NCERT Class 10 Maths – Exercise 5.2 Solutions

NCERT Class 10 Maths

Chapter 5 – Arithmetic Progressions | Exercise 5.2

(Rationalized Syllabus 2025-26)

💡 Key Formula

The $n$-th term $a_n$ of an AP with first term $a$ and common difference $d$ is given by:

$$a_n = a + (n – 1)d$$

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) $a=7, d=3, n=8, a_n=\dots$

$$a_n = a + (n-1)d$$

$$a_8 = 7 + (8-1)3 = 7 + 21 = 28$$

✔️ 28

(ii) $a=-18, d=\dots, n=10, a_n=0$

$$0 = -18 + (10-1)d \Rightarrow 18 = 9d \Rightarrow d = 2$$

✔️ 2

(iii) $a=\dots, d=-3, n=18, a_n=-5$

$$-5 = a + (18-1)(-3)$$

$$-5 = a – 51 \Rightarrow a = 51 – 5 = 46$$

✔️ 46

(iv) $a=-18.9, d=2.5, n=\dots, a_n=3.6$

$$3.6 = -18.9 + (n-1)(2.5)$$

$$22.5 = (n-1)(2.5) \Rightarrow n-1 = \frac{22.5}{2.5} = 9 \Rightarrow n = 10$$

✔️ 10

(v) $a=3.5, d=0, n=105, a_n=\dots$

$$a_n = 3.5 + (105-1)(0) = 3.5$$

✔️ 3.5

Choose the correct choice in the following and justify:

(i) 30th term of the AP: 10, 7, 4, … is

Here $a=10, d = 7-10 = -3, n=30$.

$$a_{30} = 10 + (30-1)(-3) = 10 + 29(-3) = 10 – 87 = -77$$

✔️ (C) -77

(ii) 11th term of the AP: -3, -1/2, 2, … is

Here $a=-3, d = -\frac{1}{2} – (-3) = 2.5 (\text{or } \frac{5}{2}), n=11$.

$$a_{11} = -3 + (11-1)(\frac{5}{2}) = -3 + 10(\frac{5}{2}) = -3 + 25 = 22$$

✔️ (B) 22

In the following APs, find the missing terms in the boxes:

(i) 2, $\square$, 26

The middle term is the arithmetic mean of adjacent terms.

$$x = \frac{2 + 26}{2} = \frac{28}{2} = 14$$

✔️ 14

(ii) $\square$, 13, $\square$, 3

Let AP be $a, a+d, a+2d, a+3d$.

$$a+d=13, \quad a+3d=3$$

Subtracting first from second: $2d = -10 \Rightarrow d = -5$.

So $a = 13 – (-5) = 18$. Next term is $13 – 5 = 8$.

✔️ 18, 8

(iii) 5, $\square$, $\square$, $9\frac{1}{2}$

Here $a=5, a_4 = 9.5$.

$$5 + 3d = 9.5 \Rightarrow 3d = 4.5 \Rightarrow d = 1.5$$

Terms: $5 + 1.5 = 6.5$ ($6\frac{1}{2}$) and $6.5 + 1.5 = 8$.

✔️ $6\frac{1}{2}$, 8

(iv) -4, $\square$, $\square$, $\square$, $\square$, 6

$a=-4, a_6=6$.

$$-4 + 5d = 6 \Rightarrow 5d = 10 \Rightarrow d = 2$$

Terms: -2, 0, 2, 4.

✔️ -2, 0, 2, 4

(v) $\square$, 38, $\square$, $\square$, $\square$, -22

$a+d=38$ and $a+5d=-22$.

Subtracting: $4d = -60 \Rightarrow d = -15$.

$a = 38 – (-15) = 53$.

Terms: $53$, $38-15=23$, $23-15=8$, $8-15=-7$.

✔️ 53, 23, 8, -7

Which term of the AP: 3, 8, 13, 18, … is 78?

Here $a=3, d=5, a_n=78$.

$$78 = 3 + (n-1)5$$

$$75 = 5(n-1) \Rightarrow 15 = n-1 \Rightarrow n = 16$$

✔️ 16th term

Find the number of terms in each of the following APs:

(i) 7, 13, 19, …, 205

$a=7, d=6, a_n=205$.

$$205 = 7 + (n-1)6$$

$$198 = 6(n-1) \Rightarrow 33 = n-1 \Rightarrow n = 34$$

✔️ 34 terms

(ii) 18, $15\frac{1}{2}$, 13, …, -47

$a=18, d = 15.5 – 18 = -2.5, a_n=-47$.

$$-47 = 18 + (n-1)(-2.5)$$

$$-65 = (n-1)(-2.5) \Rightarrow n-1 = \frac{65}{2.5} = 26 \Rightarrow n = 27$$

✔️ 27 terms

Check whether -150 is a term of the AP: 11, 8, 5, 2…

$a=11, d=-3$. Check if $a_n = -150$ for integer $n$.

$$-150 = 11 + (n-1)(-3)$$

$$-161 = (n-1)(-3) \Rightarrow n-1 = \frac{161}{3}$$

Since 161 is not divisible by 3 (sum of digits is 8), $n$ is not an integer.

