Sequences and Series VBQs Class 11 Mathematics

VBQs for Class 11

Question. If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is :
(a) 1/6
(b) 1/5
(c) 1/4
(d) 1/7

Answer

A

Question. If nC4, nC5 and nC6 are in A.P., then n can be :
(a) 9
(b) 14
(c) 11
(d) 12

Answer

B

Question. The sum of all natural numbers ‘n’ such that 100 < n < 200 and H.C.F. (91, n) > 1 is : 
(a) 3203
(b) 3303
(c) 3221
(d) 3121

Answer

D

Question. The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is:
(a) 1256
(b) 1465
(c) 1365
(d) 1356

Answer

D

Question. If 19th term of a non- ero A.P. is ero, then its (49th term) : (29th term) is : 
(a) 4 : 1
(b) 1 : 3
(c) 3 : 1
(d) 2 : 1

Answer

C

Question. In the sum of the series

Sequences and Series VBQs Class 11 Mathematics

upto nth term is 488 and then nth term is negative, then :
(a) n = 60
(b) nth term is –4
(c) n = 41
(d) nth term is –4 x 2/5

Answer

B

Question. If the sum of first 11 terms of an A.P., a1 , a2 , a3 … is 0 (a1 ≠ 0), then the sum of the A.P., a1 , a3 , a5 , …, a23 is ka1, where k is equal to :
(a) –121/10
(b) 121/10
(c) 72/5
(d) 72/5

Answer

D

Question. The number of terms common to the two A.P.’s 3, 7, 11, …, 407 and 2, 9, 16, …, 709 is _____.

Answer

14

Question. If the 10th term of an A.P. is 1/20 and its 20th term is 1/10 then the sum of its first 200 terms is:
(a) 50
(b) 50 1/4
(c) 100
(d) 100 1/2

Answer

D

Question. Let ƒ : R → R be such that for all x ∈ R, (21+x + 21–x), ƒ(x) and (3x + 3–x) are in A.P., then the minimum value of f (x) is:
(a) 2
(b) 3
(c) 0
(d) 4

Answer

B

Question. Five numbers are in A.P., whose sum is 25 and product is 2520. If one of these five numbers is –1/2 , then the greatest number amongst them is:
(a) 27
(b) 7
(c) 21/2
(d) 16

Answer

D

Question. Let Sn denote the sum of the first n terms of an A.P. If S4 = 16 and S6 = –48, then S10 is equal to : 
(a) –260
(b) –410
(c) –320
(d) –380

Answer

C

Question. If a1, a2, a3, …… are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P. is :
(a) 200
(b) 280
(c) 120
(d) 150

Answer

A

Question. If a1, a2, a3, ….. an are in A.P. and a1 + a4 + a7 + ….+ a16 = 114, then a1 + a6 + a11 + a16 is equal to : 
(a) 98
(b) 76
(c) 38
(d) 64

Answer

B

Question. Let the sum of the first n terms of a non-constant A.P., a1, a2, a3, ………….. be 50n + n(n+7)/2A, where A is a constant. If d is the common difference of this A.P., then the ordered pair (d, a50) is equal to:
(a) (50, 50 + 46A)
(b) (50, 50 + 45A)
(c) (A, 50 + 45A)
(d) (A, 50 + 46A)

Answer

D

Question. Let an be the nth term of a G.P. of positive terms.

Sequences and Series VBQs Class 11 Mathematics

(a) 300
(b) 225
(c) 175
(d) 150

Answer

D

Question. 

Sequences and Series VBQs Class 11 Mathematics

(a) x(1 + y) = 1
(b) y(1 – x) = 1
(c) y(1 + x) = 1
(d) x(1 – y) = 1

Answer

B

Question.

