NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 PDF Download
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 help students master Arithmetic Progressions by explaining how to find the nth term and the sum of n terms. The step-by-step solutions build strong conceptual clarity and improve problem-solving skills for exams.
Neha Tanna23 Nov, 2025
Share
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 Arithmetic Progressions: Exercise 5.2 of Class 10 Maths focuses on key concepts of Arithmetic Progressions (APs), including finding the nth term and the sum of n terms.
These NCERT solutions help students identify patterns, apply formulas, and solve real-life sequence problems with clarity. This section is especially helpful for learners searching for Arithmetic Progressions class 10 exercise 5.2.
Arithmetic Progressions Chapter 5 Exercise 5.2 Solutions
Below is the NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 Arithmetic Progressions -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 A.P.
Solutions:
(i) Given,
First term, a = 7 Common difference, d = 3 Number of terms, n = 8, We have to find the nth term, a n = ? As we know, for an A.P., a n = a +( n −1) d Putting the values, => 7+(8 −1) 3 => 7+(7) 3 => 7+21 = 28 Hence, a n = 28(ii) Given,
First term, a = -18 Common difference, d = ? Number of terms, n = 10 Nth term, a n = 0 As we know, for an A.P., a n = a +( n −1) d Putting the values, 0 = − 18 +(10−1) d 18 = 9 d d = 18/9 = 2 Hence, common difference, d = 2
(iii) Given,
(iv) Given,
(v) Given,
2. Choose the correct choice in the following and justify:
(i) 30th term of the A.P: 10,7, 4, …, is (A) 97 (B) 77 (C) −77 (D) −87
(ii) 11 th term of the A.P. -3, -1/2, 2 …. is (A) 28 (B) 22 (C) – 38 (D)
Solutions:
(i) Given here, A.P. = 10, 7, 4, …Hence, the correct answer is option C. (ii) Given here, A.P. = -3, -1/2, ,2 … Therefore, we can find, First term a = – 3 Common difference, d = a 2 − a 1 = (-1/2) -(-3) ⇒(-1/2) + 3 = 5/2 As we know, for an A.P., a n = a +( n −1) d Putting the values; a 11 = -3+(11-1)(5/2) a 11 = -3+(10)(5/2) a 11 = -3+25 a 11 = 22
Hence, the answer is option B.
3. In the following APs, find the missing term in the boxes.
Solutions:
(i) For the given A.P., 2,2, 26.The first and third term are; a = 2 a 3 = 26
As we know, for an A.P.,a n = a +( n −1) d
Therefore, putting the values here, a 3 = 2+(3-1) d 26 = 2+2 d 24 = 2 d d = 12 a 2 = 2+(2-1)12 = 14
Therefore, 14 is the missing term.
(ii) For the given A.P., , 13, ,3
Therefore, putting the values here, a 2 = a +(2-1) d 13 = a + d ………………. (i)
a 4 = a +(4-1) d 3 = a +3 d ………….. (ii)
On subtracting equation (i) from (ii) , we get, – 10 = 2 d d = – 5
From equation (i), putting the value of d, we get 13 = a +(-5) a = 18 a 3 = 18+(3-1)(-5) = 18+2(-5) = 18-10 = 8.
Therefore, the missing terms are 18 and 8, respectively.
(iii) For the given A.P.,
As we know, for an A.P.,a n = a +( n −1) d
Therefore, putting the values here, a 6 = a+(6−1)d 6 = − 4+5 d 10 = 5 d d = 2 a 2 = a + d = − 4+2 = −2 a 3 = a +2 d = − 4+2(2) = 0 a 4 = a +3 d = − 4+ 3(2) = 2 a 5 = a +4 d = − 4+4(2) = 4
Therefore, the missing terms are −2, 0, 2, and 4, respectively.
(v) For the given A.P.,
Therefore, putting the values here, a 2 = a +(2−1) d 38 = a + d ……………………. (i)
a 6 = a +(6−1) d −22 = a +5 d …………………. (ii)
On subtracting equation (i) from (ii) , we get − 22 − 38 = 4 d −60 = 4 d d = −15 a = a 2 − d = 38 − (−15) = 53 a 3 = a + 2 d = 53 + 2 (−15) = 23 a 4 = a + 3 d = 53 + 3 (−15) = 8 a 5 = a + 4 d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7, respectively.
