 = 27 \)
Thus, the required number of terms is 27.

Question. Check whether \( - 150 \) is a term of the A.P.: 11, 8, 5, 2 ...
Answer: For the given A.P., we have
\( a = 11 \)
\( d = 8 - 11 = - 3 \)
Let \( - 150 \) is the \( n \)th term of the given A.P.
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow - 150 = 11 + (n - 1) \times (- 3) \)
\( \Rightarrow - 150 - 11 = (n - 1) \times (- 3) \)
\( \Rightarrow - 161 = (n - 1) \times (- 3) \)
\( \Rightarrow n - 1 = \frac{-161}{-3} = \frac{161}{3} \)
\( \Rightarrow n = \frac{161}{3} + 1 = \frac{164}{3} = 54 \frac{2}{3} \)
But \( n \) should be a positive integer.
Thus, \( - 150 \) is not a term of the given A.P.

Question. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Answer: Here, \( T_{31} = ? \)
\( T_{11} = 38 \)
\( T_{16} = 73 \)
If the first term \( = a \) and the common difference \( = d \).
Then,
\( a + (11 - 1) d = 38 \)
\( \Rightarrow a + 10d = 38 \) ...(1)
and \( a + (16 - 1) d = 73 \)
\( \Rightarrow a + 15d = 73 \) ...(2)
Subtracting (1) from (2), we get
\( (a + 15d) - (a + 10d) = 73 - 38 \)
\( \Rightarrow 5d = 35 \)
\( \Rightarrow d = \frac{35}{5} = 7 \)
From (1),
\( a + 10 (7) = 38 \)
\( \Rightarrow a + 70 = 38 \)
\( \Rightarrow a = 38 - 70 = - 32 \)
\( \therefore T_{31} = - 32 + (31 - 1) \times 7 \)
\( \Rightarrow T_{31} = - 32 + 30 \times 7 \)
\( \Rightarrow T_{31} = - 32 + 210 \)
\( \Rightarrow T_{31} = 178 \)
Thus, the 31st term is 178.

Question. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Answer: Here, \( n = 50 \)
\( T_3 = 12 \)
\( T_n = 106 \Rightarrow T_{50} = 106 \)
If first term \( = a \) and the common difference \( = d \)
\( \therefore T_3 = a + 2d = 12 \) ...(1)
\( T_{50} = a + 49d = 106 \) ...(2)
\( \Rightarrow T_{50} - T_3 \Rightarrow a + 49d - (a + 2d) = 106 - 12 \)
\( \Rightarrow 47d = 94 \)
\( \Rightarrow d = \frac{94}{47} = 2 \)
From (1), we have
\( a + 2 (2) = 12 \Rightarrow a + 4 = 12 \)
\( \Rightarrow a = 12 - 4 = 8 \)
Now, \( T_{29} = a + (29 - 1) d \)
\( = 8 + (28) \times 2 \)
\( = 8 + 56 = 64 \)
Thus, the 29th term is 64.

Question. If the 3rd and the 9th terms of an A.P. are 4 and \( - 8 \) respectively, which term of this A.P. is zero?
Answer: Here, \( T_3 = 4 \) and \( T_9 = - 8 \)
\( \therefore \) Using \( T_n = a + (n - 1) d \)
\( \Rightarrow T_3 = a + 2d = 4 \) ...(1)
\( T_9 = a + 8d = - 8 \) ...(2)
Subtracting (1) from (2) we get
\( (a + 8d) - (a + 2d) = - 8 - 4 \)
\( \Rightarrow 6d = - 12 \)
\( \Rightarrow d = \frac{-12}{6} = - 2 \)
Now, from (1), we have:
\( a + 2d = 4 \)
\( \Rightarrow a + 2 (- 2) = 4 \)
\( \Rightarrow a - 4 = 4 \)
\( \Rightarrow a = 4 + 4 = 8 \)
Let the nth term of the A.P. be 0.
\( \therefore T_n = a + (n - 1) d = 0 \)
\( \Rightarrow 8 + (n - 1) \times (- 2) = 0 \)
\( \Rightarrow (n - 1) \times - 2 = - 8 \)
\( \Rightarrow n - 1 = \frac{-8}{-2} = 4 \)
\( \Rightarrow n = 4 + 1 = 5 \)
Thus, the 5th term of the A.P. is 0.

Question. Check whether − 150 is a term of the A.P.: 11, 8, 5, 2 ...
Answer: Sol. For the given A.P., we have
\( a = 11 \)
\( d = 8 - 11 = - 3 \)
Let − 150 is the \( n \)th term of the given A.P.
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow - 150 = 11 + (n - 1) \times (- 3) \)
\( \Rightarrow - 150 - 11 = (n - 1) \times (- 3) \)
\( \Rightarrow - 161 = (n - 1) \times (- 3) \)
\( \Rightarrow n - 1 = \frac{-161}{-3} = \frac{161}{3} \)
\( \Rightarrow n = \frac{161}{3} + 1 = \frac{164}{3} = 54 \frac{2}{3} \)
But \( n \) should be a positive integer.
Thus, − 150 is not a term of the given A.P.

