CBSE Class 10 Arithmetic Progression Printable Worksheet Set A

Access the latest CBSE Class 10 Arithmetic Progression Printable Worksheet Set A. We have provided free printable Class 10 Mathematics worksheets in PDF format, specifically designed for Chapter 5 Arithmetic Progression. These practice sets are prepared by expert teachers following the 2025-26 syllabus and exam patterns issued by CBSE, NCERT, and KVS.

Chapter 5 Arithmetic Progression Mathematics Practice Worksheet for Class 10

Students should use these Class 10 Mathematics chapter-wise worksheets for daily practice to improve their conceptual understanding. This detailed test papers include important questions and solutions for Chapter 5 Arithmetic Progression, to help you prepare for school tests and final examination. Regular practice of these Class 10 Mathematics questions will help improve your problem-solving speed and exam accuracy for the 2026 session.

Download Class 10 Mathematics Chapter 5 Arithmetic Progression Worksheet PDF

Fill in the Blanks

DIRECTIONS : Complete the following statements with an appropriate word / term to be filled in the blank space(s).

Question. 4, 10, 16, 22, ........, .......... .
Answer: 28, 34

Question. 1, –1, –3, – 5, ....... , ......... .
Answer: –7, –9

Question. 11th term from last term of an A.P. 10, 7, 4......... , – 62, is ......... .
Answer: –32

Question. In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. Number of rows in the flower bed is ............... .
Answer: n = 10

Question. Sum of 1 + 3 + 5 + .... + 1999 is ......... .
Answer: 1000/2[2(1) (1000 1)2]

Question. The sum of 8 A.Ms between 3 and 15 is ................... .
Answer: 72 [ 8 (3 + 15 / 2 ) etc.]

Question. The sum of n terms of an A.P. is 4n2 – n. The common  difference = .................. .
Answer: 11 [S2 = 4(2)2 – 2 ⇒ 14
S1 = 4(1)2 – 1 ⇒ 3 etc.]

Question. The difference of corresponding terms of two A.P’s will be .................... .
Answer: another A.P.

Question. Sum of all the integers between 100 and 1000 which are divisible by 7 is ...................
Answer: 70336 [Hint : S = 105 + 112 + ... 994 and 105 + (n – 1)7 = 994 ⇒ 105 + 7n – 7 = 994 ⇒ n = 128 etc.]

True / False

DIRECTIONS : Read the following statements and write your answer as true or false.

Question. In an AP with first term a and common difference d, the nth term (or the general term) is given by an = a + (n – 1)d.
Answer: True

Question. The balance money ( in `) after paying 5% of the total loan of ` 1000 every month is 950, 900, 850, 800, . . . 50. represented A.P.
Answer: True

Question. 2, 4, 8, 16, ............. is not an A.P.
Answer: True

Question. 10th term of A.P. 2, 7, 12, ......... is 45.
Answer: False

Question. 301 is a term of A.P. 5, 11, 17, 23, ............. .
Answer: False

Question. The general form of an A.P. is a, a + d, a + 2d, a + 3d, .............
Answer: True

Question. In an Arithmetic progression, the first term is denoted by ‘a’ and ‘d ’ is called the common difference.
Answer: True

Question. If an+1 – an = same for all ‘n’, then the given numbers form an A.P.
Answer: True

Question. If Sn of A.P. is 3n2 + 2n, then the first term of A.P. is 3.
Answer: False

VERY SHORT ANSWER QUESTIONS(1 MARK)

Question. If the sum of first m terms of an AP is am2 + bm, find the common difference.
Answer : 
Sm= am2 + bm, S1= a+b= a1, S= 4a+2b= a1+ a2
a2 = S2- S1= 4a+2b- (a+b) = 3a +b , d= a2-a1= 3a+b-(a+b)= 2a

Question. Three numbers are in AP and their sum is 24. Find the middle term.
Answer : 
Let the three numbers of the AP be a-d, a, a+d. So a-d +a+a+d= 24 ⟹ 3a=24,a=24/3=8. Hence the middle term =8.

Question. Find the sum of first 8 multiples of 3.
Answer : 
First 8 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24 These numbers are in A.P.
where a = 3, d = 3 and n = 8 , an=24, Sn= 𝑛/2(a 1+ an), S8= 8/2(3+ 24)= 4x27=108

Question. Find the sum: 34+ 32+30+……+10
Answer : 
Given, 34 + 32 + 30 + … + 10, first term, a = 34, d = a2−a1 = 32−34 = −2, Let 10 be the nth term of this A.P., an= a +(n−1)d, 10 = 34+(n−1)(−2), −24 = (n −1)(−2), 12 = n −1, n = 13, Sn = n/2 (a +l) , l = 10, Sn = n/2 (34 + 10) = 13/2 x44= 286

Question. Find the number of terms of an AP 5, 9,13, …, 185.
Answer : 
a1=5, d=9-5=4, an=185, 185 = 5 + (n - 1)4, 185 = 5 + 4n -4,185 = 1 + 4n, 185 - 1 = 4n, 184 = 4n, n= 184/4 = 46

Question. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5
Answer : 
Since, the number is divisible by both 2 and 5, means it must be divisible by 10.AP = 110, 120, 130,..., 990, a = 110, d = 10, nth term of the AP = 990
a+(n-1)d=990, 110+(n-1)10=990, (n-1)10=990-110,(n-1)= 880/10, n-1=88, n=88+1, n=89