Sequences and Series VBQs Class 11 Mathematics

where the function ƒ satisfies ƒ(x + y) = ƒ(x) ƒ(y) for all natural numbers x, y and ƒ(a) = 2. Then the natural number ‘a’ is:
(a) 2
(b) 16
(c) 4
(d) 3

Answer

D

Question. If the sum and product of the first three terms in an A.P. are 33 and 1155, respectively, then a value of its 11th term is: 
(a) –35
(b) 25
(c) –36
(d) –25

Answer

D

Question. Let a1, a2, ….., a30 be an A.P.,

Sequences and Series VBQs Class 11 Mathematics

If a5 = 27 and S – 2T = 75, then a10 is equal to:
(a) 52
(b) 57
(c) 47
(d) 42

Answer

A

Question. Let a, b and c be the 7th, 11th and 13th terms respectively of a non-constant A.P. If these are also the three consecutive terms of a G.P., then a/c is equal to:
(a) 2
(b) 1/2
(c) 7/13
(d) 4

Answer

D

Question. If a, b and c be three distinct real numbers in G.P. and a + b + c = xb, then x cannot be:
(a) – 2
(b) – 3
(c) 4
(d) 2

Answer

D

Question.

Sequences and Series VBQs Class 11 Mathematics

(a) 3
(b) 13/8
(c) 13/4
(d) 1/8

Answer

C

Question.

Sequences and Series VBQs Class 11 Mathematics

(a) 2560
(b) 2650
(c) 3200
(d) 1600

Answer

A

Question. If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is :
(a) 2
(b) 41/3
(c) 42/3
(d) 4

Answer

A

Question. The sum of the first 20 terms common between the series 3 + 7 + 11 + 15 + ……… and 1 + 6 + 11 + 16 + ……, is
(a) 4000
(b) 4020
(c) 4200
(d) 4220

Answer

B

Question. The common difference of the A.P. b1, b2, …, bm is 2 more than the common difference of A.P. a1, a2, …, an. If a40 = – 159, a100 = – 399 and b100 = a70, then b1 is equal to:
(a) 81
(b) – 127
(c) – 81
(d) 127

Answer

C

Question. If 32sin2a–1, 14 and 34–2sin2a are the first three terms of an A.P. for some a, then the sixth term of this A.P is:
(a) 66
(b) 81
(c) 65
(d) 78

Answer

A

Question. Given an A.P. whose terms are all positive integers. The sum of its first nine terms is greater than 200 and less than 220. If the second term in it is 12, then its 4th term is:
(a) 8
(b) 16
(c) 20
(d) 24

Answer

C

Question. If a1, a2, a3,…., an, …. are in A.P. such that a4 – a7 + a10 = m, then the sum of first 13 terms of this A.P., is :
(a) 10 m
(b) 12 m
(c) 13 m
(d) 15 m

Answer

C

Question. Given sum of the first n terms of an A.P. is 2n + 3n2. Another A.P. is formed with the same first term and double of the common difference, the sum of n terms of the new A.P. is :
(a) n + 4n2
(b) 6n2 – n
(c) n2 + 4n
(d) 3n + 2n2

Answer

B

Question. Let a1, a2, a3,… be an A.P, such that

Sequences and Series VBQs Class 11 Mathematics

(a) 41/11
(b) 31/121
(c) 11/41
(d) 121/1861

Answer

B

Question. If 100 times the 100th term of an AP with non ero common difference equals the 50 times its 50th term, then the 150th term of this AP is :
(a) – 150
(b) 150 times its 50th term
(c) 150
(d) Zero

Answer

D

Question. If the A.M. between pth and qth terms of an A.P. is equal to the A.M. between rth and sth terms of the same A.P., then p + q is equal to
(a) r + s – 1
(b) r + s – 2
(c) r + s + 1
(d) r + s

Answer

D

Question. Suppose q and Φ (≠ 0) are such that sec (θ + Φ), sec q and sec (θ –Φ) are in A.P. If cos θ = Kcos(Φ/2) for some k, then k is equal to
(a) ± 2
(b) ±1
(c) ±1√2
(d) ± 2

Answer

A

Question. If three distinct numbers a, b, c are in G.P. and the equations ax2 + 2bx + c = 0 and dx2 + 2ex + ƒ = 0 have a common root, then which one of the following statements is correct?
(a) d/a , e/b , f/c are in A.P.
(b) d, e, f are in A.P.
(c) d, e, f are in G.P.
(d) d/a , e/b , f/c are in G.P.