4. Which term of the A.P. 3, 8, 13, 18, … is 78?
Solutions:
Given the A.P. series as3, 8, 13, 18, … First term, a = 3 Common difference, d = a 2 − a 1 = 8 − 3 = 5 Let the n th term of given A.P. be 78. Now as we know, a n = a +( n −1) d Therefore, 78 = 3+( n −1)5 75 = ( n −1)5 ( n −1) = 15 n = 16 Hence, 16 th term of this A.P. is 78.5. Find the number of terms in each of the following A.P.
(i) 7, 13, 19, …, 205
Solutions:
(i) Given, 7, 13, 19, …, 205 is the A.P
Therefore First term, a = 7 Common difference, d = a 2 − a 1 = 13 − 7 = 6 Let there are n terms in this A.P. a n = 205
First term, a = 18 Common difference, d = a 2 -a 1 = 
d = (31-36)/2 = -5/2 Let there are n terms in this A.P. a n = -47 As we know, for an A.P., a n = a+(n−1)d -47 = 18+(n-1)(-5/2) -47-18 = (n-1)(-5/2) -65 = (n-1)(-5/2) (n-1) = -130/-5 (n-1) = 26 n = 27
Therefore, this given A.P. has 27 terms in it.
6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, …
Solution:
For the given series, A.P. 11, 8, 5, 2.. First term, a = 11 Common difference, d = a 2 − a 1 = 8−11 = −3 Let −150 be the n th term of this A.P. As we know, for an A.P., a n = a +( n −1) d -150 = 11+( n -1)(-3) -150 = 11-3 n +3 -164 = -3 n n = 164/3 Clearly, n is not an integer but a fraction. Therefore, – 150 is not a term of this A.P.7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Solution:
Given that, 11 th term, a 11 = 38 and 16 th term, a 16 = 73 We know that, a n = a+(n−1)d a 11 = a+(11−1)d 38 = a+10d ………………………………. (i)In the same way, a 16 = a +(16−1) d 73 = a +15 d ………………………………………… (ii)
On subtracting equation (i) from (ii) , we get 35 = 5 d d = 7 From equation (i) ,
we can write, 38 = a +10×(7) 38 − 70 = a a = −32 a 31 = a +(31−1) d = − 32 + 30 (7) = − 32 + 210 = 178
Hence, 31 st term is 178.
8. An A.P. consists of 50 terms of which 3 rd term is 12 and the last term is 106. Find the 29 th term.
Solution: Given that,
3 rd term, a 3 = 12 50 th term, a 50 = 106 We know that, a n = a +( n −1) d a 3 = a +(3−1) d 12 = a +2 d ……………………………. (i)In the same way, a 50 = a +(50−1) d 106 = a +49 d …………………………. (ii)
On subtracting equation (i) from (ii) , we get 94 = 47 d d = 2 = common difference From equation (i) ,
we can write now, 12 = a +2(2) a = 12−4 = 8 a 29 = a +(29−1) d a 29 = 8+(28)2 a 29 = 8+56 = 64
Therefore, 29 th term is 64.
9. If the 3rd and the 9 th terms of an A.P. are 4 and − 8, respectively. Which term of this A.P. is zero? Solution:
Given that, 3 rd term, a 3 = 4 and 9 th term, a 9 = −8 We know that, a n = a +( n −1) d Therefore, a 3 = a +(3−1) d 4 = a +2 d ……………………………………… (i)a 9 = a +(9−1) d −8 = a +8 d ………………………………………………… (ii)
On subtracting equation (i) from (ii) , we will get here, −12 = 6 d d = −2 From equation (i) , we can write, 4 = a +2(−2) 4 = a −4 a = 8 Let n th term of this A.P. be zero. a n = a +( n −1) d 0 = 8+( n −1)(−2) 0 = 8−2 n +2 2 n = 10 n = 5
Hence, 5 th term of this A.P. is 0.
10. If 17 th term of an A.P. exceeds its 10 th term by 7. Find the common difference.
Solution:
We know that, for an A.P series; a n = a +( n −1) d a 17 = a +(17−1) d a 17 = a +16 d In the same way, a 10 = a +9 dAs it is given in the question, a 17 − a 10 = 7
Therefore, ( a +16 d )−( a +9 d ) = 7 7 d = 7 d = 1
Therefore, the common difference is 1.
11. Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54 th term?
