Class 10 Maths MCQ Questions of Arithmetic Progression with answers accessible here. The MCQ Questions for Class 10 Maths with answers have been set up according to the most recent syllabus, NCERT books, and exam pattern recommended in Standard 10 by CBSE, NCERT. Multiple Choice Questions for a significant part of exams for Grade 10 Arithmetic Progression and whenever rehearsed appropriately can assist you with getting better grades. Students can tackle  MCQ Questions with Answers to realize their preparation level.

We have given Class 10 Maths MCQs Questions of Arithmetic Progressions with Answers to assist students with understanding the ideas well indeed. All of these inquiries are vital as it assists the students with prepare for CBSE Class 10 Maths Board Examination. This point is additionally vital from the viewpoint of serious tests. Discover the CBSE Class 10 Maths Objective Questions underneath: –

1. If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is

(a) 87
(b) 88
(c) 89
(d) 90

2. In an AP, if d = –4, n = 7, $$a_n$$ = 4, then a is

(a) 6
(b) 7
(c) 20
(d) 28

3. In an AP, if a = 3.5, d = 0, n = 101, then $$a_n$$ will be

(a) 0
(b) 3.5
(c) 103.5
(d) 104.5

4. 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

5. The number of multiples lie between n and $$x^2$$ which are divisible by n is

(a) n + 1
(b) n
(c) n – 1
(d) n – 2

6. If p, q, r and s are in A.P. then r – q is

(a) s – p
(b) s – q
(c) s – r
(d) none of these

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

(a) Pythagoras
(b) Newton
(c) Gauss
(d) Euclid

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

(a) 17

(b) 137

(c) 143

(d) –143

9. If p, q, r, s, t are the terms of an A.P. with common difference -1 the relation between p and t is:​

(a) t = p – 5
(b) t = p – 4
(c) t = p – 6
(d) t = p + 4

10. If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?

(a) 30
(b) 33
(c) 37
(d) 38

11. If the common difference of an AP is 5, then what is $$a_{18} – a_{13}$$?

(a) 5
(b) 20
(c) 25
(d) 30

12. The middle most term (s) of the AP:–11, –7, –3, …, 49 is:

(a) 18, 20
(b) 19, 23
(c) 17, 21
(d) 23, 25

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

(a)4
(b)5
(c)3
(d)7

14. 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

15. 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

16. 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

17. If the sum of n terms of an A.P. be $$3n^2 + n$$ and its common difference is 6, then its first term is

(a) 2
(b) 3
(c) 1
(d) 4

18. If the sum of n terms of an A.P. is $$2n^2 + 5n$$, then its nth term is

(a) 4n – 3
(b) 3n – 4
(c) 4n + 3
(d) 3n + 4

19. If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is

(a) 13
(b) 9
(c) 21
(d) 17

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

(a)50
(b)40
(c)60
(d)30

Explanation:

Here, a7 = 34

a13 = 64

a7 = a + 6d = 34……….(1)

a13 = a + 12d = 64…………(2)

Subtracting (1) from (2)

6d = 30

d = 5

Multiplying (1) by 2

2a + 12d = 68………………….(3)

Subtracting (2) from (3)

a = 4

a18 = a + (n–1) d

a18 = 4 + (17) 5

a18 = 89

Explanation: For an A.P

$$a_n$$ = a + (n – 1)d

4 = a + (7 – 1)( −4)

4 = a + 6(−4)

4 + 24 = a

a = 28

Explanation: For an A.P

$$a_n$$ = a + (n – 1)d

= 3.5 + (101 – 1) × 0

= 3.5

4. Answer: (c) – 2, – 4, – 6, – 8

Explanation: Let the first four terms of an A.P are a, a+d, a+2d and a+3d

Given that the first termis −2 and difference is also −2, then the A.P would be:

– 2, (–2–2), [–2 + 2 (–2)], [–2 + 3(–2)]

= –2, –4, –6, –8

5. Answer: (d) n – 2

6. Answer: (c) s – r

Explanation: Gauss is the famous mathematician associated with finding the sum of the first 100 natural Numbers.

Explanation: First two terms are –3 and 4

Therefore,

a = −3

a + d = 4

⇒ d = 4 − a

⇒ d = 4 + 3

⇒ d = 7

Thus,

$$a_{21}$$ = a + (21 – 1)d

$$a_{21}$$ = –3 + (20)7

$$a_{21}$$ = 137

9. Answer: (b) t = p – 4

Explanation: Since

$$a_2$$ = 13

$$a_5$$ = 25

⇒ a + d = 13 ….(i)

⇒ a + 4d = 25 ….(ii)

Solving equations (i) and (ii), we get:

a = 9; d = 4

Therefore,

$$a_7$$ = 9 + 6 × 4

$$a_7$$ = 9 + 24

$$a_7$$ = 33

Explanation: Since, d = 5

$$a_{18} – a_{13}$$ = a + 17d – a – 12d

= 5d

= 5 × 5

= 25

Explanation: Here, a = −11

d = − 7 – (−11) = 4

And $$a_n$$ = 49

We have,

$$a_n$$ = a + (n – 1)d

⇒ 49 = −11 + (n – 1)4

⇒ 60 = (n – 1)4

⇒ n = 16

As n is an even number, there will be two middle terms which are 16/2th and [(16/2)+1]th, i.e. the 8th term and the 9th term.

$$a_8$$ = a + 7d = – 11 + 7 × 4 = 17

$$a_9$$ = a + 8d = – 11 + 8 × 4 = 21

Explanation: For an AP,

$$a_n$$ = a+(n-1)d

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

= 28+6(-4)

= 28-24

$$a_n=4$$

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

15. Answer: (c) 3 and -2

Explanation: First term, a = 3

Common difference, d = Second term – First term

⇒ 1 – 3 = -2

⇒ d = -2

Explanation: $$N^{th}$$ term in AP is:

$$a_n$$ = a+(n-1)d

$$a_{17}$$ = a+(17−1)d

$$a_{17}$$ = a +16d

In the same way,

$$a_{10}$$ = a+9d

Given,

$$a_{17} − a_{10} = 7$$

Therefore,

(a +16d)−(a+9d) = 7

7d = 7

d = 1

Therefore, the common difference is 1.

Explanation: Here, $$S_n= 3n^2 + n$$
d = 6
Putting n= 1
$$S_1$$ = 3 + 1 = 4
Sum of first 1 term = first term = 4

18. Answer: (c) 4n + 3

Explanation: Here Sn = 2n2 + 5n

Sum of the A.P with 1 term = S1 = 2 + 5 = 7 = first term

Sum of the A.P with 2 terms = 8 + 10 = 18

Sum of the A.P with 3 terms = 18 + 15 = 33

a2 = S2 – S1 = 18 – 7 = 11

d = a2 –a1 = 11 – 7 = 4

nth term = a + (n–1) d

= 7 + (n–1) 4

nth term = 4n + 3

Explanation: Let 3 consecutive terms A.P is a –d, a, a + d. and the sum is 51

So, (a –d) + a + (a + d) = 51

3a –d + d = 51

3a = 51

a = 17

The product of first and third terms = 273

So, (a –d) (a + d) = 273

a2 –d2 = 273

172 –d2 = 273

289 –d2 = 273

d2 = 289 –273

d2 = 16

d = 4

Third term = a + d = 17 + 4 = 21

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