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# MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers

## MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers

Free PDF Download of CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Arithmetic ProgressionsMCQs with Answers to know their preparation level.

### Class 10 Maths MCQs Chapter 5 Arithmetic Progressions

1. The n^{th} term of an A.P. is given by a_{n} = 3 + 4n. The common difference is

(a) 7

(b) 3

(c) 4

(d) 1

**Answer/Explanation**

Answer: c

Explaination:Reason: We have an = 3 + 4n

âˆ´ a_{n+1} = 3 + 4(n + 1) = 7 + 4n

âˆ´ d = a_{n+1} - a_{n}

= (7 + 4n) - (3 + 4n)

= 7 - 3

= 4

2. 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

**Answer/Explanation**

Answer: c

Explaination:Reason: Since p, q, r, s are in A.P.

âˆ´ (q - p) = (r - q) = (s - r) = d (common difference)

3. If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are

(a) 2, 4, 6

(b) 1, 5, 3

(c) 2, 8, 4

(d) 2, 3, 4

**Answer/Explanation**

Answer: d

Explaination:Reason: Let three numbers be a - d, a, a + d

âˆ´ a - d +a + a + d = 9

â‡’ 3a = 9

â‡’ a = 3

Also (a - d) . a . (a + d) = 24

â‡’ (3 -d) .3(3 + d) = 24

â‡’ 9 - dÂ² = 8

â‡’ dÂ² = 9 - 8 = 1

âˆ´ d = Â± 1

Hence numbers are 2, 3, 4 or 4, 3, 2

4. The (n - 1)^{th} term of an A.P. is given by 7,12,17, 22,... is

(a) 5n + 2

(b) 5n + 3

(c) 5n - 5

(d) 5n - 3

**Answer/Explanation**

Answer: d

Explaination:Reason: Here a = 7, d = 12-7 = 5

âˆ´ a_{n-1} = a + [(n - 1) - l]d = 7 + [(n - 1) -1] (5) = 7 + (n - 2)5 = 7 + 5n - 10 = 5M - 3

5. The n^{th} term of an A.P. 5, 2, -1, -4, -7 ... is

(a) 2n + 5

(b) 2n - 5

(c) 8 - 3n

(d) 3n - 8

**Answer/Explanation**

Answer: c

Explaination:Reason: Here a = 5, d = 2 - 5 = -3

a_{n} = a + (n - 1)d = 5 + (n - 1) (-3) = 5 - 3n + 3 = 8 - 3n

6. The 10^{th} term from the end of the A.P. -5, -10, -15,..., -1000 is

(a) -955

(b) -945

(c) -950

(d) -965

**Answer/Explanation**

Answer: a

Explaination:Reason: Here l = -1000, d = -10 - (-5) = -10 + 5 = - 5

âˆ´ 10^{th} term from the end = l - (n - 1 )d = -1000 - (10 - 1) (-5) = -1000 + 45 = -955

7. Find the sum of 12 terms of an A.P. whose nth term is given by a_{n} = 3n + 4

(a) 262

(b) 272

(c) 282

(d) 292

**Answer/Explanation**

Answer: a

Explaination:Reason: Here a_{n} = 3n + 4

âˆ´ a_{1} = 7, a_{2} - 10, a_{3} = 13

âˆ´ a= 7, d = 10 - 7 = 3

âˆ´ S_{12} =
[2 Ã— 7 + (12 - 1) Ã—3] = 6[14 + 33] = 6 Ã— 47 = 282

8. The sum of all two digit odd numbers is

(a) 2575

(b) 2475

(c) 2524

(d) 2425

**Answer/Explanation**

Answer: b

Explaination:Reason: All two digit odd numbers are 11,13,15,... 99, which are in A.P.

Since there are 90 two digit numbers of which 45 numbers are odd and 45 numbers are even

âˆ´ Sum =
[11 + 99] =
Ã— 110 = 45 Ã— 55 = 2475

9. The sum of first n odd natural numbers is

(a) 2nÂ²

(b) 2n + 1

(c) 2n - 1

(d) nÂ²

**Answer/Explanation**

Answer: d

Explaination:Reason: Required Sum = 1 + 3 + 5 + ... + upto n terms.

Here a = 1, d = 3 - 1 = 2

Sum =
[2 Ã— 1 + (n - 1) Ã— 2] =
[2 + 2n - 2] =
Ã— 2n = nÂ²Reason: All two digit odd numbers are 11,13,15,... 99, which are in A.P.