Question. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Answer: Sol. Here, \( T_{31} = ? \)
\( T_{11} = 38 \)
\( T_{16} = 73 \)
If the first term \( = a \) and the common difference \( = d \).
Then,
\( a + (11 - 1) d = 38 \)
\( \Rightarrow a + 10d = 38 \) ...(1)
and \( a + (16 - 1) d = 73 \)
\( \Rightarrow a + 15d = 73 \) ...(2)
Subtracting (1) from (2), we get
\( (a + 15d) - (a + 10d) = 73 - 38 \)
\( \Rightarrow 5d = 35 \)
\( \Rightarrow d = \frac{35}{5} = 7 \)
From (1),
\( a + 10 (7) = 38 \)
\( \Rightarrow a + 70 = 38 \)
\( \Rightarrow a = 38 - 70 = - 32 \)
\( \therefore T_{31} = - 32 + (31 - 1) \times 7 \)
\( \Rightarrow T_{31} = - 32 + 30 \times 7 \)
\( \Rightarrow T_{31} = - 32 + 210 \)
\( \Rightarrow T_{31} = 178 \)
Thus, the 31st term is 178.

Question. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Answer: Sol. Here, \( n = 50 \)
\( T_3 = 12 \)
\( T_n = 106 \Rightarrow T_{50} = 106 \)
If first term \( = a \) and the common difference \( = d \)
\( \therefore T_3 = a + 2d = 12 \) ...(1)
\( T_{50} = a + 49d = 106 \) ...(2)
\( \Rightarrow T_{50} - T_3 \Rightarrow a + 49d - (a + 2d) = 106 - 12 \)
\( \Rightarrow 47d = 94 \)
\( \Rightarrow d = \frac{94}{47} = 2 \)
From (1), we have
\( a + 2 (2) = 12 \Rightarrow a + 4 = 12 \)
\( \Rightarrow a = 12 - 4 = 8 \)
Now, \( T_{29} = a + (29 - 1) d \)
\( = 8 + (28) \times 2 \)
\( = 8 + 56 = 64 \)
Thus, the 29th term is 64.

Question. If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively, which term of this A.P. is zero ?
Answer: Sol. Here, \( T_3 = 4 \) and \( T_9 = - 8 \)
\( \therefore \) Using \( T_n = a + (n - 1) d \)
\( \Rightarrow T_3 = a + 2d = 4 \) ...(1)
\( T_9 = a + 8d = - 8 \) ...(2)
Subtracting (1) from (2) we get
\( (a + 8d) - (a + 2d) = - 8 - 4 \)
\( \Rightarrow 6d = - 12 \)
\( \Rightarrow d = \frac{-12}{6} = - 2 \)
Now, from (1), we have:
\( a + 2d = 4 \)
\( \Rightarrow a + 2 (- 2) = 4 \)
\( \Rightarrow a - 4 = 4 \)
\( \Rightarrow a = 4 + 4 = 8 \)
Let the nth term of the A.P. be 0.
\( \therefore T_n = a + (n - 1) d = 0 \)
\( \Rightarrow 8 + (n - 1) \times (- 2) = 0 \)
\( \Rightarrow (n - 1) \times - 2 = - 8 \)
\( \Rightarrow n - 1 = \frac{-8}{-2} = 4 \)
\( \Rightarrow n = 4 + 1 = 5 \)
Thus, the 5th term of the A.P. is 0.

Question. The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Answer: Sol. Let \( a \) be the first term and \( d \) be the common difference of the given A.P.
Now, using \( T_n = a + (n - 1) d \)
\( T_{17} = a + 16d \)
\( T_{10} = a + 9d \)
According to the condition,
\( T_{10} + 7 = T_{17} \)
\( \Rightarrow (a + 9d) + 7 = a + 16d \)
\( \Rightarrow a + 9d - a - 16d = - 7 \)
\( \Rightarrow - 7d = - 7 \Rightarrow d = 1 \)
Thus, the common difference is 1.

Question. Which term of the A.P.: 3, 15, 27, 39, ... will be 132 more than its 54th term?
Answer: Sol. Here, \( a = 3 \)
\( d = 15 - 3 = 12 \)
Using \( T_n = a + (n - 1) d \), we get
\( T_{54} = a + 53d \)
\( = 3 + 53 \times 12 \)
\( = 3 + 636 = 639 \)
Let \( a_n \) be 132 more than its 54th term.
\( \therefore a_n = T_{54} + 132 \)
\( \Rightarrow a_n = 639 + 132 = 771 \)
Now \( a_n = a + (n - 1) d = 771 \)
\( \Rightarrow 3 + (n - 1) \times 12 = 771 \)
\( \Rightarrow (n - 1) \times 12 = 771 - 3 = 768 \)
\( \Rightarrow (n - 1) = \frac{768}{12} = 64 \)
\( \Rightarrow n = 64 + 1 = 65 \)
Thus, 132 more than 54th term is the 65th term.

Question. Two A.Ps. have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Answer: Sol. Let for the 1st A.P., the first term \( = a \)
\( \therefore T_{100} = a + 99d \)
And for the 2nd A.P., the first term \( = a' \)
\( \therefore T'_{100} = a' + 99d \)
According to the condition, we have:
\( T_{100} - T'_{100} = 100 \)
\( \Rightarrow a + 99d - (a' + 99d) = 100 \)
\( \Rightarrow a - a' = 100 \)
Let, \( T_{1000} - T'_{1000} = x \)
\( \therefore a + 999d - (a' + 999d) = x \)
\( \Rightarrow a - a' = x \Rightarrow x = 100 \)
\( \therefore \) The difference between the 1000th terms is 100.

