Arithmetic Progressions – Exercise 5.2 | SJMaths

Arithmetic Progressions – Exercise 5.2

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This page provides comprehensive Arithmetic Progressions – Exercise 5.2. Free NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5-2. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.

NCERT Exercise Practice questions with detailed step-by-step solutions

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Basics (Q1-3) Terms (Q4-10) Properties (Q11-15) Word Problems (Q16-20)

Basics: Tables & Missing Terms

1. Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the nth term of the AP:

(i) $a=3, d=2, n=10, a_n=\dots$

(ii) $a=-10, d=\dots, n=10, a_n=8$

(iii) $a=\dots, d=-3, n=15, a_n=-5$

(iv) $a=-10, d=2.5, n=\dots, a_n=0$

(v) $a=5, d=0, n=100, a_n=\dots$


2. Choose the correct choice in the following:

(i) 20th term of the AP: $2, 5, 8 \dots$ is?

(ii) 10th term of the AP: $-2, 0, 2 \dots$ is?


3. Find the missing terms in the boxes:

(i) $4, \square, 12$

(ii) $\square, 10, \square, 0$

(iii) $2, \square, \square, 11$

(iv) $-3, \square, \square, \square, 5$

(v) $\square, 20, \square, \square, \square, -20$

Question 1
(i) $a_{10} = 3 + (10-1)(2) = 3 + 18 = 21$.
(ii) $8 = -10 + 9d \Rightarrow 18 = 9d \Rightarrow d = 2$.
(iii) $-5 = a + 14(-3) \Rightarrow -5 = a - 42 \Rightarrow a = 37$.
(iv) $0 = -10 + (n-1)(2.5) \Rightarrow 10 = 2.5(n-1) \Rightarrow n-1 = 4 \Rightarrow n = 5$.
(v) Since $d=0$, all terms are same. $a_n = 5$.

Question 2
(i) $a=2, d=3$. $a_{20} = 2 + 19(3) = 2 + 57 = 59$.
(ii) $a=-2, d=2$. $a_{10} = -2 + 9(2) = -2 + 18 = 16$.

Question 3
(i) Middle term = $\frac{4+12}{2} = 8$.
(ii) $a_2=10, a_4=0 \Rightarrow a+d=10, a+3d=0$. Subtract: $2d=-10 \Rightarrow d=-5$. $a=15$. Terms: $15, 10, 5, 0$.
(iii) $a=2, a_4=11 \Rightarrow 2+3d=11 \Rightarrow 3d=9 \Rightarrow d=3$. Terms: $2, 5, 8, 11$.
(iv) $a=-3, a_5=5 \Rightarrow -3+4d=5 \Rightarrow 4d=8 \Rightarrow d=2$. Terms: $-3, -1, 1, 3, 5$.
(v) $a_2=20, a_6=-20 \Rightarrow a+d=20, a+5d=-20$. Subtract: $4d=-40 \Rightarrow d=-10$. $a=30$. Terms: $30, 20, 10, 0, -10, -20$.

Finding Terms & Conditions

4. Which term of the AP: $5, 10, 15, 20 \dots$ is 105?

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

(i) $5, 10, \dots, 100$    (ii) $20, 19.5, \dots, -10$

6. Check whether 50 is a term of the AP: $2, 6, 10 \dots$

7. Find the 31st term of an AP whose 10th term is 20 and the 20th term is 40.

8. An AP consists of 50 terms. The 3rd term is 5 and the last term is 99. Find the 29th term.

9. If the 3rd term is 4 and 9th term is -8, which term is zero?

10. The 15th term of an AP exceeds its 10th term by 25. Find the common difference.

4. $a=5, d=5, a_n=105$.
$105 = 5 + (n-1)5 \Rightarrow 100 = 5(n-1) \Rightarrow 20 = n-1 \Rightarrow n=21$.
✅ 21st term.

5. (i) $100 = 5 + (n-1)5 \Rightarrow 95 = 5(n-1) \Rightarrow 19 = n-1 \Rightarrow n=20$.
(ii) $a=20, d=-0.5$. $-10 = 20 + (n-1)(-0.5) \Rightarrow -30 = -0.5(n-1) \Rightarrow 60 = n-1 \Rightarrow n=61$.

6. $a=2, d=4$. $50 = 2 + (n-1)4 \Rightarrow 48 = 4(n-1) \Rightarrow 12 = n-1 \Rightarrow n=13$.
✅ Yes, it is the 13th term.

