CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set 03

Read and download the CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set 03 in PDF format. We have provided exhaustive and printable Class 10 Mathematics worksheets for Chapter 5 Arithmetic Progression, designed by expert teachers. These resources align with the 2026-27 syllabus and examination patterns issued by NCERT, CBSE, and KVS, helping students master all important chapter topics.

Chapter-wise Worksheet for Class 10 Mathematics Chapter 5 Arithmetic Progression

Students of Class 10 should use this Mathematics practice paper to check their understanding of Chapter 5 Arithmetic Progression as it includes essential problems and detailed solutions. Regular self-testing with these will help you achieve higher marks in your school tests and final examinations.

Class 10 Mathematics Chapter 5 Arithmetic Progression Worksheet with Answers

Case Study Based Questions

I. Your friend Veer wants to participate in a 200 m race. Presently, he can run 200 m in 51 seconds and during each day practice it takes him 2 seconds less. He wants to do in 31 seconds.

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Question. Which of the following terms are in AP for the given situation?
(a) 51, 53, 55, ...
(b) 51, 49, 47, ...
(c) –51, –53, –55, ...
(d) 51, 55, 59, …
Answer : B

Question. What is the minimum number of days he needs to practice till his goal is achieved?
(a) 10
(b) 12
(c) 11
(d) 9
Answer : C

Question. Which of the following term is not in the AP of the above given situation?
(a) 41
(b) 30
(c) 37
(d) 39
Answer : B

Question. If nth term of an AP is given by an = 2n + 3 then common difference of an AP is
(a) 2
(b) 3
(c) 5
(d) 1
Answer : A

Question. The value of x, for which 2x, x + 10, 3x + 2 are three consecutive terms of an AP is
(a) 6
(b) – 6
(c) 18
(d) –18
Answer : A
 

II. India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.

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Question. The production during first year is
(a) 3000 TV sets
(b) 5000 TV sets
(c) 7000 TV sets
(d) 10000 TV sets
Answer : B

Question. The production during 8th year is
(a) 10500
(b) 11900
(c) 12500
(d) 20400
Answer : D

Question. The production during first 3 years is
(a) 12800
(b) 19300
(c) 21600
(d) 25200
Answer : C

Question. In which year, the production is 29,200?
(a) 10th year
(b) 12th year
(c) 15th year
(d) 18th year
Answer : B

Question. The difference of the production during 7th year and 4th year is
(a) 6600
(b) 6800
(c) 5400
(d) 7200
Answer : A

Assertion-Reason Type Questions

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion (A): Common difference of the AP: –5, –1, 3, 7, ... is 4.
Reason (R): Common difference of the AP : a, a + d, a + 2d, ... is given by d = 2nd term – 1st term.
Answer : A

Question. Assertion (A): If nth term of an AP is 7 – 4n, then its common difference is – 4.
Reason (R): Common difference of an AP is given by d = an+1 – an.
Answer : A

Question. Assertion (A): Common difference of an AP in which a21 – a7 = 84 is 14.
Reason (R): nth term of an AP is given by an = a + (n – 1) d.
Answer : D

Very Short Answer Type Questions

Question. If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Answer : 
200

Question. Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
Answer : 
16th term

Question. Find the sum of the first 15 multiples of 8.
Answer : 
960

Question. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Answer : 
4, 10, 16, 22, . . .

Question. Find the number of terms in the AP : 7, 13, 19, . . . , 205
Answer : 
34

Question. 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 : 
–13, –8, –3

Arithmetic Progression

Question. If 5 times the 5th term of an AP is equal to 10 times the 10th term, show that its 15th term is zero.
Answer : 
Let 1st term = a and common difference = d.
a5 = a + 4d, a10 = a + 9d
According to the question, 5 × a5 = 10 × a10
=> 5(a + 4d) = 10(a + 9d)
=> 5a + 20d = 10a + 90d a = -14d
Now, a15 = a + 14d
=> a15 = -14d + 14d = 0. 

 
Question. Write the first term a and the common difference d of A.P. -1.1, - 3.1, -5.1, - 7.1,... 
Answer : 1.1, - 3.1, -5.1, - 7.1,...First term
(a) = -1.1
We know that common difference is difference between any two consecutive terms of an A.P.
So, common difference(d) = (-3.1) - (-1.1)
= -3.1 + 1.1
= -2 
 
Question. For what value of n are the nth term of the following two AP's are same 13, 19, 25, .... and 69, 68, 67 ....  
Answer : nth term of 13, 19, 25, ............ = nth term of 69, 68, 67, .........
13 + (n - 1) 6 = 69 + (n - 1) (-1)
13 + 6n - 6 = 69 - n + 1
n + 6n = 70 - 7
7n = 63
n = 9
Therefore, n = 9
 
Question. Find the 6th term from the end of the A.P. 17,14,11,..., - 40 
Answer :  A.P. is 17,14,11,..., - 40
We have,
l = Last term = -40 , a = 17 and, d = Common difference= 14 - 17 = - 3
6th term from the end = l - (n -1)d
= l - (6-1) d
= -40 - 5 (-3 )
= -40 + 15
= -25
So, 6th term of given A.P. is -25.
 
