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## 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

∴ a_{n+1} = 3 + 4(n + 1) = 7 + 4n

∴ 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.

∴ (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

∴ a – d +a + a + d = 9

⇒ 3a = 9

⇒ a = 3

Also (a – d) . a . (a + d) = 24

⇒ (3 -d) .3(3 + d) = 24

⇒ 9 – d² = 8

⇒ d² = 9 – 8 = 1

∴ d = ± 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

∴ 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

∴ 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

∴ a_{1} = 7, a_{2} – 10, a_{3} = 13

∴ a= 7, d = 10 – 7 = 3

∴ S_{12} = \(\frac{12}{2}\)[2 × 7 + (12 – 1) ×3] = 6[14 + 33] = 6 × 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

∴ Sum = \(\frac{45}{2}\)[11 + 99] = \(\frac{45}{2}\) × 110 = 45 × 55 = 2475

9. The sum of first n odd natural numbers is

(a) 2n²

(b) 2n + 1

(c) 2n – 1

(d) n²

**Answer/Explanation**

Answer: d

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

Here a = 1, d = 3 – 1 = 2

Sum = \(\frac{n}{2}\)[2 × 1 + (n – 1) × 2] = \(\frac{n}{2}\)[2 + 2n – 2] = \(\frac{n}{2}\) × 2n = n²Reason: 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

∴ Sum = \(\frac{45}{2}\)[11 + 99] = \(\frac{45}{2}\) × 110 = 45 × 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

∴ a_{p + q} = m

a_{p – q} = n

⇒ a + (p + q – 1)d = m = …(i)

⇒ a + (p – q – 1)d = m = …(ii)

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

2a + (2p – 2)d = m + n

⇒ a + (p -1)d = \(\frac{m+n}{2}\) …[Dividing by 2

∴ a_{n} = \(\frac{m+n}{2}\)

11. If a, b, c are in A.P. then \(\frac{a-b}{b-c}\) is equal to

**Answer/Explanation**

Answer: a

Explaination:Reason: Since a, b, c are in A.P.

∴ b – a = c – b

⇒ \(\frac{b-a}{c-b}\) = 1

⇒ \(\frac{a-b}{b-c}\) = 1

12. The number of multiples lie between n and n² which are divisible by n is

(a) n + 1

(b) n

(c) n – 1

(d) n – 2

**Answer/Explanation**

Answer: d

Explaination:Reason: Multiples of n from 1 to n² are n × 1, n × 2, n × 3, …, m× n

∴ There are n numbers

Thus, the number of mutiples of n which lie between n and n² is (n – 2) leaving first and last in the given list: Total numbers are (n – 2).

13. If a, b, c, d, e are in 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 common difference of A.P. be x

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

Given equation n-4b + 6c-4d + c

= a – 4(a + x) + 6(A + 2r) – 4(n + 3x) + (o + 4.v)

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

14. The next term of the sequence

**Answer/Explanation**

Answer: a

Explaination:

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

∴ 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.

∴ 2(x + 10) = 2x + (3x + 2)

⇒ 2x + 20 – 5x + 2

⇒ 2x – 5x = 2 – 20

⇒ 3x = 18

⇒ 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

∴ a_{n} = a + (n – 1 )d

⇒ 99 = 3 + (n – 1) x 6

⇒ 99 = 3 + 6n – 6

⇒ 6n = 102

⇒ n = 17

Required Sum = \(\frac{n}{2}\)[a + a_{n}] = \(\frac{17}{2}\)[3 + 99] = \(\frac{17}{2}\) × 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

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

∴ 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}

⇒ 7[a + (7 – 1)d] = 11[a + (11 – 1 )d]

⇒ 7(a + 6d) = 11(a + 10d)

⇒ 7a + 42d = 11a + 110d

⇒ 4a = -68d

⇒ a = -17d

∴ a_{18} = a + (18 – 1)d = a + 17d = -17d + 17d = 0

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