CBSE Class 10 Mathematics Chapter 5 Arithmetic Progression Objective Questions can be accessed by students to enhance their logical thinking skills, which are crucial from the exam point of view. You will find only the Multiple Choice Questions (MCQs) framed from various important topics in the chapter. Each question is provided with a detailed solution, which the CBSE students can refer to before the board examinations.
All these questions are crucial as it helps the students to prepare for CBSE Class 10 Mathematics Board Examination. As per the latest modified exam pattern, the question papers will contain more of objectivetype questions. This topic is also very important from the standpoint of competitive exams. Find the CBSE Class 1o Maths Objective Questions below:
List of Subtopics Covered in Chapter 5
5.1 General Term of AP (7 MCQs from This Topic)
5.2 Introduction to AP (6 MCQs from This Topic)
5.3 Sum of Terms in AP (7 MCQs from This Topic)
Download Free CBSE Class 10 Maths Chapter 5 – Arithmetic Progression Objective Questions PDF
General Term of AP

 Find the number of terms in each of the following APs: (i) 7, 13, 19…., 205, (ii) 18, 31/2, 13,…−47
 26, 35
 27, 34
 35, 26
 34, 27
 Find the number of terms in each of the following APs: (i) 7, 13, 19…., 205, (ii) 18, 31/2, 13,…−47
Answer: (d) 34, 27
Solution: (i) 7, 13, 19…., 205
The first term, a = 7
The common difference, d=13−7=6
a_{n}=205
Using formula an=a+ (n−1) d to find the n^{th} term of arithmetic progression, we get
205=7+ (n−1)6
⇒205=6n+1
⇒204=6n
⇒n=204/6=34
Therefore, there are 34 terms in the given arithmetic progression.
(ii) 18, 31/2, 13,…−47
The first term, a=18
Common difference, d= (31/2) −18=−5/2
a_{n} = − 47
Using formula a_{n}=a+ (n−1) d to find the n^{th }term of arithmetic progression, we get
−47=18 + (n−1) (−5/2)
⇒−94=36−5n+5
⇒5n=135
⇒n=135/5=27
Therefore, there are 27 terms in the given arithmetic progression.

 Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
 185
 210
 178
 150
 Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Answer: (c) 178
Solution: We are given that a_{11}=38 and a_{16}=73 where a_{11 }is the 11^{th} term and a_{16} is the 16^{th} term of an AP.
Using formula a_{n}=a+ (n−1) d to find the nth term of arithmetic progression, we get
38=a+10d….. (i)
73=a+15d …… (ii)
equation (ii) – equation (i) gives,
35 = 5d
d = 7….. (iii)
Substituting (iii) in (i), we get a = 32
a_{31}=−32+(31−1)(7)
⇒−32+210=178
Therefore, the 31^{st} term of AP is 178.

 If the third and the ninth terms of an AP are 4 and 8, respectively, which term of this AP is zero?
 5^{th}
 4^{th}
 3^{rd}
 6^{th}
 If the third and the ninth terms of an AP are 4 and 8, respectively, which term of this AP is zero?
Answer: (a) 5^{th}
Solution: It is given that the 3rd and 9th terms of AP are 4 and 8, respectively.
It means a_{3}=4 and a_{9}=−8
Where a_{3} and a_{9} are the third and ninth terms, respectively.
Using formula a_{n}=a+(n−1)d to find the n^{th} term of arithmetic progression, we get
4=a+(3−1)d
⇒4 = a+2d…(i)
−8 = a+(9−1)d
⇒−8=a+8d…(ii)
From equation (i), we have a=4−2d
Substituting in equation (ii), we get
−8=4−2d+8d
⇒−12 = 6d
⇒d =−12/6 =−2
Solving for (a), we get
⇒−8 = a−16
⇒a = 8
Therefore, first term a = 8 and common difference d =−2
We know a_{n} = a+ (n−1) d (where a_{n} is the n^{th} term)
Finding the value of n where an=0
0=8+ (n−1) (−2)
⇒0 = 8−2n+2
⇒0 = 10−2n
⇒2n = 10
⇒n = 10/2=5
Therefore, the 5th term is equal to 0.

 Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54^{th} term?
 70
 65
 80
 55
 Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54^{th} term?
Answer: (b) 65
Solution: Let’s first calculate the 54^{th} of the given AP.
First term = a = 3
Common difference = d =15 – 3 = 12
Using formula a_{n}=a+ (n−1) d, to find the n^{th }term of arithmetic progression, we get
a_{54}=a+ (54−1) d
a_{54} = 3 + 53 (12) =3 + 636=639
We want to find which term is 132 more than its 54^{th} term. Let’s suppose it is the n^{th} term which is 132 more than the 54^{th} term.
Therefore, we can say that
a_{n} = a_{54}+132
a_{n}=a+ (n−1) d=3+ (n−1) (12)}
⇒3+ (n−1)12=639+132
⇒3+12n−12=771
⇒12n−9=771
⇒12n=780
⇒n= 780/12=65
Therefore, the 65^{th} term is 132 more than its 54^{th} term.

 Two APs have the same common difference. The difference between their 100^{th} terms is 100; what is the difference between their 1000^{th} terms?
 200
 150
 100
 55
 Two APs have the same common difference. The difference between their 100^{th} terms is 100; what is the difference between their 1000^{th} terms?
Answer: (c) 100
Solution: Let the first term of the first AP = a
Let the first term of the 2nd AP = a′
It is given that their common difference is the same. Let their common difference be d.
It is given that difference between their 100^{th} terms is 100. Using formula a_{n} = a + (n−1) d, to find the n^{th} term of arithmetic progression, we can say that
a+ (100−1) d− (a′ +(100−1)d)=a+99d−a′−99d=100
⇒a−a′=100 …….. (1)
We want to find the difference between their 1000^{th} terms, which means we want to calculate the following:
a+ (1000−1) d−(a′+(1000−1)d)=a+999d−a′−999d=a−a′
Putting equation (1) in the above equation, we get,
a+ (1000−1) d−(a′+(1000−1)d)=a+999d−a′999d=a−a′=100
Therefore, the difference between their 1000^{th} terms would be equal to 100

 How many threedigit numbers are divisible by 7?
 112
 114
 128
 110
 How many threedigit numbers are divisible by 7?
Answer: (c) 128
Solution: We have an AP starting at 105 because it is the first threedigit number divisible by 7.
AP will end at 994 because it is the last threedigit number divisible by 7.
Therefore, we have an AP 105,112,119…994
The first term, a = 105
Common difference, d = 112 – 105 = 7
Using formula a_{n}=a+ (n−1) d to find the n^{th} term of arithmetic progression, we can say that
994=105+ (n−1) (7)
⇒994=105+ (n−1) (7)
⇒889=7(n−1)
⇒n−1=889/7⇒n=127+1=128
994 is the 128^{th} term of AP. Therefore, there are 128 terms in AP. In other words, we can also say that there are 128 threedigit numbers divisible by 7.

 For what value of n are the n^{th} terms of two AP’s: 63, 65, 67 …. and 3, 10, 17,.. equal?
 11
 14
 12
 13
 For what value of n are the n^{th} terms of two AP’s: 63, 65, 67 …. and 3, 10, 17,.. equal?
Answer: (d) 13
Solution: Let’s first consider AP: 63, 65, 67…..
First term =a=63
Common difference =d=65−63=2
Using formula a_{n}=a+ (n−1) d, to find the n^{th} term of arithmetic progression, we can say that
a_{n}=63+ (n−1) (2) (1)
Now, consider second AP 3, 10, 17…
First term = a = 3
Common difference = d = 10 3 = 7
Using formula a_{n}=a+ (n−1) d, to find the n^{th} term of arithmetic progression, we can say that
a_{n}=3+(n−1)(7) (2)
According to the given condition, we can write
(1) = (2)
⇒63+ (n−1) (2) =3+ (n−1) (7)
⇒63+2n−2=3+7n−7
⇒65=5n
⇒n=65/5=13
Therefore, the 13th terms of both APs are equal.
Introduction to AP

 If (x+1), 3x and (4x+2) are the first three terms of an AP, then its 5th term is:
 14
 19
 24
 28
 If (x+1), 3x and (4x+2) are the first three terms of an AP, then its 5th term is:
Answer: (d) 28
Solution: Given (x+1), 3x, (4x+2) are in AP.
Hence, the difference between two consecutive terms will be the same.
Hence, 3x–(x+1) = (4x+2)–3x
⇒2x−1=x+2
⇒x=3
So, the first term,
a=(x+1) =4.
The common difference,
d=9–4=5.
The n^{th} term of an AP is given by
t_{n }=a+ (n−1) d
⇒t_{5}=4+4(5) =24.