7. $a+9d=20, a+19d=40$. Subtract: $10d=20 \Rightarrow d=2$.
$a + 18 = 20 \Rightarrow a=2$.
$a_{31} = 2 + 30(2) = 62$.

8. $a_3=5 \Rightarrow a+2d=5$. Last term $a_{50}=99 \Rightarrow a+49d=99$.
Subtract: $47d=94 \Rightarrow d=2$. $a+4=5 \Rightarrow a=1$.
$a_{29} = 1 + 28(2) = 57$.

9. $a+2d=4, a+8d=-8$. Subtract: $6d=-12 \Rightarrow d=-2$. $a=8$.
$a_n=0 \Rightarrow 8 + (n-1)(-2) = 0 \Rightarrow 8 = 2(n-1) \Rightarrow n=5$.

10. $a_{15} - a_{10} = 25 \Rightarrow (a+14d) - (a+9d) = 25 \Rightarrow 5d=25 \Rightarrow d=5$.

Properties & Counts

11. Which term of the AP: $3, 15, 27, 39 \dots$ will be 120 more than its 20th 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?

13. How many three-digit numbers are divisible by 6?

14. How many multiples of 4 lie between 12 and 248?

15. For what value of $n$, are the nth terms of two APs: $10, 12, 14 \dots$ and $5, 8, 11 \dots$ equal?

11. $a=3, d=12$. We need $a_n = a_{20} + 120$.
$a + (n-1)d = (a + 19d) + 120$.
$(n-1)12 = 19(12) + 120$. Divide by 12: $n-1 = 19 + 10 = 29 \Rightarrow n=30$.
✅ 30th term.

12. Difference between nth terms of two APs with same $d$ is constant ($a-b$).
Given $a_{100} - b_{100} = 100$, so $a-b=100$.
Thus, $a_{1000} - b_{1000} = 100$.

13. 3-digit numbers divisible by 6: $102, 108, \dots, 996$.
$a=102, d=6, a_n=996$.
$996 = 102 + (n-1)6 \Rightarrow 894 = 6(n-1) \Rightarrow 149 = n-1 \Rightarrow n=150$.

14. Multiples of 4 between 12 and 248 (inclusive): $12, 16, \dots, 248$.
$248 = 12 + (n-1)4 \Rightarrow 236 = 4(n-1) \Rightarrow 59 = n-1 \Rightarrow n=60$.

15. AP1: $10, 12 \dots (d=2)$. AP2: $5, 8 \dots (d=3)$.
$10 + (n-1)2 = 5 + (n-1)3$.
$10 + 2n - 2 = 5 + 3n - 3 \Rightarrow 8 + 2n = 2 + 3n \Rightarrow n=6$.

Word Problems & Construction

16. Determine the AP whose 3rd term is 5 and the 7th term exceeds the 5th term by 10.

17. Find the 10th term from the last term of the AP: $2, 4, 6, \dots, 100$.

18. The sum of the 3rd and 7th terms of an AP is 20 and the sum of the 5th and 9th terms is 32. Find the AP.

19. A person started work at an annual salary of ₹ 10,000 with an increment of ₹ 500 each year. In which year did his income reach ₹ 20,000?

20. Priya saved ₹ 10 in the first week and increased her savings by ₹ 5 weekly. If her savings become ₹ 110 in the nth week, find $n$.

16. $a_7 - a_5 = 10 \Rightarrow 2d = 10 \Rightarrow d=5$.
$a_3 = 5 \Rightarrow a + 2d = 5 \Rightarrow a + 10 = 5 \Rightarrow a = -5$.
AP: $-5, 0, 5, 10 \dots$

17. Reverse the AP: $100, 98, \dots, 2$.
$a=100, d=-2$.
$a_{10} = 100 + 9(-2) = 100 - 18 = 82$.

18. $(a+2d) + (a+6d) = 20 \Rightarrow 2a+8d=20$.
$(a+4d) + (a+8d) = 32 \Rightarrow 2a+12d=32$.
Subtract: $4d=12 \Rightarrow d=3$.
$2a + 24 = 20 \Rightarrow 2a = -4 \Rightarrow a=-2$.
AP: $-2, 1, 4 \dots$

19. $a=10000, d=500, a_n=20000$.
$20000 = 10000 + (n-1)500 \Rightarrow 10000 = 500(n-1) \Rightarrow 20 = n-1 \Rightarrow n=21$.
✅ 21st year.

20. $a=10, d=5, a_n=110$.
$110 = 10 + (n-1)5 \Rightarrow 100 = 5(n-1) \Rightarrow 20 = n-1 \Rightarrow n=21$.
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