Answer

A

Question. The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is :
(a) 36
(b) 32
(c) 24
(d) 28

Answer

D

Question. The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 27/19 Then the common ratio of this series is :
(a) 1/3
(b) 2/3
(c) 2/9
(d) 4/9

Answer

B

Question. Let Sn = 1 + q + q2 + …. + qn and

Sequences and Series VBQs Class 11 Mathematics

where q is a real number and q ≠ 1. If 101C1 + 101C2 .S1 + …. + 101C101 .S100 = aT100, then a is equal to : 
(a) 299
(b) 202
(c) 200
(d) 2100

Answer

D

Question.

Sequences and Series VBQs Class 11 Mathematics

(a) α – β
(b) α – β /100
(c) β – α
(d) α – β /200

Answer

B

Question. A man saves ₹ 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ₹ 40 more than the saving of immediately previous month. His total saving from the start of service will be ₹ 11040 after 
(a) 19 months
(b) 20 months
(c) 21 months
(d) 18 months

Answer

C

Question. A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a1 = a2 = … = a10 = 150 and a10, a11, … are in an AP with common difference –2, then the time taken by him to count all notes is
(a) 34 minutes
(b) 125 minutes
(c) 135 minutes
(d) 24 minutes

Answer

A

Question. Let a1, a2, a3, …., an, be in A.P. If a3 + a7 + a11 + a15 = 72, then the sum of its first 17 terms is equal to :
(a) 306
(b) 204
(c) 153
(d) 612

Answer

A

Question. Let a and b be the roots of equation px2 + qx + r = 0, p ¹ 0. If p, q, r are in A.P and 1/α + 1/β = 4, then the value of | α – β| is:
(a) 34/9
(b) 2√13/9
(c) 61/9
(d) 2√17/9

Answer

B

Question. Let a1, a2 , a3………… be terms on A.P. If

Sequences and Series VBQs Class 11 Mathematics

(a) 11/41
(b) 2/7
(c) 7/2
(d) 41/11

Answer

D

Question. If the coefficients of rth, (r + 1)th, and (r + 2)th terms in the the binomial expansion of (1+y)m are in A.P., then m and r satisfy the equation
(a) m2 – m (4r – 1) + 4r2 – 2 = 0
(b) m2 – m (4r + 1) + 4r2 + 2 = 0
(c) m2 – m (4r + 1) + 4r2 – 2 = 0
(d) m2 – m (4r – 1) + 4r2 + 2 = 0

Answer

C

Question. Let Tr be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m,n, m ≠ n, Tm = 1/n and Tn = 1/m then a – d equals
(a) 1/m + 1/n
(b) 1
(c) 1/mn
(d) 0

Answer

D

Question. If 1, log9 (31–x + 2), log3 (4.3x – 1) are in A.P. then x equals
(a) log3 4
(b) 1 – log3 4
(c) 1 – log4 3
(d) log4 3

Answer

B

Question. If f (x + y) = f (x) f (y) and x=1 f(x) = 2 , x, y Î N, where N is the set of all natural numbers, then the value of f(4)/f(2) is :
(a) 2/3
(b) 1/9
(c) 1/3
(d) 4/9

Answer

D

Question. Let a1,a2,a3,…,a49 be in A.P. such that

Sequences and Series VBQs Class 11 Mathematics

(a) 68
(b) 34
(c) 33
(d) 66

Answer

B

Question. For any three positive real numbers a, b and c, 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c). Then :
(a) a, b and c are in G.P.
(b) b, c and a are in G.P.
(c) b, c and a are in A.P.
(d) a, b and c are in A.P

Answer

C

Question. Let a, b, c, d and p be any non ero distinct real numbers such that (a2 + b2 + c2)p2 – 2 (ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :
(a) a, c, p are in A.P.
(b) a, c, p are in G.P.
(c) a, b, c, d are in G.P.
(d) a, b, c, d are in A.P.

Answer

C

Question. Suppose that a function ƒ : R→R satisfies ƒ (x + y) = ƒ(x)ƒ(y) for all

Sequences and Series VBQs Class 11 Mathematics

then n is equal to _________ .