 The sum of the first ten terms of an A.P. is four times the sum of its first five terms, then the ratio of the first term and the common difference is
 ½
 ¼
 4
 1
 The sum of the first ten terms of an A.P. is four times the sum of its first five terms, then the ratio of the first term and the common difference is
Answer: (a) ½
Solution: Let S_{10} be the sum of the first 10 terms and S5 be the sum of the first 5 terms.
We know,
S_{n = (n/2) [2a + (n1) d] }
Given,
S_{10}=4S_{5}
⇒ (10/2) [2a + (101) d] = 4 × (5/2) [2a + (51) d]
⇒ (10/2) [2a+ 9d] = 4 × (5/2) [2a+ 4d]
⇒ 2a+ 9d = 4a+ 8d
⇒ a/d= ½

 Find the common difference (d) in the following APs, respectively.
(i) 20, 40, 60, 80, 100,…
(ii) 5, 0, 5, 10, 15, …
(iii) 2, 2, 2, 2, 2..


 20, 5, 0
 20, 5, 0
 20, 5, 0
 20, 5, 0

Answer: (d) 20, 5, 0
Solution: The common difference is
(i) 20, 40, 60, 80, 100,
d= 4020 =20
(ii) 5, 0, 5, 10, 15, …
d= 05 = 5
(iii) 2, 2, 2, 2, 2….
d= 22 =0

 Which of the following sequences form an AP?
(a) 2, 4, 8, 16……..
(b) 2, 3, 5, 7, 11…….
(c) 1, 1.25, 1.5, 1.75……
(d) 1, 1, 3, 5, 7…………
Answer: (ic) 1, 1.25, 1.5, 1.75…… and
(d) 1, 1, 3, 5, 7…………
Solution: Consider each list of numbers:
(a) 2, 4, 8, 16……..
Difference between the first two terms = 4 – 2 = 2
Difference between the third and second term = 8 – 4 = 4
Since a_{2}–a_{1}≠a_{3}–a_{2}, this sequence is not an AP
(b) 2, 3, 5, 7, 11……..
Difference between the first two terms = 3 – 2 = 1
Difference between the third and second term = 5 – 3 = 2
Since a_{2}–a_{1}≠a_{3}–a_{2}, this sequence is not an AP
(c) 1, 1.25, 1.5, 1.75…….
Difference between the first two terms = 1.25 – (1) = 0.25
Difference between the third and second term = 1.5 – (1.25) = 0.25
Difference between the fourth and third term = 1.75 – (1.5) = 0.25
Since a_{2}–a_{1}=a_{3}−a_{2}=a_{4}−a_{3}, this sequence is an AP
(d) 1, 1, 3, 5, 7…………
Difference between the first two terms = 1 – 1 = 2
Difference between the third and second term = 3 – (1) = 2
Difference between the fourth and third term = 5 – (3) = 2
Since a_{2}–a_{1}=a_{3}−a_{2}=a_{4}−a_{3}, this sequence is an AP

 What are the conditions for a sequence to be an AP?
 The sum of two consecutive terms should be constant.
 The product of two consecutive numbers should be constant.
 The difference between two consecutive terms should be constant.
 The ratio of two consecutive terms should be constant.
 What are the conditions for a sequence to be an AP?
Answer: (c) The difference between two consecutive terms should be constant.
Solution: Let a1, a_{2}, a_{3}, a_{4},_{ }a_{5}, a_{6,} a_{7,} a_{8}… be a sequence.
For this sequence to be an AP, the difference between any two consecutive terms should be constant.
a_{2}–a_{1}=a_{3}–a_{2}=a_{4}–a_{3}=d
This difference is called the common difference of the AP and is denoted by d.
So an AP can also be represented in this form as well a, a+d, a+2d…