Question. How many terms of the AP 17,15,13,11,... must be added to get the sum 72? 
Answer : Given A.P. is 17, 15, 13, 11........
Here, 1st term (a) = 17 and common difference (d ) = (15 - 17) = -2
Let the sum of n terms be 72. Then,
Sn = 72
=>n/2{2a + (n - 1)d} = 72
=>n {2 × 17 + (n - 1)(-2)} = 144
=>n(36 - 2n)=144
=>2n2- 36n + 144 = 0
=>n2 - 18n + 72 = 0
=>n2 - 12n - 6n + 72 = 0
=>n(n - 12) - 6(n - 12) = 0
=>(n - 12)(n - 6) = 0
=>n = 6 or n = 12.
sum of first 6 terms = sum of first 12 terms = 72.
∴ This means that the sum of all terms from 7th to 12th is zero.
 
Question. The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25th term of this A.P. 
Answer : According to the question,
Sum of n terms of the A.P. Sn = 3n2 + 4n
S1 = 3 × 12 + 4 × 1 = 7 = t1 ....(i)
S2 = 3 × 22 + 4 × 2 = 20 = t1 + t2 ....(i)
S3 = 3 × 32 + 4×  3 = 39 = t1 + t2 + t3 .... (iii)
From (i), (ii), (iii)
t1 = 7, t2 = 13, t3 = 19
Comm on difference, d = 13 - 7 = 6
25th of the term of this A.P., t = 7 + (25 - 1)6
= 7 + 144 = 151
∴ The 25th term of the A.P. is 151.
 
Question. Find the sum of first‘n’ terms of an A.P.? 
a. 7n – 8
b. S =n/2 [2a + (n - 1)d]
c. 2n + 3
d n 2 + 2
Answer : b. S = n/2 [2a + (n - 1)d]
Explanation: let a = 1st term, d = common difference,
Sn = sum of 1st n terms of an AP
then Sn = (a) + (a + d) + (a + 2d) + .................{a + (n - 3) d} + { a + (n - 2)d} + {a + (n - 1)d} .......... (i)
Now Rewrite Sn as follows
Sn = {a +( n-1)d} + {a +( n-2)d }+ {a + (n - 3)d} ..........................(a + 3d) + (a + 2d) + (a + d) + a ............. (ii)
adding the terms i and ii vertically
adding 1st term of both we get (a) + {a + (n-1)d} = 2a + (n-1)d
adding 2nd term of both (a + d) + {a + (n - 2)d} = 2a + (n-1)d
adding 3rd terms of both (a + 2d) + {a + (n-3)d} = 2a + (n-1)d
since there are n terms in each of the equations i and ii , adding both the
equations we get
2Sn = n{2a + (n-1)d}
Sn = n/2 {2a + (n-1)d}.
 
Question. If a, b and c are in A. P., then the value of  a-b/b-c  is 
a. a/b
b. 1
c. c/a
d. b/c
Ans- b. 1
Explanation: According to question,
Given that the 17th term of an A.P exceeds its 10th term by 7 .
d = ?
=> a + 16d = a + 9d + 7
=> 16d - 9d = 7
=> 7d = 7
=> d = 7/7= 1
Therefore , common difference = 1.
 
Question. Find the common difference of the A.P. and write the next two terms of A.P. 119,136,153,170,..... 
Answer : Given A.P is
119, 136, 153, 170........
We know that common differnce is difference between any consecutive terms of an A.P.
So, common difference = 136 - 119 = 17
5th term = 170 + 17 = 187 ( a5 = a + 4d)
6th term = 187 + 17 = 204. ( a6 = a + 5d)

 

Please click the link below to download CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set C

CBSE Mathematics Class 10 Chapter 5 Arithmetic Progression Worksheet

Students can use the practice questions and answers provided above for Chapter 5 Arithmetic Progression to prepare for their upcoming school tests. This resource is designed by expert teachers as per the latest 2026 syllabus released by CBSE for Class 10. We suggest that Class 10 students solve these questions daily for a strong foundation in Mathematics.

Chapter 5 Arithmetic Progression Solutions & NCERT Alignment

Our expert teachers have referred to the latest NCERT book for Class 10 Mathematics to create these exercises. After solving the questions you should compare your answers with our detailed solutions as they have been designed by expert teachers. You will understand the correct way to write answers for the CBSE exams. You can also see above MCQ questions for Mathematics to cover every important topic in the chapter.

Class 10 Exam Preparation Strategy

Regular practice of this Class 10 Mathematics study material helps you to be familiar with the most regularly asked exam topics. If you find any topic in Chapter 5 Arithmetic Progression difficult then you can refer to our NCERT solutions for Class 10 Mathematics. All revision sheets and printable assignments on studiestoday.com are free and updated to help students get better scores in their school examinations.

Where can I download the 2026-27 CBSE printable worksheets for Class 10 Mathematics Chapter Chapter 5 Arithmetic Progression?

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Are these Chapter Chapter 5 Arithmetic Progression Mathematics worksheets based on the new competency-based education (CBE) model?

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