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 their confidence level. The 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.

Click here to download the PDF of additional MCQs for Practice on Arithmetic Progression, Chapter 5 of Class 10 Maths along with answer key:

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Students can also get access to Arithmetic Progression Class 10 Notes here.

#### Below are the MCQs for Arithmetic Progression

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

(a) 4

(b) 5

(c) 3

(d) 7

Answer: **(a) 4**

Explanation: For an AP,

a_{n} = a+(n-1)d

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

= 28+6(-4)

= 28-24

a_{n}=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

Answer:** (b) 10, 20, 30, 40**

Explanation: a = 10, d = 10

a_{1} = a = 10

a_{2} = a_{1}+d = 10+10 = 20

a_{3} = a_{2}+d = 20+10 = 30

a_{4} = a_{3}+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

Answer: **(c) 3 and -2**

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

Answer: **(c) -77**

Explanation: Given,

A.P. = 10, 7, 4, …

First term, a = 10

Common difference, d = a_{2} − a_{1} = 7−10 = −3

As we know, for an A.P.,

a_{n} = a +(n−1)d

Putting the values;

a_{30} = 10+(30−1)(−3)

a_{30} = 10+(29)(−3)

a_{30} = 10−87 = −77

**5. 11th term of the A.P. -3, -1/2, 2 …. Is**

(a) 28

(b) 22

(c) -38

(d) -48

Answer: **(b) 22**

Explanation: A.P. = -3, -1/2, 2 …

First term a = – 3

Common difference, d = a_{2} − a_{1} = (-1/2) -(-3)

⇒(-1/2) + 3 = 5/2

Nth term;

a_{n} = a+(n−1)d

a_{11} = 3+(11-1)(5/2)

a_{11} = 3+(10)(5/2)

a_{11} = -3+25

a_{11} = 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

Answer: **(c)**

Explanation: a_{2} = 13 and

a_{4} = 3

The nth term of an AP;

a_{n} = a+(n−1) d

a_{2} = a +(2-1)d

13 = a+d ………………. (i)

a_{4} = 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)

a_{3 }= 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

Answer:** (d) (d) 16th**

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

First term, a = 3

Common difference, d = a_{2} − a_{1} = 8 − 3 = 5

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

a_{n} = 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

Answer:** (b) 137**

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

a = -3

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

a_{21}=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

Answer: **(a) 1**

Explanation: Nth term in AP is:

a_{n }= a+(n-1)d

a_{17} = a+(17−1)d

a_{17} = 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

Answer: **(c) 60**

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, a_{n} = 248

As we know,

a_{n} = 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

Answer: **(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,

a_{20} = a+(20−1)d

a_{20} = 253+(19)(−5)

a_{20} = 253−95

a_{20} = 158

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

(a) 45

(b) 55

(c) 65

(d) 75

Answer:** (a) 45**

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

a=3 and d=3

n=5

Sum, S_{n} = n/2[2a+(n-1)d]

S_{5} = 5/2[2(3)+(5-1)3]

=5/2[6+12]

=5/2[18]

=5 x 9

= 45

**13. The 10th term of the AP: 5, 8, 11, 14, … is**

(a) 32

(b) 35

(c) 38

(d) 185

Answer:** (a) 32**

Explanation:

Given AP: 5, 8, 11, 14,….

First term = a = 5

Common difference = d = 8 – 5 = 3

nth term of an AP = a_{n} = a + (n – 1)d

Now, 10th term = a_{10} = a + (10 – 1)d

= 5 + 9(3)

= 5 + 27

= 32

**14. In an AP, if d = -4, n = 7, a _{n}**

**= 4, then a is**

(a) 6

(b) 7

(c) 20

(d) 28

Answer: (**d) 28**

Solution;

Given,

d = -4, n = 7, an = 4

We know that,

a_{n} = a + (n – 1)d

4 = a + (7 – 1)(-4)

4 = a + 6(-4)

4 = a – 24

⇒ a = 4 + 24 = 28

**15. The list of numbers –10, –6, –2, 2,… is**

(a) an AP with d = –16

(b) an AP with d = 4

(c) an AP with d = –4

(d) not an AP

Answer: **(b) an AP with d = 4**

Explanation:

–10, –6, –2, 2,…

Let a_{1} = -10, a_{2} = -6, a_{3} = -3, a_{4} = 2

a_{2} – a_{1} = -6 – (-10) = 4

a_{3} – a_{2} = -2 – (-6) = 4

a_{4} – a_{3} = 2 – (-2) = 4

The given list of numbers is an AP with d = 4.

**16. If the 2nd term of an AP is 13 and the 5th term is 25, then its 7th term is**

(a) 30

(b) 33

(c) 37

(d) 38

Answer: **(b) 33**

Explanation:

Given,

a_{2} = 13

a + d = 13

a = 13 – d….(i)

a_{5} = 25

a + 4d = 25….(ii)

Substituting (i) in (ii),

13 – d + 4d = 25

3d = 12

d = 4

So, a = 13 – 4 = 9

a_{7} = a + 6d = 9 + 6(4) = 9 + 24 = 33

**17. Which term of the AP: 21, 42, 63, 84,… is 210?**

(a) 9th

(b) 10th

(c) 11th

(d) 12th

Answer: **(b) 10th**

Explanation:

Given AP:

21, 42, 63, 84,…

a = 21

d = 42 – 21 = 21

a_{n} = 210

a + (n – 1)d = 210

21 + (n – 1)(21) = 210

21 + 21n – 21 = 210

21n = 210

n = 10

**18. What is the common difference of an AP in which a _{18}**

**– a**

_{14}**= 32?**

(a) 8

(b) -8

(c) -4

(d) 4

Answer: **(a) 8**

Explanation:

Given,

a_{18} – a_{14} = 32

We know that, a_{n} = a + (n – 1)d

So,

a + 17d – (a + 13d) = 32

17d – 13d = 32

4d = 32

d = 8

**19. The famous mathematician associated with finding the sum of the first 100 natural numbers is**

(a) Pythagoras

(b) Newton

(c) Gauss

(d) Euclid

Answer: **(c) Gauss**

Explanation:

The famous mathematician associated with finding the sum of the first 100 natural numbers is Gauss.

**20. The sum of first 16 terms of the AP: 10, 6, 2,… is**

(a) –320

(b) 320

(c) –352

(d) –400

Answer: **(a) -320**

Explanation:

Given AP: 10, 6, 2,…

Here, a = 10, d = -4

Sum of first n terms = S_{n} = (n/2)[2a + (n – 1)d]

The sum of first 16 terms = S_{16} = (16/2)[2(10) + (16 – 1)(-4)]

= 8[20 + 15(-4)]

= 8(20 – 60)

= 8(-40)

= -320

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

Ah yes those ones are hard and come for exams

EXCELENT

NEED MORE QUESTIONS LIKE THIS

KINDLY INCLUDE THEM

THANK YOU