 # Class 10 Maths Chapter 5 Arithmetic Progression MCQs

Class 10 Maths MCQs for Chapter 5 (Arithmetic Progression) is made available online here for students to score better marks in exams. These objective questions are provided here with answers and detailed explanations as per the CBSE syllabus and NCERT guidelines. The chapter-wise multiple choice questions for Class 10 Maths are given here.

## Class 10 Maths MCQs for Arithmetic Progressions

Students of 10th standard can practice these questions to develop their problem-solving skills and increase the confidence level. Arithmetic progression chapter teaches us about the arrangement of numbers or objects in Maths and in real-life situations. It has huge applications. Get important questions for class 10 Maths here as well.

#### Below are the MCQs for Arithmetic Progression

1.In an Arithmetic Progression, if a=28, d=-4, n=7, then an is:

(a)4

(b)5

(c)3

(d)7

Explanation: For an AP,

an = a+(n-1)d

= 28+(7-1)(-4)

= 28+6(-4)

= 28-24

an=4

2.If a=10 and d=10, then first four terms will be:

(a)10,30,50,60

(b)10,20,30,40

(c)10,15,20,25

(d)10,18,20,30

Explanation: a = 10, d = 10

a1 = a = 10

a2 = a1+d = 10+10 = 20

a3 = a2+d = 20+10 = 30

a4 = a3+d = 30+10 = 40

3.The first term and common difference for the A.P. 3,1,-1,-3 is:

(a)1 and 3

(b)-1 and 3

(c)3 and -2

(d)2 and 3

Explanation: First term, a = 3

Common difference, d = Second term – First term

⇒ 1 – 3 = -2

⇒ d = -2

4.30th term of the A.P: 10,7, 4, …, is

(a)97

(b)77

(c)-77

(d)-87

Explanation: Given,

A.P. = 10, 7, 4, …

First term, a = 10

Common difference, d = a2 − a1 = 7−10 = −3

As we know, for an A.P.,

an = a +(n−1)d

Putting the values;

a30 = 10+(30−1)(−3)

a30 = 10+(29)(−3)

a30 = 10−87 = −77

5.11th term of the A.P. -3, -1/2, ,2 …. Is

(a)28

(b)22

(c)-38

(d)-48

Explanation: A.P. = -3, -1/2, ,2 …

First term a = – 3

Common difference, d = a2 − a1 = (-1/2) -(-3)

⇒(-1/2) + 3 = 5/2

Nth term;

an = a+(n−1)d

Putting the values;

a11 = 3+(11-1)(5/2)

a11 = 3+(10)(5/2)

a11 = -3+25

a11 = 22

6.The missing terms in AP: __, 13, __, 3 are:

(a)11 and 9

(b)17 and 9

(c)18 and 8

(d)18 and 9

Explanation: a2 = 13 and

a4 = 3

The nth term of an AP;

an = a+(n−1) d

a2 = a +(2-1)d

13 = a+d ………………. (i)

a4 = a+(4-1)d

3 = a+3d ………….. (ii)

Subtracting equation (i) from (ii), we get,

– 10 = 2d

d = – 5

Now put value of d in equation 1

13 = a+(-5)

a = 18 (first term)

a3 = 18+(3-1)(-5)

= 18+2(-5) = 18-10 = 8 (third term).

7. Which term of the A.P. 3, 8, 13, 18, … is 78?

(a)12th

(b)13th

(c)15th

(d)16th

Explanation: Given, 3, 8, 13, 18, … is the AP.

First term, a = 3

Common difference, d = a2 − a1 = 8 − 3 = 5

Let the nth term of given A.P. be 78. Now as we know,

an = a+(n−1)d

Therefore,

78 = 3+(n −1)5

75 = (n−1)5

(n−1) = 15

n = 16

8.The 21st term of AP whose first two terms are -3 and 4 is:

(a)17

(b)137

(c)143

(d)-143

Explanation: First term = -3 and second term = 4

a = -3

d = 4-a = 4-(-3) = 7

a21=a+(21-1)d

=-3+(20)7

=-3+140

=137

9. If 17th term of an A.P. exceeds its 10th term by 7. The common difference is:

(a)1

(b)2

(c)3

(d)4

Explanation: Nth term in AP is:

an = a+(n-1)d

a17 = a+(17−1)d

a17 = a +16d

In the same way,

a10 = a+9d

Given,

a17 − a10 = 7

Therefore,

(a +16d)−(a+9d) = 7

7d = 7

d = 1

Therefore, the common difference is 1.

10. The number of multiples of 4 between 10 and 250 is:

(a)50

(b)40

(c)60

(d)30

Explanation: The multiples of 4 after 10 are:

12, 16, 20, 24, …

So here, a = 12 and d = 4

Now, 250/4 gives remainder 2. Hence, 250 – 2 = 248 is divisible by 2.

12, 16, 20, 24, …, 248

So, nth term, an = 248

As we know,

an = a+(n−1)d

248 = 12+(n-1)×4

236/4 = n-1

59 = n-1

n = 60

11. 20th term from the last term of the A.P. 3, 8, 13, …, 253 is:

(a)147

(b)151

(c)154

(d)158

Explanation: Given, A.P. is 3, 8, 13, …, 253

Common difference, d= 5.

In reverse order,

253, 248, 243, …, 13, 8, 5

So,

a = 253

d = 248 − 253 = −5

n = 20

By nth term formula,

a20 = a+(20−1)d

a20 = 253+(19)(−5)

a20 = 253−95

a20 = 158

12. The sum of the first five multiples of 3 is:

(a)45

(b)55

(c)65

(d)75

Explanation: The first five multiples of 3 is 3, 6, 9, 12 and 15

a=3 and d=3

n=5

Sum, Sn = n/2[2a+(n-1)d]

S5 = 5/2[2(3)+(5-1)3]

=5/2[6+12]

=5/2

=5 x 9

= 45

1. Apoorva S Chandra

Nice questions and answer. It would have been even better if you had included more application based questions

1. Leraux

Ah yes those ones are hard and come for exams

2. A

EXCELENT
NEED MORE QUESTIONS LIKE THIS
KINDLY INCLUDE THEM
THANK YOU