Answer

5.00

Question. If 210 + 29×31 + 28×32 + . . . . + 2×39 + 310 = S – 211 then S is equal to:
(a) 311 – 212
(b) 311
(c) 311/2 + 210
d. 2×311

Answer

B

Question. If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is:
(a) 1 /26 (349–1)
(b) 1 /26 (350–1)
c. 2 /13 (350–1)
d. 1 /13 (350–1)

Answer

B

Question. If the sum of the first 20 terms of the series log(71/2) x + log(71/3) + log(71/4)x + … is 460, then x is equal to : 
(a) 72
(b) 71/2
(c) e2
(d) 746/21

Answer

A

Question. Let a1, a2 , ,….., an be a given A.P. whose common difference is an integer and Sn = a1 + a2 +…. + an If a1 = 1 , an = 300 and 15 ≤ n ≤ 50, then the ordered pair (Sn-4 , an-4) is equal to :
(a) (2490, 249)
(b) (2480, 249)
(c) (2480, 248)
(d) (2490, 248)

Answer

D

Question. Let a and b be the roots of x2 -3x + p = 0 and g and d be the roots of 2 x – 6x + q = 0. If a, b, g, d form a geometric progression. Then ratio (2q + p) : (2q – p) is :
(a) 3 : 1
(b) 9 : 7
(c) 5 : 3
(d) 33 : 31

Answer

B

Question. The value of (0.16)log2.5 (1/3 + 1/32 + 1/33 +……..to ∞) is equal to _________ .

Answer

4

Question. The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie in : 
(a) (-∞, – 9] ∪ [3, ∞)
(b) [-3, ∞)
(c) (-∞, – 3] ∪ [9, ∞)
(d) (-∞, 9]

Answer

C

Question. If | x |< 1, | y |< 1 and x ¹ y, then the sum to infinity of the following series (x + y) + (x2 + xy + y2 ) + (x3 + x2y + xy2 + y3 ) +…. is :

Sequences and Series VBQs Class 11 Mathematics
Answer

C

Question. The product 21/4 . 41/16 . 81/48 .161/128 … to ∞ is equal to:
(a) 21/2
(b) 21/4
(c) 1
(d) 2

Answer

A

Question. The greatest positive integer k, for which 49k + 1 is a factor of the sum 49125 + 49124 + … + 492 + 49 + l, is:
(a) 32
(b) 63
(c) 60
(d) 65

Answer

B

Question. Let a1, a2, a3, … be a G. P. such that a1 < 0, a1 + a2 = 4 and a3 + a4 = 16. If 

Sequences and Series VBQs Class 11 Mathematics

(a) –513
(b) –171
(c) 171
(d) 511/3

Answer

B

Question. The coefficient of x7 in the expression (1 + x)10 + x(l + x)9 + x2(l + x)8 + … + x10 is:
(a) 210
(b) 330
(c) 120
(d) 420

Answer

B

Question. If a, b and g are three consecutive terms of a nonconstant G.P. such that the equations αx2 + 2βx + γ = 0 and x2 + x – 1 = 0 have a common root, then a (b+ g) is equal to : 
(a) 0
(b) αβ
(c) αγ
(d) βγ

Answer

D

Question. Let a, b and c be in G.P. with common ratio r, where a≠ 0 and 0 r ≤ 1/2. If 3a, 7b and 15c are the first three terms of an A.P., then the 4th term of this A.P. is :
(a) 2/3 a
(b) 5 a
(c) 7/3 a
(d) a

Answer

D

Question. Let a and b be the roots of the quadratic equation x2 sinθ – x (sinθ cosθ + 1) + cosθ = 0 (0 < q < 450 ), and

Sequences and Series VBQs Class 11 Mathematics
Answer

C

Question. Let a1, a2, …, a10 be a G.P. If a3/a1=25, then a9/a5 equals :
(a) 54
(b) 4(52)
(c) 53
(d) 2(52

Answer

A

Question. Let S be the sum of the first 9 terms of the series :
{x + ka}+{x2 + (k + 2)a}+{x3 + (k + 4)a} +{x4 + (k + 6)a}+… where a ≠ 0 and x ≠ 1.
S = x10–x+45a(x–1) / x-1 , then k is equal to :
(a) – 5
(b) 1
(c) – 3
(d) 3

Answer

C

Question. If a, b, c, d and p are distinct real numbers such that (a2 + b2 + c2) p2 – 2p (ab + bc + cd) + (b2 + c2 + d2) ≤ 0, then
(a) a, b, c, d are in A.P.
(b) ab = cd
(c) ac = bd
(d) a, b, c, d are in G.P.

