CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set 03

Welcome! Check out CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set 03 right here. Get free printable Class 10 Mathematics worksheets in PDF for Chapter 5 Arithmetic Progression. Created by experienced teachers, these practice sheets match the 2026-27 exam guidelines set by CBSE, NCERT, and KVS.

Practice Paper: Class 10 Mathematics Chapter 5 Arithmetic Progression

Every student of Class 10 Mathematics should practice with these chapter-wise worksheets. They provide great questions and solutions for Chapter 5 Arithmetic Progression to boost your confidence. Practicing these Class 10 Mathematics sheets regularly makes you faster and more accurate for your 2026-27 exams.

Download Class 10 Mathematics Chapter 5 Arithmetic Progression Worksheet PDF

Question. Which of the following form of an AP?
(a) − 1, − 1, − 1, − 1, ...
(b) 0, 2, 0, 2, …
(c) 1, 1, 2, 2, 3, 3
(d) 1/2, 1/3, 1/4
Answer : A

Question. The common difference of an AP, whose nth term is an =(3n + 7), is
(a) 3
(b) 7
(c) 10
(d) 6
Answer : A

Question. The first four terms of an AP whose first term is − 2 and the common difference is −2, are
(a) − 2, 0, 2, 4
(b) − 2, 4, − 8, 16
(c) − 2, − 4, − 6, − 8
(d) − 2, − 4, − 8, − 16
Answer : C

Question. Which of the following is not an AP?
(a) −12 . , 0.8, 2.8, ...
(b) 3, 3 + √2, 3 +2 √2, 3 + 3√2, …
(c) 4/3, 7/3, 9/3, 12/3 .... 
(d) −1/5, −2/5, −3/5 .... 
Answer : C

Question. If − (5/7), a,2 are consecutive terms in an Arithmetic Progression, then the value of ‘a’ is
(a) 9/7
(b) 9/14
(c) 19/7
(d) 19/14
Answer : B

Question.  Let a be a sequence defined by a1 = 1, a2 = 1 and an = an − 1 + an − 2 for all n > 2, then the value of a4/a3 is
(a) 2/3
(b) 5/4
(c) 4/5
(d) 3/2
Answer : D

Question. The 11th term of an AP −5, −5/2, 0, 5/2, , , ,...
(a) − 20
(b) 20
(c) −30
(d) 30
Answer : B

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

Question.  The value of p for which (2p + 1), 10 and (5p + 5) are three consecutive terms of an AP is
(a) − 1
(b) − 2
(c) 1
(d) 2
Answer : C

Question. The 21st term of an AP whose first two terms are − 3 and 4, is
(a) 17
(b) 137
(c) 143
(d) − 143
Answer : B

Question.  If the common difference of an AP is 5, then what is a18 − a13?
(a) 5
(b) 20
(c) 25
(d) 30
Answer : C

Question. If an AP have 8 as the first term and −5 as the common difference and its first three terms are 8, A,B, then (A +B) is equal to
(a) 0
(b) −1
(c) 1
(d) 2
Answer : C

Question. In an AP, if d = −4, n = 7 and an = 4, then a is equal to
(a) 6
(b) 7
(c) 20
(d) 28
Answer : D

Question. Which term of an AP : 21, 42, 63, 84, ... is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th
Answer : B

Question. If the first term of an AP is − 5 and the common difference is 2, then the sum of the first 6 terms is
(a) 0
(b) 5
(c) 6
(d) 15
Answer : A

Question.  In an AP, if a = 1, an = 20 and Sn = 399, then n is equal to
(a) 19
(b) 21
(c) 38
(d) 42
Answer : C

Question.  What is the common difference of an AP in which a18 − a14 = 32 ?
(a) 8
(b) − 8
(c) − 4
(d) 4
Answer : A

Question. Which term of the AP 5, 15, 25, ... will be 130 more than its 31st term?
(a) 42
(b) 44
(c) 46
(d) 48
Answer : B

Question.  If the 2nd term of an AP is 13 and 5th term is 25, what is its 7th term?
(a) 30
(b) 33
(c) 37
(d) 38
Answer : B

Question. The number of terms of an AP 5, 9, 13, ..., 185 is
(a) 31
(b) 51
(c) 41
(d) 46
Answer : D

Question. The sum of AP, sequence −37, − 33, − 29, . . . . . . . . . upto 12 term is
(a) 180
(b) −180
(c) 170
(d) −170
Answer : B

Question. Two APs have the same common difference. The first term of one of these is − 1 and that of the other is − 8. The difference between their 4th terms is [NCERT Exemplar]
(a) − 1
(b) − 8
(c) 7
(d) − 9
Answer : C

Question. If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be
(a) 7
(b) 11
(c) 18
(d) 0
Answer : D

Question. The 4th term from the end of an AP − 11 , − 8, − 5, . . . , 49 is
(a) 37
(b) 40
(c) 43
(d) 58
Answer : B

Question. The sum of first 16 terms of the AP 10, 6, 2, ... is
(a) − 320
(b) 320
(c) − 352
(d) − 400
Answer : A

Question.  The sum of first five multiples of 3 is
(a) 45
(b) 55
(c) 65
(d) 75
Answer : A

Answer the following:

Question. Find the sum of first 10 terms of the AP: 2, 7, 12, ... 
Answer : 245

Question. If the sum of first m terms of an AP is 2 m2 + 3 m, then what is its second term? 
Answer : 9

Question. Find the sum of first 10 multiples of 6.
Answer : First 10 multiples of 6 are 6, 12, 18, ....., 60.
This is an AP in which a = 6, n = 10 and d = 6.
∴ Sum of first 10 multiples of 6 = S10
⇒ S10 = n/2 [2a + (n - 1) d]
= 10/2 [2×6 + ]10 – 1)6]
= 5 (12 + 54)
= 5 × 66 = 330

Question. What is the sum of five positive integers divisible by 6? 
Answer : 90

Question. If the sum of the first q terms of an AP is 2q + 3q2, what is its common difference? 
Answer : Given that,
Sq = 2q + 3q2
S1 = 2 + 3 = 5 = T1 = First term            [put q = 1]
S2 = 4 + 3(4) = 16                                 [put q = 2]
S3 = 6 + 3(9) = 33                                 [put q = 3
∴ 2nd term,
T2 = S2 – S1 = 16 – 5 = 11
∴ 3rd term,
T3 = S3 – S2 = 33 – 16 = 17
Common difference
= T3 – T2 = 17 – 11 = 6

Question. If nth term of an AP is (2n + 1), what is the sum of its first three terms? 
Answer : a1 = 3, a3 = 7, S3 = 3/2 (3 + 7) = 15

Question. Find the sum of first 100 natural numbers.
Answer : Natural numbers are 1, 2, 3, 4, ...
The sum of first 100 natural numbers is given by
Sn = n(n + 1)/2 = 100 x (100 + 1)/2
= 100 x 101 / 2
= 50 × 101 = 5050

Short Answer type Questions

Question. The sum of three numbers in A.P. is –3, and their product is 8. Find the numbers.
Answer : Let the numbers be (a – d), a, (a + d). Then,
Sum = – 3 =>(a – d) + a (a + d) = – 3
=>3a = – 3
=>a = – 1
Product = 8
=> (a – d) (a) (a + d) = 8
=> a (a2 – d2) = 8
=> (–1) (1 – d2) = 8
=> d2 = 9 => d = ± 3
If d = 3, the numbers are –4, –1, 2. If d = – 3,the numbers are 2, – 1, –4.
Thus, the numbers are –4, –1, 2, or 2, – 1, – 4.
 
Question. Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.
Answer : Let the numbers be (a – 3d), (a – d), (a + d),(a + 3d), Then
Sum = 20 
=>(a – 3d) + (a – d) + (a + d) + (a + 3d) = 20 
=> 4a = 20 
=> a = 5
Sum of the squares = 120
(a – 3d)2 + (a – d)2 + (a + d)2 + (a + 3d)2 = 120
=> 4a2 + 20d2 = 120
=> a2 + 5d2 = 30
=> 25 + 5d2 = 30       [a = 5]
=> 5d2 = 5    => d = ± 1
If d = 1, then the numbers are 2, 4, 6, 8.
If d = – 1, then the numbers are 8, 6, 4, 2.
Thus, the numbers are 2, 4, 6, 8 or 8, 6, 4, 2.

Question. If m times the mth term of an Arithmetic Progression is equal to n times its nth term and m ≠ n, show that the (m + n)th term of the AP is zero.
Answer : We know that an = a + (n – 1)d
From the given conditions,
m[a + (m – 1) d] = n[a + (n – 1)d]
⇒ m[a + (md – d)] = n[a + nd – d]
⇒ am + m2d – md = an + n2d – nd (1)
⇒ am – an + m2d – n2d – md + nd = 0
⇒ a(m – n) + d(m2 – n2) – d(m –n) = 0
⇒ a(m – n) + (m + n) (m – n)d – (m – n)d = 0 (1)
⇒ (m – n) [a + (m + n) d – d] = 0
⇒ a + md + nd – d = 0 (1)
⇒ a + (m + n –1)d = 0
Since, m ≠ n, it is clear that (m + n)th term of the AP is zero

Question. If 4 times the 4th term of an AP is equal to 18 times the 18th term, then find the 22nd term.
Answer : Let a1, a2, a3, ... an, ... be the AP with its first term a and
common difference d.
It is given that
4a4 = 18a18 (1)
⇒ 4(a + 3d) = 18(a + 17d) (1)
⇒ 4a + 12d = 18a + 306d (1)
⇒ 14a + 294d = 0 ⇒ 14(a + 21d) = 0 (1)
⇒ a + 21d = 0 ⇒ a + (22 – 1)d = 0
⇒ a22 = 0
Thus, 22nd term is 0.

Question. Find the sum of 20 terms of the A.P. 1, 4,7, 10, ......
Answer : Let a be the first term and d be the common difference of the given A.P. Then, we have
a = 1 and d = 3.
We have to find the sum of 20 terms of the given A.P.
Putting a = 1, d = 3, n = 20 in
Sn = n/2 [2a + (n – 1) d], we get
S20 = 20/2 [2 × 1 + (20 – 1) × 3]
= 10 × 59 = 590
 
CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set C 1
CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set C 2

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