Question. How many three-digit numbers are divisible by 7?
Answer: Ans. The first three digit number divisible by 7 is 105.
The last such three digit number is 994.
\( \therefore \) The A.P. is 105, 112, 119, ....., 994
Here, \( a = 105 \) and \( d = 7 \)
Let \( n \) be the required number of terms.
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow 994 = 105 + (n - 1) \times 7 \)
\( \Rightarrow (n - 1) \times 7 = 994 - 105 = 889 \)
\( \Rightarrow (n - 1) = \frac{889}{7} = 127 \)
\( \Rightarrow n = 127 + 1 = 128 \)
Thus, 128 numbers of 3-digit are divisible by 7.

Question. How many multiples of 4 lie between 10 and 250?
Answer: Sol. \( \because \) The first multiple of 4 beyond 10 is 12.
The multiple of 4 just below 250 is 248.
\( \therefore \) The A.P. is given by:
12, 16, 20, ....., 248
Here, \( a = 12 \) and \( d = 4 \)
Let the number of terms \( = n \)
\( \therefore \) Using \( T_n = a + (n - 1) d \), we get
\( \therefore T_n = 12 + (n - 1) \times 4 \)
\( \Rightarrow 248 = 12 + (n - 1) \times 4 \)
\( \Rightarrow (n - 1) \times 4 = 248 - 12 = 236 \)
\( \Rightarrow n - 1 = \frac{236}{4} = 59 \)
\( \Rightarrow n = 59 + 1 = 60 \)
Thus, the required number of terms \( = 60 \).

Question. For what value of n, are the nth terms of two A.Ps.: 63, 65, 67, ... and 3, 10, 17, ... equal?
Answer: Sol. For the 1st A.P.
\( \because a = 63 \) and \( d = 65 - 63 = 2 \)
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow T_n = 63 + (n - 1) \times 2 \)
For the 2nd A.P.
\( \because a = 3 \) and \( d = 10 - 3 = 7 \)
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow T_n = 3 + (n - 1) \times 7 \)
Now, according to the condition,
\( 3 + (n - 1) \times 7 = 63 + (n - 1) \times 2 \)
\( \Rightarrow (n - 1) \times 7 - (n - 1) \times 2 = 63 - 3 \)
\( \Rightarrow 7n - 7 - 2n + 2 = 60 \)
\( \Rightarrow 5n - 5 = 60 \)
\( \Rightarrow 5n = 60 + 5 = 65 \)
\( \Rightarrow n = \frac{65}{5} = 13 \)
Thus, the 13th terms of the two given A.Ps. are equal.

Question. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Answer: Sol. Let the first term \( = a \) and the common difference \( = d \).
\( \therefore \) Using \( T_n = a + (n - 1) d \), we have:
\( T_3 = a + 2d \)
\( \Rightarrow a + 2d = 16 \) ...(1)
And \( T_7 = a + 6d, T_5 = a + 4d \)
According to the condition,
\( T_7 - T_5 = 12 \)
\( \Rightarrow (a + 6d) - (a + 4d) = 12 \)
\( \Rightarrow a + 6d - a - 4d = 12 \)
\( \Rightarrow 2d = 12 \)
\( \Rightarrow d = \frac{12}{2} = 6 \) ...(2)
Now, from (1) and (2), we have:
\( a + 2 (6) = 16 \)
\( \Rightarrow a + 12 = 16 \)
\( \Rightarrow a = 16 - 12 = 4 \)
\( \therefore \) The required A.P. is
4, [4 + 6], [4 + 2 (6)], [4 + 3 (6)], .....
or 4, 10, 16, 22, .....

Question. Find the 20th term from the last term of the A.P.: 3, 8, 13, ..., 253.
Answer: Sol. We have, the last term \( l = 253 \)
Here, \( d = 8 - 3 = 5 \)
Since, the nth term before the last term is given by \( l - (n - 1) d \),
\( \therefore \) We have
20th term from the end \( = l - (20 - 1) \times 5 \)
\( = 253 - 19 \times 5 \)
\( = 253 - 95 = 158 \)

Question. The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.
Answer: Sol. Let the first term \( = a \)
And the common difference \( = d \)
\( \therefore \) Using \( T_n = a + (n - 1) d \),
\( T_4 + T_8 = 24 \)
\( \Rightarrow (a + 3d) + (a + 7d) = 24 \)
\( \Rightarrow 2a + 10d = 24 \)
\( \Rightarrow a + 5d = 12 \) ...(1)
And \( T_6 + T_{10} = 44 \)
\( \Rightarrow (a + 5d) + (a + 9d) = 44 \)
\( \Rightarrow 2a + 14d = 44 \)
\( \Rightarrow a + 7d = 22 \) ...(2)
Now, subtracting (1) from (2), we get
\( (a + 7d) - (a + 5d) = 22 - 12 \)
\( \Rightarrow 2d = 10 \)
\( \Rightarrow d = \frac{10}{2} = 5 \)
\( \therefore \) From (1), \( a + 5 \times 5 = 12 \)
\( \Rightarrow a + 25 = 12 \)
\( \Rightarrow a = 12 - 25 = - 13 \)
Now, the first three terms of the A.P. are given by:
\( a, (a + d), (a + 2d) \)
or \( - 13, (- 13 + 5), [- 13 + 2 (5)] \)
or \( - 13, - 8, - 3 \)

Question. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs  200 each year. In which year did his income reach Rs  7000?
Answer: Sol. Here, \( a = Rs 5000 \) and \( d = Rs 200 \)
Say, in the nth year he gets Rs 7000.
\( \therefore \) Using \( T_n = a + (n - 1) d \), we get
\( 7000 = 5000 + (n - 1) \times 200 \)
\( \Rightarrow (n - 1) \times 200 = 7000 - 5000 = 2000 \)
\( \Rightarrow n - 1 = \frac{2000}{200} = 10 \)
\( \Rightarrow n = 10 + 1 = 11 \)
Thus, his income becomes Rs 7000 in 11 years.