Solution:
Given A.P. is 3, 15, 27, 39, … first term, a = 3 common difference, d = a 2 − a 1 = 15 − 3 = 12We know that, a n = a +( n −1) d Therefore, a 54 = a +(54−1) d ⇒3+(53)(12) ⇒3+636 = 639 a 54 = 639+132=771
We have to find the term of this A.P. which is 132 more than a 54, i.e.771. Let n th term be 771. a n = a +( n −1) d 771 = 3+( n −1)12 768 = ( n −1)12 ( n −1) = 64 n = 65
Therefore, 65 th term was 132 more than 54 th term.
Or another method is;
Let n th term be 132 more than 54 th term. n = 54 + 132/2 = 54 + 11 = 65 th term
12. Two APs have the same common difference. The difference between their 100 th term is 100, what is the difference between their 1000 th terms?
Solution:
Let, the first term of two APs be a 1 and a 2 respectively And the common difference of these APs be d .For the first A.P., we know, a n = a +( n −1) d Therefore, a 100 = a 1 +(100−1) d = a 1 + 99d a 1000 = a 1 +(1000−1) d a 1000 = a 1 +999 d For second A.P., we know, a n = a +( n −1) d
Therefore, a 100 = a 2 +(100−1) d = a 2 +99 d a 1000 = a 2 +(1000−1) d = a 2 +999 d Given that, difference between 100 th term of the two APs = 100
Therefore, ( a 1 +99 d ) − ( a 2 +99 d ) = 100 a 1 − a 100……………………………………………………………….. (i)
Difference between 1000 th terms of the two APs ( a 1 +999 d ) − ( a 2 +999 d ) = a 1 − a 2 From equation (i) ,
This difference, a 1 − a 2 = 100
Hence, the difference between the 1000th terms of the two A.P. will be 100.
13. How many three-digit numbers are divisible by 7?
Solution:
First three-digit numbers that is divisible by 7 are:First number = 105 Second number = 105+7 = 112 Third number = 112+7 =119 Therefore, 105, 112, 119, …
All are three digit numbers are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
As we know, the largest possible three-digit number is 999. When we divide 999 by 7, the remainder will be 5.
Therefore, 999-5 = 994 is the maximum possible three-digit number that is divisible by 7.
Now the series is as follows. 105, 112, 119, …, 994 Let 994 be the nth term of this A.P. first term, a = 105 common difference, d = 7 a n = 994 n = ?
As we know, a n = a+(n−1)d 994 = 105+(n−1)7 889 = (n−1)7 (n−1) = 127 n = 128
Therefore, 128 three-digit numbers are divisible by 7.
14. How many multiples of 4 lie between 10 and 250?
Solution:
The first multiple of 4 that is greater than 10 is 12.Next multiple will be 16.
Therefore, the series formed as; 12, 16, 20, 24, … All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows, now; 12, 16, 20, 24, …, 248 Let 248 be the n th term of this A.P. first term, a = 12 common difference, d = 4 a n = 248
As we know, a n = a+(n−1)d 248 = 12+( n -1)×4 236/4 = n-1 59 = n-1 n = 60 Therefore, 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, and 3, 10, 17, … equal?
Solution:
Given two APs as; 63, 65, 67,… and 3, 10, 17,….Taking first AP,
63, 65, 67, … First term, a = 63 Common difference, d = a 2 −a 1 = 65−63 = 2 We know, n th term of this A.P. = a n = a+(n−1)d a n = 63+( n −1)2 = 63+2 n −2 a n = 61+2 n ………………………………………. (i)Taking second AP,
3, 10, 17, … First term, a = 3 Common difference, d = a 2 − a 1 = 10 − 3 = 7 We know that, n th term of this A.P. = 3+( n −1)7 a n = 3+7 n −7 a n = 7 n −4 ……………………………………………………….. (ii) Given, n th term of these A.P.s are equal to each other. Equating both these equations, we get, 61+2 n = 7 n −4 61+4 = 5 n 5 n = 65 n = 13 Therefore, 13 th terms of both these A.P.s are equal to each other.
16. Determine the A.P. whose third term is 16 and the 7 th term exceeds the 5 th term by 12.
Solutions:
Given, Third term, a 3 = 16 As we know, a +(3−1) d = 16 a +2 d = 16 ………………………………………. (i) It is given that, 7 th term exceeds the 5 th term by 12. a 7 − a 5 = 12 [ a +(7−1) d ]−[ a +(5−1) d ]= 12 ( a +6 d )−( a +4 d ) = 12 2 d = 12 d = 6From equation (i), we get, a +2(6) = 16 a +12 = 16 a = 4 Therefore, A.P. will be4, 10, 16, 22, …
17. Find the 20 th term from the last term of the A.P. 3, 8, 13, …, 253.
Solution:
Given A.P. is3, 8, 13, …, 253 Common difference, d= 5. Therefore, we can write the given AP in reverse order as; 253, 248, 243, …, 13, 8, 5Now for the new AP, first term, a = 253 and
common difference, d = 248 − 253 = −5 n = 20
Therefore, using nth term formula, we get a 20 = a +(20−1) d a 20 = 253+(19)(−5) a 20 = 253−95 a = 158
Therefore, 20 th term from the last term of the AP 3, 8, 13, …, 253 . is 158.
