1. If nth term of an AP is 7 – 4n, then its common difference is
ANSWER=
D. 11
2. If the sum of first n terms of an AP is An + Bn² where A and B are constants. The common difference of AP will be
ANSWER=
D. 2B
3. If p – 1, p + 3, 3p – 1 are in AP, then p is equal to
ANSWER=
A. 4
4. The sum of the first 2n terms of the AP: 2, 5, 8, …. is equal to sum of the first n terms of the AP: 57, 59, 61, … then n is equal to.
ANSWER=
B. 11
5. If the sum of the first m terms of an AP is n and the sum of its n terms is m, then the sum of its (m + n) terms will be
ANSWER=
B. -(m + n)
6. If nth term of an AP is given by fn = 3n + 4, find the common difference of the AP
ANSWER=
A. 3
7. The sum of first ten natural number is
ANSWER=
A. 55
8. Find the 15th term of an AP -2, -5, -8, ….
ANSWER=
B. -44
9. The 6th term from the end of the AP: 5, 2, -1, -4, …., -31, is
ANSWER=
D. -16
10. Which term of the AP: 27, 24, 21, ……… is zero?
ANSWER=
B. 10th
11. The sum of first n terms of the series a, 3a, 5a, …….. is
ANSWER=
C. n²a
12. Which term of the AP: 92, 88, 84, 80 … is 0?
ANSWER=
D. 24
13. The common difference of the AP … -4, -2, 0, 2, …. is
ANSWER=
A. 2
14. The nth term of an A.P. is given by an = 3 + 4n. The common difference is
ANSWER=
C. 4
Explaination:Reason: We have an = 3 + 4n
∴ an+1 = 3 + 4(n + 1) = 7 + 4n
∴ d = an+1 – an
= (7 + 4n) – (3 + 4n)
= 7 – 3
= 4
15. If p, q, r and s are in A.P. then r – q is
ANSWER=
C. s – r
Explaination:Reason: Since p, q, r, s are in A.P.
∴ (q – p) = (r – q) = (s – r) = d (common difference)
16. If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are
ANSWER=
D. 2, 3, 4
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
17. The (n – 1)th term of an A.P. is given by 7,12,17, 22,… is
ANSWER=
D. 5n – 3
Explaination:Reason: Here a = 7, d = 12-7 = 5
∴ an-1 = a + [(n – 1) – l]d = 7 + [(n – 1) -1] (5) = 7 + (n – 2)5 = 7 + 5n – 10 = 5M – 3
18. The nth term of an A.P. 5, 2, -1, -4, -7 … is
ANSWER=
C. 8 – 3n
Explaination:Reason: Here a = 5, d = 2 – 5 = -3
an = a + (n – 1)d = 5 + (n – 1) (-3) = 5 – 3n + 3 = 8 – 3n
19. The 10th term from the end of the A.P. -5, -10, -15,…, -1000 is
ANSWER=
A. -955
Explaination:Reason: Here l = -1000, d = -10 – (-5) = -10 + 5 = – 5
∴ 10th term from the end = l – (n – 1 )d = -1000 – (10 – 1) (-5) = -1000 + 45 = -955
20. Find the sum of 12 terms of an A.P. whose nth term is given by an = 3n + 4
ANSWER=
B. 2475
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 = 452[11 + 99] = 452 × 110 = 45 × 55 = 2475
22. The sum of first n odd natural numbers is
ANSWER=
D. n²
Explaination:Reason: Required Sum = 1 + 3 + 5 + … + upto n terms.
Here a = 1, d = 3 – 1 = 2
Sum = n2[2 × 1 + (n – 1) × 2] = n2[2 + 2n – 2] = n2 × 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 = 452[11 + 99] = 452 × 110 = 45 × 55 = 2475
23. The number of multiples lie between n and n² which are divisible by n is
ANSWER=
D. n – 2
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).
24. nth term of the sequence a, a + d, a + 2d,… is
ANSWER=
A. a + nd
Explaination:Reason: an = a + (n – 1)d
25. The 10th term from the end of the A.P. 4, 9,14, …, 254 is
ANSWER=
A. 209
Explaination:Reason: Here l – 254, d = 9-4 = 5
∴ 10th term from the end = l – (10 – 1 )d = 254 -9d = 254 = 9(5) = 254 – 45 = 209
26. If 2x, x + 10, 3x + 2 are in A.P., then x is equal to
ANSWER=
D. 6
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
27. The sum of all odd integers between 2 and 100 divisible by 3 is
ANSWER=
B. 867
Explaination:Reason: The numbers are 3, 9,15, 21, …, 99
Here a = 3, d = 6 and an = 99
∴ an = a + (n – 1 )d
⇒ 99 = 3 + (n – 1) x 6
⇒ 99 = 3 + 6n – 6
⇒ 6n = 102
⇒ n = 17
Required Sum = n2[a + an] = 172[3 + 99] = 172 × 102 = 867
28. If the numbers a, b, c, d, e form an A.P., then the value of a – 4b + 6c – 4d + e is
ANSWER=
A. 0
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
29. If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, then 18th term is
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