Question. Ramkali saved Rs 5 in the first week of a year and then increased weekly savings by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.
Answer: Sol. Here, \( a = Rs 5 \) and \( d = Rs 1.75 \)
\( \because \) In the nth week her savings become Rs 20.75.
\( \therefore T_n = Rs 20.75 \)
\( \therefore \) Using \( T_n = a + (n - 1) d \), we have
\( 20.75 = 5 + (n - 1) \times (1.75) \)
\( \Rightarrow (n - 1) \times 1.75 = 20.75 - 5 \)
\( \Rightarrow (n - 1) \times 1.75 = 15.75 \)
\( \Rightarrow n - 1 = \frac{15.75}{1.75} = 9 \)
\( \Rightarrow n = 9 + 1 = 10 \)
Thus, the required number of years \( = 10 \).

Sum of First n Terms of an A.P.

• (i) If the first term of an A.P. is ‘\( a \)’ and the common difference is ‘\( d \)’ then the sum of its first n terms is given by:
  \( S_n = \frac{n}{2} [2a + (n - 1) d] \)
• (ii) If the last term of the A.P. is \( l \) then
  \( S_n = \frac{n}{2} (a + l) \)

Remember,
The sum of first n positive integers is given by:
\( S_n = \frac{n(n + 1)}{2} \)

Question. Find the sum of the following A.Ps.:
(i) 2, 7, 12, ..., to 10 terms.
(ii) − 37, − 33, − 29, ..., to 12 terms.
(iii) 0.6, 1.7, 2.8, ..., to 100 terms.
(iv) \( \frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots \), to 11 terms.
Answer: (i) Here, \( a = 2 \), \( d = 7 - 2 = 5 \), \( n = 10 \).
Since, \( S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( \therefore S_{10} = \frac{10}{2} [2 \times 2 + (10 - 1) \times 5] \)
\( \Rightarrow S_{10} = 5 [4 + 9 \times 5] \)
\( \Rightarrow S_{10} = 5 [49] = 245 \)
Thus, the sum of first 10 terms is 245.

(ii) We have: \( a = - 37 \), \( d = - 33 - (- 37) = 4 \), \( n = 12 \).
\( \therefore S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( \Rightarrow S_{12} = \frac{12}{2} [2 (- 37) + (12 - 1) \times 4] \)
\( = 6 [- 74 + 11 \times 4] \)
\( = 6 [- 74 + 44] \)
\( = 6 \times [- 30] = - 180 \)
Thus, sum of first 12 terms = −180.

(iii) Here, \( a = 0.6 \), \( d = 1.7 - 0.6 = 1.1 \), \( n = 100 \).
\( \therefore S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( S_{100} = \frac{100}{2} [2 (0.6) + (100 - 1) \times 1.1] \)
\( = 50 [1.2 + 99 \times 1.1] \)
\( = 50 [1.2 + 108.9] \)
\( = 50 [110.1] = 5505 \)
Thus, the required sum of first 100 terms is 5505.

(iv) Here, \( a = \frac{1}{15} \), \( d = \frac{1}{12} - \frac{1}{15} = \frac{1}{60} \), \( n = 11 \).
\( \therefore S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( S_{11} = \frac{11}{2} [ (2 \times \frac{1}{15}) + (11 - 1) \times \frac{1}{60} ] \)
\( = \frac{11}{2} [\frac{2}{15} + 10 \times \frac{1}{60}] \)
\( = \frac{11}{2} [\frac{2}{15} + \frac{1}{6}] \)
\( = \frac{11}{2} [\frac{4 + 5}{30}] \)
\( = \frac{11}{2} \times \frac{9}{30} = \frac{99}{60} = \frac{33}{20} \)
Thus, the required sum of first 11 terms = \( \frac{33}{20} \).

Question. Find the sums given below:
(i) \( 7 + 10 \frac{1}{2} + 14 + \dots + 84 \)
(ii) \( 34 + 32 + 30 + \dots + 10 \)
(iii) \( - 5 + (- 8) + (- 11) + \dots + (- 230) \)
Answer: (i) Here, \( a = 7 \), \( d = 10 \frac{1}{2} - 7 = 3 \frac{1}{2} = \frac{7}{2} \), \( l = 84 \).
Let \( n \) be the number of terms.
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow 84 = 7 + (n - 1) \times \frac{7}{2} \)
\( \Rightarrow (n - 1) \times \frac{7}{2} = 84 - 7 = 77 \)
\( \Rightarrow n - 1 = 77 \times \frac{2}{7} = 22 \)
\( \Rightarrow n = 22 + 1 = 23 \)
Now, \( S_n = \frac{n}{2} (a + l) \)
\( \Rightarrow S_{23} = \frac{23}{2} (7 + 84) \)
\( = \frac{23}{2} \times 91 = \frac{2093}{2} = 1046 \frac{1}{2} \)
Thus, the required sum = \( 1046 \frac{1}{2} \).