Since there are 90 two digit numbers of which 45 numbers are odd and 45 numbers are even

âˆ´ Sum =
[11 + 99] =
Ã— 110 = 45 Ã— 55 = 2475

10. If (p + q)^{th} term of an A.P. is m and (p - q)^{tn} term is n, then pth term is

**Answer/Explanation**

Answer: d

Explaination:Reason: Let a is first term and d is common difference

âˆ´ a_{p + q} = m

a_{p - q} = n

â‡’ a + (p + q - 1)d = m = ...(i)

â‡’ a + (p - q - 1)d = m = ...(ii)

On adding (i) and (if), we get

2a + (2p - 2)d = m + n

â‡’ a + (p -1)d =
...[Dividing by 2

âˆ´ a_{n} =

15. n^{th} term of the sequence a, a + d, a + 2d,... is

(a) a + nd

(b) a - (n - 1)d

(c) a + (n - 1)d

(d) n + nd

**Answer/Explanation**

Answer: a

Explaination:Reason: an = a + (n - 1)d

16. The 10th term from the end of the A.P. 4, 9,14, ..., 254 is

(a) 209

(b) 205

(c) 214

(d) 213

**Answer/Explanation**

Answer: a

Explaination:Reason: Here l - 254, d = 9-4 = 5

âˆ´ 10^{th} term from the end = l - (10 - 1 )d = 254 -9d = 254 = 9(5) = 254 - 45 = 209

17. If 2x, x + 10, 3x + 2 are in A.P., then x is equal to

(a) 0

(b) 2

(c) 4

(d) 6

**Answer/Explanation**

Answer: d

Explaination:Reason: Since 2x, x + 10 and 3x + 2 are in A.P.

âˆ´ 2(x + 10) = 2x + (3x + 2)

â‡’ 2x + 20 - 5x + 2

â‡’ 2x - 5x = 2 - 20

â‡’ 3x = 18

â‡’ x = 6

18. The sum of all odd integers between 2 and 100 divisible by 3 is

(a) 17

(b) 867

(c) 876

(d) 786

**Answer/Explanation**

Answer: b

Explaination:Reason: The numbers are 3, 9,15, 21, ..., 99

Here a = 3, d = 6 and a_{n} = 99

âˆ´ a_{n} = a + (n - 1 )d

â‡’ 99 = 3 + (n - 1) x 6

â‡’ 99 = 3 + 6n - 6

â‡’ 6n = 102

â‡’ n = 17

Required Sum =
[a + a_{n}] =
[3 + 99] =
Ã— 102 = 867

19. If the numbers a, b, c, d, e form an A.P., then the value of a - 4b + 6c - 4d + e is

(a) 0

(b) 1

(c) -1

(d) 2

**Answer/Explanation**

Answer: a

Explaination:Reason: Let x be the common difference of the given AP

âˆ´ b = a + x, c = a + 2x, d = a + 3x and e = a + 4x

âˆ´ a - 4b + 6c - 4d + e = a - 4 (a + x) + 6(a + 2x) - 4(a + 3x) + (a + 4x)

= a - 4a - 4x + 6a + 12x - 4a - 12x + a + 4x = 8a - 8a + 16x - 16x = 0

20. If 7 times the 7^{th} term of an A.P. is equal to 11 times its 11^{th} term, then 18^{th} term is

(a) 18

(b) 9

(c) 77

(d) 0

**Answer/Explanation**

Answer: d

Explaination:Reason: We have 7a_{7} = 11a_{11}

â‡’ 7[a + (7 - 1)d] = 11[a + (11 - 1 )d]

â‡’ 7(a + 6d) = 11(a + 10d)

â‡’ 7a + 42d = 11a + 110d

â‡’ 4a = -68d

â‡’ a = -17d

âˆ´ a_{18} = a + (18 - 1)d = a + 17d = -17d + 17d = 0

21. If p, q, r are in AP, then p^{3} + r^{3} - 8q^{3} is equal to

(a) 4pqr

(b) -6pqr

(c) 2pqr

(d) 8pqr

**Answer/Explanation**

Answer: b

Explaination:

âˆµ p, q, r are in AP.

