NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2#
Welcome to our NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2. This exercise develops students’ ability to find the $n$th term of an AP using the formula, to find which term a given number is, and to solve problems using the $n$th term formula. These solutions are essential for CBSE Class 10 board exam preparation.
$n$th Term of an AP: The $n$th term (general term) of an AP with first term $a$ and common difference $d$ is:
$$a_n = a + (n-1)d$$This can also be written as $a_n = a + (n-1)d = dn + (a-d)$, which is a linear function of $n$.
1. 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 AP: #
(i) $a = 7$, $d = 3$, $n = 8$, $a_n = ?$
(ii) $a = -18$, $d = ?$, $n = 10$, $a_n = 0$
(iii) $a = ?$, $d = -3$, $n = 18$, $a_n = -5$
(iv) $a = -18.9$, $d = 2.5$, $n = ?$, $a_n = 3.6$
(v) $a = 3.5$, $d = 0$, $n = 105$, $a_n = ?$
2. Choose the correct choice in the following and justify: #
(i) 30th term of the AP: $10, 7, 4, \ldots$ is (A) 97 (B) 77 (C) −77 (D) −87
$a = 10$, $d = 7-10 = -3$, $n = 30$.
$$a_{30} = 10 + (30-1)(-3) = 10 - 87 = -77$$∴ Answer: (C) −77
(ii) 11th term of the AP: $-3, -\dfrac{1}{2}, 2, \ldots$ is (A) 28 (B) 22 (C) −38 (D) $-48\dfrac{1}{2}$
$a = -3$, $d = -\dfrac{1}{2}-(-3) = \dfrac{5}{2}$, $n = 11$.
$$a_{11} = -3 + (11-1)\left(\frac{5}{2}\right) = -3 + 25 = 22$$∴ Answer: (B) 22
3. In the following APs, find the missing terms in the boxes: #
(i) $2, \square, 26$
$a = 2$, $a_3 = 26$. So $26 = 2 + 2d \implies d = 12$.
Missing term $= 2 + 12 = \mathbf{14}$
The AP: $2, 14, 26$
(ii) $\square, 13, \square, 3$
Let $a_1 = a$, $a_2 = 13$, $a_4 = 3$.
$d = \dfrac{3-13}{4-2} = \dfrac{-10}{2} = -5$
$a = 13 - (-5) = 13 + 5 = \mathbf{18}$
$a_3 = 13 + (-5) = \mathbf{8}$
The AP: $18, 13, 8, 3$
(iii) $5, \square, \square, 9\dfrac{1}{2}$
$a = 5$, $a_4 = 9.5$. $d = \dfrac{9.5-5}{3} = \dfrac{4.5}{3} = 1.5$
$a_2 = 5 + 1.5 = \mathbf{6.5}$, $a_3 = 6.5 + 1.5 = \mathbf{8}$
The AP: $5, 6.5, 8, 9.5$
(iv) $-4, \square, \square, \square, \square, 6$
$a = -4$, $a_6 = 6$. $d = \dfrac{6-(-4)}{5} = 2$
$a_2 = -2$, $a_3 = 0$, $a_4 = 2$, $a_5 = 4$
The AP: $-4, -2, 0, 2, 4, 6$
(v) $\square, 38, \square, \square, \square, -22$
$a_2 = 38$, $a_6 = -22$. $d = \dfrac{-22-38}{4} = \dfrac{-60}{4} = -15$
$a = 38 - (-15) = \mathbf{53}$
$a_3 = 38 + (-15) = \mathbf{23}$, $a_4 = \mathbf{8}$, $a_5 = \mathbf{-7}$
The AP: $53, 38, 23, 8, -7, -22$
4. Which term of the AP: $3, 8, 13, 18, \ldots$ is 78? #
Answer
$a = 3$, $d = 5$, $a_n = 78$.
$$78 = 3 + (n-1) \times 5$$$$75 = 5(n-1)$$$$n - 1 = 15 \implies n = 16$$∴ 78 is the 16th term of the AP.
5. Find the number of terms in each of the following APs: #
(i) $7, 13, 19, \ldots, 205$
$a = 7$, $d = 6$, $a_n = 205$.
$$205 = 7 + (n-1) \times 6$$$$198 = 6(n-1)$$$$n-1 = 33 \implies n = \mathbf{34}$$(ii) $18, 15\dfrac{1}{2}, 13, \ldots, -47$
$a = 18$, $d = -\dfrac{5}{2}$, $a_n = -47$.
$$-47 = 18 + (n-1)\left(-\frac{5}{2}\right)$$$$-65 = -\frac{5}{2}(n-1)$$$$n-1 = 26 \implies n = \mathbf{27}$$6. Check whether $-150$ is a term of the AP: $11, 8, 5, 2, \ldots$ #
Answer
$a = 11$, $d = -3$, $a_n = -150$.
$$-150 = 11 + (n-1)(-3)$$$$-161 = -3(n-1)$$$$n-1 = \frac{161}{3} = 53.67 \ldots$$Since $n$ is not a natural number (not an integer), $-150$ is not a term of the given AP.