(ii) Here, \( a = 34 \), \( d = 32 - 34 = - 2 \), \( l = 10 \).
Let the number of terms be \( n \).
\( \therefore T_n = 10 \)
\( \Rightarrow 10 = 34 + (n - 1) \times (- 2) \)
\( \Rightarrow (n - 1) \times (- 2) = 10 - 34 = - 24 \)
\( \Rightarrow n - 1 = \frac{-24}{-2} = 12 \)
\( \Rightarrow n = 13 \)
Now, \( S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( \Rightarrow S_{13} = \frac{13}{2} [2 \times 34 + (13 - 1) \times (- 2)] \)
\( = \frac{13}{2} [68 + 12 \times (- 2)] = \frac{13}{2} [68 - 24] \)
\( = \frac{13}{2} [44] = 13 \times 22 = 286 \)
OR
\( S_{13} = \frac{n}{2} (a + l) = \frac{13}{2} (34 + 10) = \frac{13}{2} \times 44 = 286 \)
Thus, the required sum is 286.

(iii) Here, \( a = - 5 \), \( d = - 8 - (- 5) = - 3 \), \( l = - 230 \).
Let \( n \) be the number of terms.
\( \therefore T_n = - 230 \)
\( \Rightarrow - 230 = - 5 + (n - 1) \times (- 3) \)
\( \Rightarrow (n - 1) \times (- 3) = - 230 + 5 = - 225 \)
\( \Rightarrow n - 1 = \frac{-225}{-3} = 75 \)
\( \Rightarrow n = 75 + 1 = 76 \)
Now, \( S_{76} = \frac{76}{2} [(- 5) + (- 230)] \)
\( = 38 \times (- 235) = - 8930 \)
\( \therefore \) The required sum = − 8930.

Question. In an A.P.:
(i) given \( a = 5, d = 3, a_n = 50 \), find \( n \) and \( S_n \).
(ii) given \( a = 7, a_{13} = 35 \), find \( d \) and \( S_{13} \).
(iii) given \( a_{12} = 37, d = 3 \), find \( a \) and \( S_{12} \).
(iv) given \( a_3 = 15, S_{10} = 125 \), find \( d \) and \( a_{10} \).
(v) given \( d = 5, S_9 = 75 \), find \( a \) and \( a_9 \).
(vi) given \( a = 2, d = 8, S_n = 90 \), find \( n \) and \( a_n \).
(vii) given \( a = 8, a_n = 62, S_n = 210 \), find \( n \) and \( d \).
(viii) given \( a_n = 4, d = 2, S_n = - 14 \), find \( n \) and \( a \).
(ix) given \( a = 3, n = 8, S = 192 \), find \( d \).
(x) given \( l = 28, S = 144 \), and there are total 9 terms. Find \( a \).
Answer: (i) Here, \( a = 5, d = 3 \) and \( a_n = 50 = l \)
\( \because a_n = a + (n - 1) d \)
\( \therefore 50 = 5 + (n - 1) \times 3 \)
\( \Rightarrow 50 - 5 = (n - 1) \times 3 \)
\( \Rightarrow (n - 1) \times 3 = 45 \)
\( \Rightarrow (n - 1) = \frac{45}{3} = 15 \)
\( \Rightarrow n = 15 + 1 = 16 \)
Now \( S_n = \frac{n}{2} (a + l) = \frac{16}{2} (5 + 50) = 8 (55) = 440 \)
Thus, \( n = 16 \) and \( S_n = 440 \).

(ii) Here, \( a = 7 \) and \( a_{13} = 35 = l \)
\( \therefore a_n = a + (n - 1) d \)
\( \Rightarrow 35 = 7 + (13 - 1) d \)
\( \Rightarrow 35 - 7 = 12d \)
\( \Rightarrow 28 = 12d \)
\( \Rightarrow d = \frac{28}{12} = \frac{7}{3} \)
Now, using \( S_n = \frac{n}{2} (a + l) \)
\( S_{13} = \frac{13}{2} (7 + 35) = \frac{13}{2} \times 42 = 13 \times 21 = 273 \)
\( S_n = 273 \) and \( d = \frac{7}{3} \).

(iii) Here, \( a_{12} = 37 = l \) and \( d = 3 \)
Let the first term of the A.P. be ‘a’.
Now \( a_{12} = a + (12 - 1) d \)
\( \Rightarrow 37 = a + 11d \)
\( \Rightarrow 37 = a + 11 \times 3 \)
\( \Rightarrow 37 = a + 33 \)
\( \Rightarrow a = 37 - 33 = 4 \)
Now, \( S_n = \frac{n}{2} (a + l) \)
\( \Rightarrow S_{12} = \frac{12}{2} (4 + 37) = 6 \times (41) = 246 \)
Thus, \( a = 4 \) and \( S_{12} = 246 \).

(iv) Here, \( a_3 = 15 = l \), \( S_{10} = 125 \)
Let first term of the A.P. be ‘a’ and the common difference = \( d \)
\( \therefore a_3 = a + 2d \Rightarrow a + 2d = 15 \) ...(1)
Again \( S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( \Rightarrow S_{10} = \frac{10}{2} [2a + (10 - 1) d] \)
\( \Rightarrow 125 = 5 [2a + 9d] \)
\( \Rightarrow 2a + 9d = \frac{125}{5} = 25 \) ...(2)
Multiplying (1) by 2 and subtracting (2) from it, we get
\( 2 [a + 2d = 15] - [2a + 9d = 25] \)
\( \Rightarrow 2a + 4d - 2a - 9d = 30 - 25 \)
\( \Rightarrow - 5d = 5 \Rightarrow d = \frac{5}{-5} = - 1 \)
\( \therefore \) From (1), \( a + 2 (- 1) = 15 \Rightarrow a = 15 + 2 \Rightarrow a = 17 \)
Now, \( a_{10} = a + (10 - 1) d = 17 + 9 \times (- 1) = 17 - 9 = 8 \)
Thus, \( d = - 1 \) and \( a_{10} = 8 \).