Question. The fourth term of an AP is 11 and the eleventh term is 25. Determine the first term and common difference.
Answer : 
a + 3d = 11….(1) a + 10d = 25…..(2)
Subtracting equation (1) from equation (2 )
a + 10d – (a + 3d) = 25 – 11, 7d = 14, d = 2,
Putting value of d=2 in the equation 2,
a + 10x2 = 25, a + 20 = 25, a = 25 – 20, a = 5

Question. In an AP a = 15, d = -3, an = 0, then find the value of n.
Answer : 
First term (a) =15,Common difference (d) = -3,
Last term(an) = 0, 0 = 15 + (n - 1)-3 -15 = -3n + 3, -15 - 3 = -3n, -18 = -3n, n=6

Question. If the nth term of a progression be a linear expression in n, then prove that this progression is an AP.
Answer : Let the nth term of a given progression be given by
Tn = an + b, where a and b are constants.
Then, Tn–1 = a(n – 1) + b = [(an + b) – a]
(Tn – Tn–1) = (an + b) – [(an + b) – a] = a,
which is a constant.
Hence, the given progression is an AP.
 
Question. Write the first three terms in each of the sequences defined by the following -
(i) an = 3n + 2    (ii) an = n2 + 1
Answer : (i) We have,
an = 3n + 2
Putting n = 1, 2 and 3, we get
a1 = 3 × 1 + 2 = 3 + 2 = 5,
a2 = 3 × 2 + 2 = 6 + 2 = 8,
a3 = 3 × 3 + 2 = 9 + 2 = 11
Thus, the required first three terms of the sequence defined by an = 3n + 2 are 5, 8, and 11.
(ii) We have,
an = n2 + 1
Putting n = 1, 2, and 3 we get
a1 = 12 + 1 = 1 + 1 = 2
a2 = 22 + 1 = 4 + 1 = 5
a3 = 32 + 1 = 9 + 1 = 10
Thus, the first three terms of the sequence
defined by an = n2 + 1 are 2, 5 and 10.
 
Question. The nth term of a sequence is 3n – 2. Is the sequence an A.P. ? If so, find its 10th term.
Answer : We have an = 3n – 2
Clearly an is a linear expression in n. So, the given sequence is an A.P. with common difference 3.
Putting n = 10, we get
a10 = 3 × 10 – 2 = 28
 
Question. Find the 12th, 24th and nth term of the A.P. given by 9, 13, 17, 21, 25, .........
Answer : We have,
a = First term = 9 and,
d = Common difference = 4
[ 13 – 9 = 4, 17 – 13 = 4, 21 – 7 = 4 etc.]
We know that the nth term of an A.P. with first term a and common difference d is given by
an = a + (n – 1) d
Therefore,
a12 = a + (12 – 1) d
= a + 11d = 9 + 11 × 4 = 53
a24 = a + (24 – 1) d
= a + 23 d = 9 + 23 × 4 = 101
and, an = a + (n – 1) d
= 9 + (n – 1) × 4 = 4n + 5
a12 = 53, a24 = 101 and an = 4n + 5
Sequences
Arithmetic Progression
A sequence of terms is said to be in arithmetic progression (A.P) when the difference between any term and its preceeding term is a fixed constant. This constant is called the common difference (c.d) of the A.P.
If a is first term and d is the common difference of the A.P., then its nth term tn is given by tn = a + ( n – 1)d.
 
CBSE Class 10 Arithmetic Progression Printable Worksheet Set A
 
CBSE Class 10 Arithmetic Progression Printable Worksheet Set A-
 
Geometric Progression (G.P.)
 
A sequence is said to be in G.P. when its first term is non-zero and each term is r times the preceeding term,where r is a non-zero constant. The fixed number r is known as the common ratio of the G.P.
 
CBSE Class 10 Arithmetic Progression Printable Worksheet Set A-1
CBSE Class 10 Arithmetic Progression Printable Worksheet Set A-2
CBSE Class 10 Arithmetic Progression Printable Worksheet Set A-3
CBSE Class 10 Arithmetic Progression Printable Worksheet Set A-4
CBSE Class 10 Arithmetic Progression Printable Worksheet Set A-5
CBSE Class 10 Arithmetic Progression Printable Worksheet Set A-6
 
 
 
 
 
Please click the link below to download CBSE Class 10 Arithmetic Progression Printable Worksheet Set A

Chapter 5 Arithmetic Progression CBSE Class 10 Mathematics Worksheet

Students can use the Chapter 5 Arithmetic Progression practice sheet provided above to prepare for their upcoming school tests. This solved questions and answers follow the latest CBSE syllabus for Class 10 Mathematics. You can easily download the PDF format and solve these questions every day to improve your marks. Our expert teachers have made these from the most important topics that are always asked in your exams to help you get more marks in exams.

NCERT Based Questions and Solutions for Chapter 5 Arithmetic Progression

Our expert team has used the official NCERT book for Class 10 Mathematics to create this practice material for students. After solving the questions our teachers have also suggested to study the NCERT solutions  which will help you to understand the best way to solve problems in Mathematics. You can get all this study material for free on studiestoday.com.

Extra Practice for Mathematics

To get the best results in Class 10, students should try the Mathematics MCQ Test for this chapter. We have also provided printable assignments for Class 10 Mathematics on our website. Regular practice will help you feel more confident and get higher marks in CBSE examinations.

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