 Which of the following list of numbers forms an AP?
(a) 4, 4 + √3 , 4 + 2√3, 4 + 3√3, 4 + 4√3
(b) 0.3, 0.33, 0.333, 0.3333, 0.33333
(c) 3/5, 6/5, 9/5, 12/5, 3
(d) 1/5, 1/5, 1/5, 1/5, 1/5
Answer: (a), (c) and (d)
Solution: Consider each series (a) 4, 4 + √3, 4 + 2 √3, 4 + 3 √3, 4 + 4 √3
Difference between first two consecutive terms = 4 + √3 – 4 = √3
Difference between third and second consecutive terms = 4 + 2 √3 – 4 + √3 = √3
Difference between fourth and third consecutive terms = 4 + 3 √3– 4 + 2 √3 = √3
Since a_{2}–a_{1} = a_{3}−a_{2} = a_{4}−a_{3}
This series is an AP
(b) 0.3, 0.33, 0.333, 0.3333, 0.33333
Difference between first two consecutive terms = 0.330.3 = 0.03
Difference between third and second consecutive terms = 0.333 – 0.33 = 0.003
Since a_{2}–a_{1} ≠ a_{3}−a_{2}
This series is not an AP
(c) 3/5, 6/5, 9/5, 12/5, 3
Difference between first two consecutive terms = (6/5)–(3/5) = 3/5
Difference between third and second consecutive terms = (9/5)–(6/5) = 3/5
Difference between fourth and third consecutive terms = (12/5) – (9/5) = 3/5
Since a_{2}–a_{1} = a_{3}–a_{2} = a_{4}–a_{3}
This series is an AP
(d) 1/5, 1/5, 1/5, 1/5, 1/5
Difference between first two consecutive terms = 1/5 – ( 1/5) = 0
Difference between third and second consecutive terms = 1/5 – (1/5) = 0
Difference between fourth and third consecutive terms = 1/5 – (1/5) = 0
Since a_{2}–a_{1} = a_{3}–a_{2} = a_{4}–a_{3}
This series is an AP.
Sum of Terms in AP

 The first term of an AP is 5, the last term is 50, and the sum is 440. Find the number of terms and the common difference.
 17; 3
 16; 2
 17; 2
 16; 3
 The first term of an AP is 5, the last term is 50, and the sum is 440. Find the number of terms and the common difference.
Answer: (d) 16; 3
Solution: First term, a=5
Last term, l=50
S_{n}=440
Applying the formula S_{n}= n/2 (a+l) to find the sum of n terms of AP, we get
440= n/2 (5+50)
⇒440/55=n/2
⇒8=n/ 2
⇒n=16
Applying the formula, S_{n}=n/2(2a+ (n−1) d) to find the sum of n terms of AP and putting the value of n, we get
440=16/2(2(5) + (16−1) d)
⇒440=8(10+15d)
⇒10+15d=55
⇒15d=45
⇒d=45/15=3

 The first and the last terms of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there, and what is their sum?
 38; 7114
 32; 7114
 38; 6973
 32; 6973
 The first and the last terms of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there, and what is their sum?
Answer: (c) 38; 6973
Solution: First term = a = 17
Last term = l = 350
Common difference = d = 9
Using formula a_{n }= a + (n−1) d to find the nth term of arithmetic progression, we can say that
350=17+ (n−1) (9)
⇒350=17+9n−9
⇒342=9n
⇒n=342/9=38
Applying the formula, S_{n}=n/2(2a+ (n−1) d) to find the sum of n terms of AP and putting the value of n, we get
S_{38}=38/2(34+ (38−1) (9))
⇒S_{38}=19(34+333) =6973
Therefore, there are 38 terms, and their sum is equal to 6973.

 Find the sum of the first 22 terms of an AP in which d = 7 and the 22^{nd} term is 149.
 1623
 1712
 1542
 1661
 Find the sum of the first 22 terms of an AP in which d = 7 and the 22^{nd} term is 149.
Answer: (d) 1661
Solution: It is given that the 22nd term is equal to 149.
It means a_{22}=149
Using formula a_{n}=a+ (n−1) d, to find the n^{th} term of AP, we can say that
149=a+ (22−1) (7)
⇒149=a+147
⇒a=2
Applying the formula, S_{n} = n/2(2a+ (n−1) d) to find the sum of n terms of AP and putting the value of a, we get
S_{22}= 22/2(4+ (22−1) (7))
⇒S_{22}=11(4+147)
⇒S_{22}=1661
Therefore, the sum of the first 22 terms of AP is equal to 1661.

 Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18, respectively.
 5610
 5840
 5320
 5000
 Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18, respectively.
Answer: (a) 5610
Solution: It is given that the second and third terms of AP are 14 and 18, respectively.
Using formula a_{n}=a+ (n−1) d, to find the n^{th} term of arithmetic progression, we can say that
14=a+ (2−1) d
⇒14=a+ d (1)
And, 18=a+ (3−1) d
⇒18=a+2d (2)
These are equations consisting of two variables. We can solve them by the method of substitution.
Using equation (1), we can say that a=14−d
Putting the value of a in equation (2), we can say that
18=14−d+2d
⇒d=4
Therefore, the common difference d=4
Putting the value of d in equation number (1), we can say that
18=a+2(4)
⇒a=10
Applying the formula, S_{n}= n/2(2a+ (n−1) d) to find the sum of n terms of AP, we get
S_{15}=51/2(20+ (51−1) 4) =51/2(20+200) =51×110 =5610
Therefore, the sum of the first 51 terms of an AP is equal to 5610.

 The sum of four consecutive numbers in an A.P. with d > 0 is 20. The sum of their square is 120, then the middle terms are
 8, 10
 6, 8
 4, 6
 2, 4
 The sum of four consecutive numbers in an A.P. with d > 0 is 20. The sum of their square is 120, then the middle terms are
Answer: (c) 4, 6
Solution: Let the numbers be a−3d,a−d,a+d,a+3d
Given: a−3d+a—d+a+d+a+3d=20
4a=20
a=5 and
(a−3d)^{2}+ (a−d)^{2} + (a+3d)^{2} + (a+3d)^{2}=120
(4)(a)^{2} + 20 d^{2}=120
(4)(5)^{2}+20 d^{2}=120
d^{2}=1
d=+1 or−1
Hence, the numbers are 2, 4, 6, and 8

 Find the sum of all the nonnegative terms of the following sequence: 100, 97, 94, ….
 1717
 1719
 1721
 1723
 Find the sum of all the nonnegative terms of the following sequence: 100, 97, 94, ….
Answer: (a) 1717
Solution: Here, t_{1} = 100, common difference, d = t_{2} – t_{1} = 97 – 100 = 3
t_{n} = t_{1 }+ (n1) d ⇒100 + (n1)(3) = 100 – 3(n1) = 103 – 3n.
Let t_{n} be the first negative term. i.e.,
t_{n} < 0 ⇒ 103 – 3n < 0 ⇒ n > 103/3 > 34
That is, the 35^{th} term will be negative.
The sum of the first 34 terms = (34/2) (first term + last term)
= 17(100+100+ (34−1) (−3)) =1717

 What is the sum of all 3 digit numbers that leave a remainder of ‘2’ when divided by 3?
 149700
 164749
 164850
 897
 What is the sum of all 3 digit numbers that leave a remainder of ‘2’ when divided by 3?
Answer: (c) 164850
Solution: The smallest 3digit number that will leave a remainder of 2 when divided by 3 is 101.
The next number that will leave a remainder of 2 when divided by 3 is 104, 107,…
The largest 3digit number that will leave a remainder of 2 when divided by 3 is 998.
So, it is an AP with the first term being 101 and the last term being 998, and the common difference being 3.
Sum of an AP = n/2(a+l)
we know that in an A.P., the nth term a_{n}=a_{1}+ (n−1)×d
In this case, therefore,
998 = 101 + (n – 1) × 3
⇒897= (n−1) ×3
⇒299= (n−1)
⇒n=300
Therefore, the sum of the AP will be (101+198)/2 ×300= 164850
CBSE Class 10 Maths Chapter 5 Extra MCQs
1. In an AP, if d = –4, n = 7, a_{n} = 4. Calculate a.
(a) 20
(b) 7
(c) 28
(d) 6
2. What is the sum of the first five multiples of 3?
(a)75
(b) 55
(c) 65
(d) 45
In this chapter, the concepts discussed include patterns in succeeding terms obtained by adding a fixed number to the preceding terms. Students will understand how to find nth terms and the sum of n consecutive terms. Also, they will learn arithmetic progression effectively when they solve daily life problems.
This chapter has Arithmetic Progression Derivation of the nth term and sum of the first n terms of an A.P. and their application in solving daily life problems.
Keep visiting BYJU’S to get complete assistance for CBSE Class 10 Board exams. At BYJU’S, students can obtain several sample papers, question papers, notes, textbooks, videos, animations and effective preparation tips, which can help them to score good grades in the Class 10 board exams.
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