(v) Here, \( d = 5, S_9 = 75 \)
Let the first term of the A.P. is ‘a’.
\( \therefore S_9 = \frac{9}{2} [2a + (9 - 1) \times 5] \)
\( \Rightarrow 75 = \frac{9}{2} [2a + 40] \)
\( \Rightarrow 75 \times \frac{2}{9} = 2a + 40 \)
\( \Rightarrow \frac{50}{3} = 2a + 40 \)
\( \Rightarrow 2a = \frac{50}{3} - 40 = \frac{- 70}{3} \)
\( \Rightarrow a = \frac{- 70}{3} \times \frac{1}{2} = \frac{- 35}{3} \)
Now, \( a_9 = a + (9 - 1) d = \frac{- 35}{3} + (8 \times 5) = \frac{- 35}{3} + 40 = \frac{- 35 + 120}{3} = \frac{85}{3} \)
Thus, \( a = \frac{- 35}{3} \) and \( a_9 = \frac{85}{3} \).

(vi) Here, \( a = 2, d = 8 \) and \( S_n = 90 \)
\( \because S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( \therefore 90 = \frac{n}{2} [2 \times 2 + (n - 1) \times 8] \)
\( \Rightarrow 90 \times 2 = 4n + n (n - 1) \times 8 \)
\( \Rightarrow 180 = 4n + 8n^2 - 8n \)
\( \Rightarrow 180 = 8n^2 - 4n \)
\( \Rightarrow 45 = 2n^2 - n \)
\( \Rightarrow 2n^2 - n - 45 = 0 \)
\( \Rightarrow 2n^2 - 10n + 9n - 45 = 0 \)
\( \Rightarrow 2n (n - 5) + 9 (n - 5) = 0 \)
\( \Rightarrow (2n + 9) (n - 5) = 0 \)
\( \therefore \) Either \( 2n + 9 = 0 \Rightarrow n = - \frac{9}{2} \) or \( n - 5 = 0 \Rightarrow n = 5 \)
But \( n = - \frac{9}{2} \) is not required. \( \therefore n = 5 \)
Now, \( a_n = a + (n - 1) d \Rightarrow a_5 = 2 + (5 - 1) \times 8 = 2 + 32 = 34 \)
Thus, \( n = 5 \) and \( a_5 = 34 \).

(vii) Here, \( a = 8, a_n = 62 = l \) and \( S_n = 210 \)
Let the common difference = \( d \)
Now, \( S_n = 210 \Rightarrow \frac{n}{2} (a + l) = 210 \)
\( \Rightarrow \frac{n}{2} (8 + 62) = 210 \Rightarrow \frac{n}{2} \times 70 = 210 \Rightarrow 35n = 210 \)
\( \therefore n = \frac{210}{35} = 6 \)
Again \( a_n = a + (n - 1) d \Rightarrow 62 = 8 + (6 - 1) \times d \)
\( \Rightarrow 62 - 8 = 5d \Rightarrow 54 = 5d \Rightarrow d = \frac{54}{5} \)
Thus, \( n = 6 \) and \( d = \frac{54}{5} \).

(viii) Here, \( a_n = 4, d = 2 \) and \( S_n = - 14 \)
Let the first term be ‘a’.
\( \because a_n = 4 \Rightarrow a + (n - 1) 2 = 4 \Rightarrow a + 2n - 2 = 4 \Rightarrow a = 6 - 2n \) ...(1)
Also \( S_n = - 14 \Rightarrow \frac{n}{2} (a + l) = - 14 \Rightarrow \frac{n}{2} (a + 4) = - 14 \Rightarrow n (a + 4) = - 28 \) ...(2)
Substituting the value of \( a \) from (1) into (2),
\( n [6 - 2n + 4] = - 28 \Rightarrow n [10 - 2n] = - 28 \Rightarrow 2n [5 - n] = - 28 \)
\( \Rightarrow n (5 - n) = - 14 \) [Dividing throughout by 2]
\( \Rightarrow 5n - n^2 + 14 = 0 \Rightarrow n^2 - 5n - 14 = 0 \Rightarrow n^2 - 7n + 2n - 14 = 0 \)
\( \Rightarrow n (n - 7) + 2 (n - 7) = 0 \Rightarrow (n - 7) (n + 2) = 0 \)
\( \therefore \) Either \( n - 7 = 0 \Rightarrow n = 7 \) or \( n + 2 = 0 \Rightarrow n = - 2 \)
But \( n \) cannot be negative, \( \therefore n = 7 \)
Now, from (1), we have \( a = 6 - 2 \times 7 \Rightarrow a = - 8 \)
Thus, \( a = - 8 \) and \( n = 7 \).

(ix) Here, \( a = 3, n = 8 \) and \( S_n = 192 \)
Let the common difference = \( d \).
\( \because S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( \therefore 192 = \frac{8}{2} [2 (3) + (8 - 1) d] \)
\( \Rightarrow 192 = 4 [6 + 7d] \Rightarrow 192 = 24 + 28d \)
\( \Rightarrow 28d = 192 - 24 = 168 \Rightarrow d = \frac{168}{28} = 6 \)
Thus, \( d = 6 \).

(x) Here, \( l = 28 \) and \( S_9 = 144 \)
Let the first term be ‘a’.
Then \( S_n = \frac{n}{2} (a + l) \Rightarrow S_9 = \frac{9}{2} (a + 28) \)
\( \Rightarrow 144 = \frac{9}{2} (a + 28) \Rightarrow a + 28 = 144 \times \frac{2}{9} = 16 \times 2 = 32 \)
\( \Rightarrow a = 32 - 28 = 4 \)
Thus, \( a = 4 \).