18. The sum of 4 th and 8 th terms of an A.P. is 24 and the sum of the 6 th and 10 th terms is 44. Find the first three terms of the A.P .
Solution:
We know that, the nth term of the AP is; a n = a +( n −1) d a 4 = a +(4−1) d a 4 = a +3 d In the same way, we can write, a 8 = a +7 d a 6 = a +5 d a 10 = a +9 d Given that, a 4 +a 8 = 24 a+3d+a+7d = 24 2a+10d = 24 a+5d = 12 …………………………………………………… (i) a 6 +a 10 = 44 a +5d+a+9d = 44 2a+14d = 44 a+7d = 22 …………………………………….. (ii) On subtracting equation (i) from (ii) , we get, 2d = 22 − 12 2d = 10 d = 5 From equation (i) , we get, a +5 d = 12 a +5(5) = 12 a +25 = 12 a = −13 a 2 = a + d = − 13+5 = −8 a 3 = a 2 + d = − 8+5 = −3 Therefore, the first three terms of this A.P. are −13, −8, and −3.
19. 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?
Solution:
It can be seen from the given question, that the incomes of Subba Rao increases every year by Rs.200 and hence, forms an AP.Therefore, after 1995, the salaries of each year are; 5000, 5200, 5400, … Here, first term, a = 5000 and common difference, d = 200
Let aftern th year, his salary be Rs 7000.
Therefore, by the n th term formula of AP, a n = a +( n −1) d 7000 = 5000+( n −1)200 200( n −1)= 2000 ( n −1) = 10 n = 11
Therefore, in 11th year, his salary will be Rs 7000.
20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the n th week, her weekly savings become Rs 20.75, find n.
Solution:
Given that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75.Hence, First term, a = 5 and common difference, d = 1.75 Also given, a n = 20.75 Find, n = ?
As we know, by the n th term formula, a n = a +( n −1) d
Therefore, 20.75 = 5+( n -1)×1.75 15.75 = ( n -1)×1.75 ( n -1) = 15.75/1.75 = 1575/175 = 63/7 = 9 n -1 = 9 n = 10
Hence, n is 10.
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 PDF
The PDF provided includes detailed solutions for all questions in Exercise 5.2, covering both the nth term and sum of an AP.
The explanations simplify complex ideas and make revision easier for exam preparation. This resource is ideal for students looking for class 10 Arithmetic Progressions exercise 5.2 or guidance for Arithmetic Progressions chapter 5 ex 5.2
NCERT solutions Class 10 Maths PDF
Check More Related Chapters
|
|
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 FAQs
What is covered in NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2?
Exercise 5.2 focuses on Arithmetic Progressions (AP), including how to find the nth term and the sum of the first n terms. The solutions provide clear, step-by-step explanations.
How do these solutions help in understanding Arithmetic Progressions?
The solutions simplify formulas and methods, helping students grasp concepts like the common difference, nth term formula, and AP sum. This boosts accuracy and speed in solving sums.
Is Exercise 5.2 important for board exams?
Yes. Many CBSE questions are directly based on Class 10 Arithmetic Progressions Exercise 5.2. Mastering this exercise is crucial for scoring well in Chapter 5.
Do these solutions include the nth term formula explanation?
Absolutely. The nth term formula an=a+(n−1)da_n = a + (n-1)dan=a+(n−1)d is explained with examples to help students understand how to apply it in Arithmetic Progressions Class 10 Exercise 5.2.
Free Learning Resources
PW Books
Notes (Class 10-12)
PW Study Materials
Notes (Class 6-9)
Ncert Solutions
Govt Exams
Class 6th to 12th Online Courses
Govt Job Exams Courses
UPSC Coaching
Defence Exam Coaching
Gate Exam Coaching
Other Exams
Know about Physics Wallah
Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.
We Stand Out because
We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.
Our Key Focus Areas
Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.
What Makes Us Different
Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others
Copyright © 2025 Physicswallah Limited All rights reserved.
Get App