âˆ´ 2q = p + r

â‡’ p + r - 2q = 0

âˆ´ p^{3} + r^{3} + (-2p)^{3} = 3 Ã— p Ã— r Ã— -2q

[Using ifa + 6 + c = 0 â‡’ a^{3} + b^{3} + c^{3} = 3 abc]

â‡’ p^{3} + r^{3} - 8q^{3} = -6pqr.

22. In an AP, if a = 3.5, d = 0, n = 101, then a will be [NCERT Exemplar Problems]

(a) 0

(b) 3.5

(c) 103.5

(d) 104.5

**Answer/Explanation**

Answer: b

Explaination: (b) a_{101} = 3.5 + 0(100) = 3.5

23. The list of numbers -10, -6, -2, 2, ... is [NCERT Exemplar Problems]

(Ð°) an AP with d = -16

(b) an AP with d = 4

(c) an AP with d = -4

(d) not an AP

**Answer/Explanation**

Answer: b

Explaination: (b) An AP with d = 4.

24. Two APs have the same common difference. . The first term of one of these is -1 and that of the other is -8. Then the difference between their 4th terms is [NCERT Exemplar Problems]

(a) -1

(b) -8

(c) 7

(d) -9

**Answer/Explanation**

Answer: c

Explaination:

a_{4} - b_{4} = (a_{1} + 3d) - (b_{1} + 3d)

= a_{1} - b_{1}= - 1 - (-8) = 7

25. In an AP, if d = -2, n = 5 and an = 0, the value of a is

(a) 10

(b) 5

(c) -8

(d) 8

**Answer/Explanation**

Answer: d

Explaination:

d = - 2, n = 5, a_{n} = 0

âˆµ a_{n} = 0

â‡’ a + (n - 1)d=0

â‡’ a + (5 - 1)(- 2) = 0

â‡’ a = 8

Correct option is (d).

26. If the common difference of an AP is 3, then a_{20} - a_{15} is

(a) 5

(b) 3

(c) 15

(d) 20

**Answer/Explanation**

Answer: c

Explaination:

Common difference, d = 3

a_{20} - a_{15} = (a + 19d) - (a+ 14d)

= 5d=5 Ã— 3 = 15

27. The next term of the AP âˆš18, âˆš50, âˆš98, ........ is

(a) âˆš146

(b) âˆš128

(c) âˆš162

(d) âˆš200

**Answer/Explanation**

Answer: c

Explaination:

(c) âˆš18, âˆš50, âˆš98, ..... = 3âˆš2, 5âˆš2, 7âˆš2, ......

âˆ´ Next term is 9âˆš2 = âˆš162

28. The common difference of the AP

(a) p

(b) -p

(c) -1

(d) 1

**Answer/Explanation**

Answer: c

Explaination: (c) Common difference = a_{2} - a_{1}

29. If the n^{th} term of an AP is (2n +1), then the sum of its first three terms is

(a) 6n + 3

(b) 15

(c) 12

(d) 21

**Answer/Explanation**

Answer: b

Explaination:

a_{1}= 2 Ã— 1 + 1 = 3,

a_{2} = 2 Ã— 2 + 1 = 5,

a_{3} = 2 Ã— 3 + l= 7

âˆ´ Sum = 3 + 5 + 7 = 15

30. An AP consists of 31 terms. If its 16th term is m, then sum of all the terms of this AP is

(a) 16 m

(b) 47 m

(c) 31 m

(d) 52 m

**Answer/Explanation**

Answer: c

Explaination:

S_{31} =
(2a + 30d)

a_{16} = a + 15d = zw

â‡’ S_{31} =
Ã— 2(a + 15d)

â‡’ S_{31} = 31m

31. The first term of an AP of consecutive integers is pÂ² + 1. The sum of 2p + 1 terms of this AP is

(a) (p + 1)Â²

(b) (2p + 1) (p + 1)Â²

(c) (p+1)^{3}

(d) p^{3} + (p + 1)^{3}

**Answer/Explanation**

Answer: d

Explaination:

32. If the sum of first n terms of an AP is An + BnÂ² where A and B are constants, the common difference of AP will be

(a) A + B

(b) A - B

(c) 2A

(d) 2B

**Answer/Explanation**

Answer: d

Explaination:

S_{n} = An + BnÂ²

S_{1} = A Ã— 1 +B Ã— 1Â²

= A + B

âˆµ S_{1} = a_{1}

âˆ´ a_{1} = A + B ... (1)

and S_{2} = A Ã— 2 + B Ã— 2Â²

â‡’ a_{1} + a_{2} = 2A + 4B

â‡’ (A + B) + a_{2} = 2A + 4B[Using (i)]

â‡’ a_{2} = A + 3B

âˆ´ d = a_{2} - a_{1} = 2B

33. If p - 1, p + 3, 3p - 1 are in AP, then p is equal to ______ .

**Answer/Explanation**

Answer:

Explaination:

âˆµ p - 1, p + 3 and 3p - 1 are in AP.