Question. How many terms of the A.P.: 9, 17, 25, ... must be taken to give a sum of 636?
Answer: Here, \( a = 9, d = 17 - 9 = 8, S_n = 636 \)
\( \because S_n = \frac{n}{2} [2a + (n - 1) d] = 636 \)
\( \therefore \frac{n}{2} [(2 \times 9) + (n - 1) \times 8] = 636 \)
\( \Rightarrow n [18 + (n - 1) \times 8] = 1272 \)
\( \Rightarrow n (8n + 10) = 1272 \Rightarrow 8n^2 + 10n - 1272 = 0 \)
\( \Rightarrow 4n^2 + 5n - 636 = 0 \Rightarrow 4n^2 + 53n - 48n - 636 = 0 \)
\( \Rightarrow n (4n + 53) - 12 (4n + 53) = 0 \)
\( \Rightarrow (n - 12) (4n + 53) = 0 \Rightarrow n = 12 \) and \( n = - \frac{53}{4} \)
Rejecting \( n = - \frac{53}{4} \), we have \( n = 12 \).

Question. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Answer: Here, \( a = 5, l = 45 = T_n, S_n = 400 \)
\( \because T_n = a + (n - 1) d \)
\( \therefore 45 = 5 + (n - 1) d \Rightarrow (n - 1) d = 40 \) ...(1)
Also \( S_n = \frac{n}{2} (a + l) \Rightarrow 400 = \frac{n}{2} (5 + 45) \)
\( \Rightarrow 400 \times 2 = n \times 50 \Rightarrow n = \frac{400 \times 2}{50} = 16 \)
From (1), we get \( (16 - 1) d = 40 \Rightarrow 15d = 40 \Rightarrow d = \frac{40}{15} = \frac{8}{3} \).

Question. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer: We have, First term \( a = 17 \), Last term \( l = 350 = T_n \), Common difference \( d = 9 \).
Let the number of terms be ‘n’.
\( \because T_n = a + (n - 1) d \)
\( \therefore 350 = 17 + (n - 1) \times 9 \Rightarrow (n - 1) \times 9 = 350 - 17 = 333 \)
\( \Rightarrow n - 1 = \frac{333}{9} = 37 \Rightarrow n = 37 + 1 = 38 \)
Since, \( S_n = \frac{n}{2} (a + l) \)
\( \therefore S_{38} = \frac{38}{2} (17 + 350) = 19 (367) = 6973 \)
Thus, \( n = 38 \) and \( S_n = 6973 \).

Question. Find the sum of first 22 terms of an A.P. in which \( d = 7 \) and 22nd term is 149.
Answer: Here, \( n = 22, T_{22} = 149 = l, d = 7 \).
Let the first term of the A.P. be ‘a’.
\( \therefore T_n = a + (n - 1) d \Rightarrow T_{22} = a + (22 - 1) \times 7 \)
\( \Rightarrow 149 = a + 21 \times 7 \Rightarrow 149 = a + 147 \Rightarrow a = 149 - 147 = 2 \)
Now, \( S_{22} = \frac{n}{2} [a + l] \Rightarrow S_{22} = \frac{22}{2} [2 + 149] = 11 [151] = 1661 \)
Thus \( S_{22} = 1661 \).

Question. Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.
Answer: Here, \( n = 51, T_2 = 14 \) and \( T_3 = 18 \).
Let the first term of the A.P. be ‘a’ and the common difference is \( d \).
\( \therefore \) We have: \( T_2 = a + d \Rightarrow a + d = 14 \) ...(1)
\( T_3 = a + 2d \Rightarrow a + 2d = 18 \) ...(2)
Subtracting (1) from 2, we get \( a + 2d - a - d = 18 - 14 \Rightarrow d = 4 \)
From (1), we get \( a + 4 = 14 \Rightarrow a = 14 - 4 = 10 \)
Now, \( S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( \Rightarrow S_{51} = \frac{51}{2} [(2 \times 10) + (51 - 1) \times 4] = \frac{51}{2} [20 + 200] = \frac{51}{2} [220] = 51 \times 110 = 5610 \)
Thus, the sum of 51 terms is 5610.

Question. If the sum of first 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of first n terms.
Answer: Here, we have: \( S_7 = 49 \) and \( S_{17} = 289 \).
Let the first term of the A.P. be ‘a’ and ‘d’ be the common difference, then
\( S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( \Rightarrow S_7 = \frac{7}{2} [2a + (7 - 1) d] = 49 \Rightarrow 7 (2a + 6d) = 2 \times 49 = 98 \)
\( \Rightarrow 2a + 6d = \frac{98}{7} = 14 \Rightarrow 2 [a + 3d] = 14 \Rightarrow a + 3d = 7 \) ...(1)
Also, \( S_{17} = \frac{17}{2} [2a + (17 - 1) d] = 289 \Rightarrow \frac{17}{2} (2a + 16d) = 289 \)
\( \Rightarrow a + 8d = \frac{289}{17} = 17 \Rightarrow a + 8d = 17 \) ...(2)
Subtracting (1) from (2), we have: \( a + 8d - a - 3d = 17 - 7 \Rightarrow 5d = 10 \Rightarrow d = \frac{10}{5} = 2 \)
Now, from (1), we have \( a + 3 (2) = 7 \Rightarrow a = 7 - 6 = 1 \)
Now, \( S_n = \frac{n}{2} [2a + (n - 1) d] = \frac{n}{2} [2 \times 1 + (n - 1) \times 2] = \frac{n}{2} [2 + 2n - 2] = \frac{n}{2} [2n] = n \times n = n^2 \)
Thus, the required sum of n terms = \( n^2 \).