✖️ No, -150 is not a term

Q7-Q10

Miscellaneous Problems

Q7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

$$a + 10d = 38 \quad \text{…(1)}$$

$$a + 15d = 73 \quad \text{…(2)}$$

Subtracting (1) from (2): $5d = 35 \Rightarrow d = 7$.

From (1): $a + 70 = 38 \Rightarrow a = -32$.

$a_{31} = -32 + 30(7) = -32 + 210 = 178$.

✔️ 178

Q8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

$a_3 = 12 \Rightarrow a+2d=12$.

$a_{50} = 106 \Rightarrow a+49d=106$.

Subtracting: $47d = 94 \Rightarrow d = 2$.

$a + 4 = 12 \Rightarrow a = 8$.

$a_{29} = 8 + 28(2) = 8 + 56 = 64$.

✔️ 64

Q9. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

$$a+2d = 4, \quad a+8d = -8$$

Subtracting: $6d = -12 \Rightarrow d = -2$.

$a + 2(-2) = 4 \Rightarrow a = 8$.

Let $a_n = 0$: $8 + (n-1)(-2) = 0 \Rightarrow 2(n-1) = 8 \Rightarrow n=5$.

✔️ 5th term

Q10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

$$a_{17} – a_{10} = 7$$

$$(a+16d) – (a+9d) = 7 \Rightarrow 7d = 7 \Rightarrow d = 1$$

✔️ d = 1

Q11-Q14

Application Problems

Q11. Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?

$a=3, d=12$. We need $a_n = a_{54} + 132$.

Shortcut: $132 = k \times d \Rightarrow 132 = k(12) \Rightarrow k=11$.

So the term is $54 + 11 = 65$th term.

✔️ 65th term

Q12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Let APs be $a, a+d…$ and $b, b+d…$

Difference between $n$th terms is always $(a-b)$.

Since $a_{100} – b_{100} = 100 \Rightarrow a – b = 100$.

Thus, $a_{1000} – b_{1000} = a – b = 100$.

✔️ 100

Q13. How many three-digit numbers are divisible by 7?

First number: 105 ($15 \times 7$). Last number: 994 ($142 \times 7$).

$a=105, d=7, a_n=994$.

$$994 = 105 + (n-1)7$$

$$889 = 7(n-1) \Rightarrow n-1 = 127 \Rightarrow n = 128$$

✔️ 128 numbers

Q14. How many multiples of 4 lie between 10 and 250?

First multiple: 12. Last multiple: 248.

$$248 = 12 + (n-1)4$$

$$236 = 4(n-1) \Rightarrow n-1 = 59 \Rightarrow n = 60$$

✔️ 60 multiples

Q15-Q20

Advanced Problems

Q15. For what value of $n$, are the $n$th terms of two APs: 63, 65, 67… and 3, 10, 17… equal?

AP1: $63 + (n-1)2$. AP2: $3 + (n-1)7$.

$$63 + 2n – 2 = 3 + 7n – 7$$

$$61 + 2n = 7n – 4 \Rightarrow 65 = 5n \Rightarrow n = 13$$

✔️ 13th term

Q16. Determine the AP whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.

$a_7 – a_5 = 12 \Rightarrow 2d = 12 \Rightarrow d = 6$.

$a_3 = 16 \Rightarrow a + 2(6) = 16 \Rightarrow a = 4$.

✔️ AP: 4, 10, 16, 22…

Q17. Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.

Reverse the AP: $a=253, d=-5$. Find $a_{20}$.

$$a_{20} = 253 + 19(-5) = 253 – 95 = 158$$

✔️ 158

Q18. 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.

$$(a+3d) + (a+7d) = 24 \Rightarrow 2a + 10d = 24 \Rightarrow a + 5d = 12$$

$$(a+5d) + (a+9d) = 44 \Rightarrow 2a + 14d = 44 \Rightarrow a + 7d = 22$$

Subtracting: $2d = 10 \Rightarrow d = 5$. Then $a + 25 = 12 \Rightarrow a = -13$.

✔️ -13, -8, -3

Q19. Subba Rao started work at ₹5000 salary with ₹200 increment per year. In which year did his income reach ₹7000?

$a=5000, d=200, a_n=7000$.

$$7000 = 5000 + (n-1)200$$

$$2000 = 200(n-1) \Rightarrow 10 = n-1 \Rightarrow n = 11$$

✔️ 11th year

Q20. Ramkali saved ₹5 in the first week and increased her weekly saving by ₹1.75. If in the $n$th week, her weekly saving becomes ₹20.75, find $n$.

$a=5, d=1.75, a_n=20.75$.

$$20.75 = 5 + (n-1)1.75$$

$$15.75 = (n-1)1.75$$

$n-1 = \frac{1575}{175} = 9 \Rightarrow n = 10$.

✔️ n = 10

🎉 Exercise 5.2 Completed