7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73. #
Answer
$a_{11} = a + 10d = 38 \quad \cdots (1)$
$a_{16} = a + 15d = 73 \quad \cdots (2)$
Subtract (1) from (2): $5d = 35 \implies d = 7$
From (1): $a = 38 - 70 = -32$
$$a_{31} = -32 + 30 \times 7 = -32 + 210 = \mathbf{178}$$8. An AP consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term. #
Answer
$a_3 = a + 2d = 12 \quad \cdots (1)$
$a_{50} = a + 49d = 106 \quad \cdots (2)$
Subtract (1) from (2): $47d = 94 \implies d = 2$
From (1): $a = 12 - 4 = 8$
$$a_{29} = 8 + 28 \times 2 = 8 + 56 = \mathbf{64}$$9. If the 3rd and the 9th terms of an AP are 4 and −8 respectively, which term of this AP is zero? #
Answer
$a + 2d = 4 \quad \cdots (1)$
$a + 8d = -8 \quad \cdots (2)$
Subtract (1) from (2): $6d = -12 \implies d = -2$
From (1): $a = 4 + 4 = 8$
For $a_n = 0$:
$$0 = 8 + (n-1)(-2) \implies 2(n-1) = 8 \implies n-1 = 4 \implies n = 5$$∴ The 5th term is zero.
10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference. #
Answer
$a_{17} - a_{10} = 7$
$[a + 16d] - [a + 9d] = 7$
$7d = 7 \implies d = \mathbf{1}$
11. Which term of the AP: $3, 15, 27, 39, \ldots$ will be 132 more than its 54th term? #
Answer
$a = 3$, $d = 12$.
$a_{54} = 3 + 53 \times 12 = 3 + 636 = 639$
We need $a_n = 639 + 132 = 771$.
$$771 = 3 + (n-1) \times 12$$$$768 = 12(n-1)$$$$n-1 = 64 \implies n = \mathbf{65}$$∴ The 65th term will be 132 more than the 54th term.
12. Two APs have the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000th terms? #
Answer
Let the first terms be $a$ and $a'$, and common difference be $d$ (same for both).
$a_{100} - a'_{100} = [a + 99d] - [a' + 99d] = a - a' = 100$
$a_{1000} - a'_{1000} = [a + 999d] - [a' + 999d] = a - a' = \mathbf{100}$
∴ The difference between their 1000th terms is also 100.
13. How many three-digit numbers are divisible by 7? #
Answer
Three-digit numbers divisible by 7: $105, 112, 119, \ldots, 994$
$a = 105$, $d = 7$, $a_n = 994$.
$$994 = 105 + (n-1) \times 7$$$$889 = 7(n-1)$$$$n-1 = 127 \implies n = \mathbf{128}$$∴ There are 128 three-digit numbers divisible by 7.
14. How many multiples of 4 lie between 10 and 250? #
Answer
Multiples of 4 between 10 and 250: $12, 16, 20, \ldots, 248$
$a = 12$, $d = 4$, $a_n = 248$.
$$248 = 12 + (n-1) \times 4$$$$236 = 4(n-1)$$$$n - 1 = 59 \implies n = \mathbf{60}$$∴ There are 60 multiples of 4 between 10 and 250.
15. For what value of $n$, are the $n$th terms of two APs: $63, 65, 67, \ldots$ and $3, 10, 17, \ldots$ equal? #
Answer
AP 1: $a = 63$, $d = 2$. $n$th term $= 63 + (n-1)(2) = 2n + 61$
AP 2: $a = 3$, $d = 7$. $n$th term $= 3 + (n-1)(7) = 7n - 4$
Setting equal:
$$2n + 61 = 7n - 4$$$$65 = 5n \implies n = \mathbf{13}$$∴ The 13th terms of both APs are equal.
16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. #
Answer
$a_7 - a_5 = 12$
$[a+6d] - [a+4d] = 12 \implies 2d = 12 \implies d = 6$
$a_3 = a + 2d = 16 \implies a + 12 = 16 \implies a = 4$
The AP: $4, 10, 16, 22, \ldots$
17. Find the 20th term from the last term of the AP: $3, 8, 13, \ldots, 253$. #
Answer
The AP from the last term backwards has $a' = 253$ and $d' = -5$.
20th term from last $= 253 + (20-1)(-5) = 253 - 95 = \mathbf{158}$
18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP. #
Answer
$a_4 + a_8 = (a+3d) + (a+7d) = 2a + 10d = 24 \implies a + 5d = 12 \quad \cdots (1)$
$a_6 + a_{10} = (a+5d) + (a+9d) = 2a + 14d = 44 \implies a + 7d = 22 \quad \cdots (2)$
Subtract (1) from (2): $2d = 10 \implies d = 5$
From (1): $a = 12 - 25 = -13$
The first three terms: $-13, -8, -3$
19. Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000? #
Answer
$a = 5000$, $d = 200$, $a_n = 7000$.
$$7000 = 5000 + (n-1) \times 200$$$$2000 = 200(n-1)$$$$n-1 = 10 \implies n = 11$$He started in 1995, so year = $1995 + 10 = \mathbf{2005}$.
∴ His income reached ₹7000 in the year 2005.
20. Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the $n$th week, her weekly savings become ₹20.75, find $n$. #
Answer
$a = 5$, $d = 1.75$, $a_n = 20.75$.
$$20.75 = 5 + (n-1)(1.75)$$$$15.75 = 1.75(n-1)$$$$n-1 = 9 \implies n = \mathbf{10}$$∴ In the 10th week, her savings will be ₹20.75.