Answer

D

Question. The difference between the fourth term and the first term of a Geometrical Progresssion is 52. If the sum of its first three terms is 26, then the sum of the first six terms of the progression is
(a) 63
(b) 189
(c) 728
(d) 364

Answer

C

Question. Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is:
(a) 157
(b) 262
(c) 225
(d) 190

Answer

C

Question. The sum

Sequences and Series VBQs Class 11 Mathematics

is equal to :
(a) 2 – 3/217
(b) 1 – 11/220
(c) 2 – 11/219
(d) 2 – 21/220

Answer

C

Question. If a, b, c are in A.P. and a2, b2, c2 are in G.P. such that a < b < c and a + b + c = 3/4 , then the value of a is
(a) 1/4 − 1/3√2
(b) 1/4 − 1/4√2
(c) 1/4 − 1/√2
(d) 1/4 − 1/2√2

Answer

C

Question. If m is the A.M. of two distinct real numbers l and n(l, n > 1) and G1, G2 and G3 are three geometric means between and n, then G14 + 2G24 + G34 equals.
(a) 4 lmn2
(b) 4 l2m2n2
(c) 4 l2 mn
(d) 4 lm2n

Answer

Question. The least positive integer n such that 1 – 2/3 – 2/32 – …. – 2/3n-1 < 1/100 , is :
(a) 4
(b) 5
(c) 6
(d) 7

Answer

B

Question. Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal, then the last term is :
(a) 16
(b) 8
(c) 4
(d) 2

Answer

B

Question. The first two terms of a geometric progression add up to 12. the sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is 
(a) –4
(b) –12
(c) 12
(d) 4

Answer

B

Question. If b is the first term of an infinite G. P whose sum is five, then b lies in the interval.
(a) (– ∞, – 10)
(b) (10, ∞)
(c) (0, 10)
(d) (– 10, 0)

Answer

C

Question. Let An = (3/4) – (3/4)2 + (3/4)3 – … + (-1)n-1(3/4)n and Bn = 1 – An. Then, the least odd natural number p, so that Bn > An, for all n ≥ p is
(a) 5
(b) 7
(c) 11
(d) 9

Answer

B

Question. In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals
(a) √5
(b) 1/2 (√5 – 1)
(c) 1/2 (1 – √5)
(d) 1/2 √5

Answer

B

Question. The value of

Sequences and Series VBQs Class 11 Mathematics

(a) i
(b) 1
(c) – 1
(d) – i

Answer

D

Question. Let Sk = (1 + 2 + 3 + ….. + K) / 2 . If S12 + S22 + …… + S102 = (5/12)A. Then A is equal to
(a) 283
(b) 301
(c) 303
(d) 156

Answer

C

Question. If the sum of the first 15 terms of the series

Sequences and Series VBQs Class 11 Mathematics

is equal to 225 k then k is equal to :
(a) 108
(b) 27
(c) 54
(d) 9

Answer

B

Question. If the expansion in powers of x of the function 1/(1- ax)(1- bx) is a0 + a1x + a2x2 + a3x3 …… then an is
(a) (bn – an) / b – a
(b) (an – bn) / b – a
(c) (an+1 – bn+1) / b – a
(d) (bn+1 – an+1) / b – a

Answer

D

Question. Fifth term of a GP is 2, then the product of its 9 terms is
(a) 256
(b) 512
(c) 1024
(d) none of these

Answer

B

Question. If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is :
(a) 1
(b) 7/4
(c) 8/5
(d) 4/3

Answer

D

Question. Let z = 1 + ai be a complex number, a > 0, such that z3 is areal number. Then the sum 1 + z + z2 + …. + z11 is equal to:
(a) 1365√3i
(b) −1365√3i
(c) −1250√3i
(d) 1250√3i

Answer

B

Question. The sum of all values of θ ∈ (0, π/2) satisfying sin22θ + cos42θ = 3/4 is:
(a) π
(b) 5π/4
(c) π/2
(d) 3π/8

Answer

D