Question. Show that \( a_1, a_2, \dots, a_n, \dots \) form an A.P. where \( a_n \) is defined as below:
(i) \( a_n = 3 + 4n \) (ii) \( a_n = 9 - 5n \)
Also find the sum of the first 15 terms in each case.
Answer: (i) Here, \( a_n = 3 + 4n \)
Putting \( n = 1, 2, 3, 4, \dots n \), we get:
\( a_1 = 3 + 4 (1) = 7 \)
\( a_2 = 3 + 4 (2) = 11 \)
\( a_3 = 3 + 4 (3) = 15 \)
\( a_4 = 3 + 4 (4) = 19 \)
\( \dots \dots \)
\( a_n = 3 + 4n \)
\( \therefore \) The A.P. in which \( a = 7 \) and \( d = 11 - 7 = 4 \) is: 7, 11, 15, 19, ..., (3 + 4n).
Now \( S_{15} = \frac{15}{2} [(2 \times 7) + (15 - 1) \times 4] = \frac{15}{2} [14 + (14 \times 4)] = \frac{15}{2} [14 + 56] = \frac{15}{2} [70] = 15 \times 35 = 525 \).

(ii) Here, \( a_n = 9 - 5n \)
Putting \( n = 1, 2, 3, 4, \dots, n \), we get
\( a_1 = 9 - 5 (1) = 4 \)
\( a_2 = 9 - 5 (2) = - 1 \)
\( a_3 = 9 - 5 (3) = - 6 \)
\( a_4 = 9 - 5 (4) = - 11 \)
\( \dots \dots \)
\( \therefore \) The A.P. is: 4, − 1, − 6, − 11, ..... \( 9 - 5 (n) \) [having first term as 4 and \( d = - 1 - 4 = - 5 \)]
\( \therefore S_{15} = \frac{15}{2} [(2 \times 4) + (15 - 1) \times (- 5)] = \frac{15}{2} [8 + 14 \times (- 5)] = \frac{15}{2} [8 - 70] = \frac{15}{2} \times (- 62) = 15 \times (- 31) = - 465 \).

Question. If the sum of the first n terms of an A.P. is \( 4n - n^2 \), what is the first term (that is \( S_1 \))? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Answer: We have: \( S_n = 4n - n^2 \)
\( \therefore S_1 = 4 (1) - (1)^2 = 4 - 1 = 3 \Rightarrow \) First term = 3
\( S_2 = 4 (2) - (2)^2 = 8 - 4 = 4 \Rightarrow \) Sum of first two terms = 4
\( \therefore \) Second term \( (S_2 - S_1) = 4 - 3 = 1 \)
\( S_3 = 4 (3) - (3)^2 = 12 - 9 = 3 \Rightarrow \) Sum of first 3 terms = 3
\( \therefore \) Third term \( (S_3 - S_2) = 3 - 4 = - 1 \)
\( S_9 = 4 (9) - (9)^2 = 36 - 81 = - 45 \)
\( S_{10} = 4 (10) - (10)^2 = 40 - 100 = - 60 \)
\( \therefore \) Tenth term \( = S_{10} - S_9 = [- 60] - [- 45] = - 15 \)
Now, \( S_n = 4 (n) - (n)^2 = 4n - n^2 \)
Also \( S_{n - 1} = 4 (n - 1) - (n - 1)^2 = 4n - 4 - [n^2 - 2n + 1] = 4n - 4 - n^2 + 2n - 1 = 6n - n^2 - 5 \)
\( \therefore nth \text{ term} = S_n - S_{n - 1} = [4n - n^2] - [6n - n^2 - 5] = 4n - n^2 - 6n + n^2 + 5 = 5 - 2n \)
Thus, \( S_1 = 3 \) and \( a_1 = 3 \), \( S_2 = 4 \) and \( a_2 = 1 \), \( S_3 = 3 \) and \( a_3 = - 1 \), \( a_{10} = - 15 \) and \( a_n = 5 - 2n \).

Question. Find the sum of the first 40 positive integers divisible by 6.
Answer: \( \because \) The first 40 positive integers divisible by 6 are: 6, 12, 18, ....., \( (6 \times 40) \).
And, these numbers are in A.P. such that \( a = 6, d = 12 - 6 = 6 \) and \( a_n = 6 \times 40 = 240 = l \)
\( \therefore S_{40} = \frac{40}{2} [(2 \times 6) + (40 - 1) \times 6] = 20 [12 + 39 \times 6] = 20 [12 + 234] = 20 \times 246 = 4920 \)
OR
\( S_n = \frac{n}{2} [a + l] \Rightarrow S_{40} = \frac{40}{2} [6 + 240] = 20 \times 246 = 4920 \)
Thus, the sum of first 40 multiples of 6 is 4920.

Question. Find the sum of the first 15 multiples of 8.
Answer: The first 15 multiples of 8 are: 8, (8 × 2), (8 × 3), (8 × 4), ....., (8 × 15) or 8, 16, 24, 32, ....., 120.
These numbers are in A.P., where \( a = 8 \) and \( l = 120 \)
\( \therefore S_{15} = \frac{15}{2} [a + l] = \frac{15}{2} [8 + 120] = \frac{15}{2} \times 128 = 15 \times 64 = 960 \)
Thus, the sum of first positive 15 multiples of 8 is 960.