âˆ´ 2(p + 3) = p - 1 + 3p - 1

â‡’ 2p + 6 = 4/> -2.

â‡’ -2p = -8

â‡’ p = 4.

34. Write down the first four terms of the sequences whose general terms are

(i) T_{n} = 2n + 3

(ii) T_{n} =3^{n + 1}

(iii) T_{1} = 2, T_{n} = T_{n-1}+ 5, n â‰¥ 2

**Answer/Explanation**

Answer:

Explaination:

(i) T_{n}= 2n + 3

T_{1} = 2 Ã— 1 + 3 = 5,

T_{2} = 2 Ã— 2 + 3 = 7,

T_{3} = 2 Ã— 3 + 3 = 9,

T_{4} = 2 Ã— 4 + 3 = 11

âˆ´ Ist four terms are 5, 7, 9 and 11.

(ii) T_{n} = 3^{n+1}

â‡’ T_{1} = 3^{1+1} = 9,

T_{2} = 3^{2+1} = 27,

T_{3}= 3^{3+1} = 81,

T_{4} = 3^{4+1} = 243

âˆ´ Ist four terms are 9, 27, 81 and 243

(iii) T_{1} = 2, T_{n} = T_{n-1} + 5, n â‰¥ 2

â‡’ T_{2} = T_{2-1} + 5

= T_{1} + 5 = 2 + 5 = 7

T_{3} = T_{3-1} + 5

= T_{2} + 5 = 7 + 5 = 12

and T_{4}=T_{4-1} + 5

= T_{3} + 5 = 12 + 5 = 17

âˆ´ Ist four terms are 2, 7, 12 and 17.

35. Find:

The 10th term of 10.0, 10.5, 11.0, 11.5, .....

**Answer/Explanation**

Answer:

Explaination:

a=10, d = 10.5 - 10 = 0.5

a_{10} = a + 9d= 10 + 9 Ã— 0.5 = 14.5

36. In an A.P., if the common difference (d) = - 4 and the seventh term (a_{7}) is 4, then find the first term. [CBSE 2018]

**Answer/Explanation**

Answer:

Explaination:

Let a = first term,

Given, d = - 4, a_{7} = 4

â‡’ a + (7 - 1)d = 4

[ âˆµ nth term of an AP = a_{n} = a + (n - 1)d]

â‡’ a + 6d = 4

â‡’ a + 6 Ã— (-4) = 4

â‡’ a - 24 = 4.

â‡’ a = 4 + 24 = 28

âˆ´ First term, a = 28

38. Which term of the AP 21, 18, 15, ... , is zero?

**Answer/Explanation**

Answer:

Explaination:

Here, a = 21, J= 18-21 = -3

Let a_{n} = 0

â‡’ a + (n - 1 )d = 0

â‡’ 21 +(n - 1)(-3) = 0

â‡’ (n - 1)(- 3) = -21

â‡’ n - 1 =

â‡’ n = 8

âˆ´ 8th term is zero.

39. For what value ofp, are 2p+ 1, 13, 5p - 3 three consecutive terms of an AP?

**Answer/Explanation**

Answer:

Explaination:

If terms are in AP, then

13 - (2p + 1) = (5p - 3) - 13

â‡’ 13 - 2p - 1 = 5p - 3 - 13

â‡’ 28 = 7p

â‡’ p = 4.

40. What is the common difference of an A.P. in which a_{21} - a_{7} = 84? [AI 2017]

**Answer/Explanation**

Answer:

Explaination:

Let â€˜cf be the common difference of the AP whose first term is â€˜aâ€™

Now, a_{21} - a_{7} = 84

â‡’ (a + 20d) - (a + 6d)= 84

â‡’ 20d - 6d = 84

â‡’ 14d = 84

â‡’ d = 6