Class 11 Mathematics Sequences and Series MCQs Set 09

Question. \( \sum_{n=1}^{n} n(1-a)(1-2a)(1-3a)....\{1-(n-1)a\} = \)
(a) \( 1 - (1-a)(1-2a)(1-3a)....(1-na) \)
(b) \( a\left[ 1 - (1-a)(1-2a)....(1-na) \right] \)
(c) \( \frac{1}{a} \left[ 1 - (1-a)(1-2a)....(1-na) \right] \)
(d) \( \frac{1}{a} \left[ 1 - (1-a)(1-2a)(1-3a)....\{1-(n-1)a\} \right] \)
Answer: (c) \( \frac{1}{a} \left[ 1 - (1-a)(1-2a)....(1-na) \right] \)

Question. If \( \sum_{r=1}^{n} t_r = \sum_{k=1}^{n} \sum_{j=1}^{k} \sum_{i=1}^{j} 2 \), then \( \sum_{r=1}^{n} \frac{1}{t_r} = \)
(a) \( \frac{n+1}{n} \)
(b) \( \frac{n}{n+1} \)
(c) \( \frac{n-1}{n} \)
(d) \( \frac{n}{n-1} \)
Answer: (b) \( \frac{n}{n+1} \)

Question. \( S_n = \sum_{n=1}^{n} \frac{n}{1+n^2+n^4} \), then \( S_{10} \cdot S_{20} \)
(a) \( \frac{110}{111} \cdot \frac{211}{421} \)
(b) \( \frac{110}{421} \cdot \frac{111}{112} \)
(c) \( \frac{110}{111} \cdot \frac{420}{421} \)
(d) \( \frac{55}{111} \cdot \frac{210}{421} \)
Answer: (d) \( \frac{55}{111} \cdot \frac{210}{421} \)

Question. If \( b_i = 1 - a_i, na = \sum_{i=1}^{n} a_i, nb = \sum_{i=1}^{n} b_i \), then
\( \sum_{i=1}^{n} a_i b_i + \sum_{i=1}^{n} (a_i - a)^2 \)
(a) ab
(b) -nab
(c) nab
(d) \( (n+1)ab \)
Answer: (c) nab

Question. If (1 + 3 + 5 + ..... + p) + (1 + 3 + 5 + .. + q) = (1 + 3 + 5 + .... + r) where each set of parentheses contains the sum of consecutive odd integers as shown, the smallest possible value of p + q + r, (where p > 6) is
(a) 12
(b) 21
(c) 45
(d) 54
Answer: (b) 21

Question. The largest term of the sequence
\( \frac{1}{503}, \frac{4}{524}, \frac{9}{581}, \frac{16}{692}, ...... \)
(a) \( \frac{49}{16} \)
(b) \( \frac{48}{1509} \)
(c) \( \frac{49}{1529} \)
(d) \( \frac{64}{1509} \)
Answer: (c) \( \frac{49}{1529} \)

Question. Consecutive odd integers whose sum is \( 25^2 - 11^2 \) are
(a) 23, 25, 27, ...., 49
(b) 25, 27, 29, ...., 51
(c) 21, 23, 25, ...., 49
(d) 19, 21, 23, ....., 47
Answer: (a) 23, 25, 27, ...., 49

Question. If \( a_n = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} dx \), then \( \begin{vmatrix} a_1 & a_{51} & a_{101} \\ a_2 & a_{52} & a_{102} \\ a_3 & a_{53} & a_{103} \end{vmatrix} \)
(a) 1
(b) 0
(c) -1
(d) 2
Answer: (b) 0

Question. Consider the sequence 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, ..... then \( 1025^{th} \) term will be
(a) \( 2^9 \)
(b) \( 2^{11} \)
(c) \( 2^{10} \)
(d) \( 2^{12} \)
Answer: (c) \( 2^{10} \)

Question. If set of two numbers \( (\tan^{-1} x, \tan^{-1} y, \tan^{-1} z) \) and \( (x, y, z) \) are in A.P such that y does not belong to the set \( \{0, -1, 1\} \) then
(a) set \( \left\{ \frac{x}{y}, \frac{y}{z}, \frac{z}{x} \right\} \in G.P \)
(b) set of numbers \( \left\{ \frac{x}{y}, \frac{y}{z}, \frac{z}{x} \right\} \notin A.G.P \)
(c) set of numbers are not identical
(d) sum of squares of their differences taken pairwise is not equal to zero
Answer: (a) set \( \left\{ \frac{x}{y}, \frac{y}{z}, \frac{z}{x} \right\} \in G.P \)

Question. Let the sequence \( a_1, a_2, a_3, \dots, a_n \) form an A.P. \( a_1^2 - a_2^2 + a_3^2 - a_4^2 + \dots + a_{2n-1}^2 - a_{2n}^2 \) is equal to
(a) \( \frac{n}{2n-1}(a_1^2 - a_{2n}^2) \)
(b) \( \frac{1}{2n-1}(a_1^2 - a_{2n}^2) \)
(c) \( \frac{n}{n+1}(a_1^2 + a_{2n}^2) \)
(d) \( \frac{1}{2n+1}(a_1^2 - a_{2n}^2) \)
Answer: (a) \( \frac{n}{2n-1}(a_1^2 - a_{2n}^2) \)

Question. The sum to 101 terms of an A.P. is 1212. The middle term is
(a) 6
(b) 12
(c) 24
(d) 26
Answer: (b) 12

Question. If \( \log 2, \log(2^x - 1) \) and \( \log(2^x + 3) \) are in AP, then the value of \( x \) is given by
(a) \( \frac{5}{2} \)
(b) \( \log_2 5 \)
(c) \( \log_3 5 \)
(d) \( \log_5 3 \)
Answer: (b) \( \log_2 5 \)

Question. If in AP, \( a_7 = 9 \) and if \( a_1.a_2.a_7 \) is least, then common difference is
(a) \( \frac{11}{30} \)
(b) \( \frac{13}{10} \)
(c) \( \frac{32}{33} \)
(d) \( \frac{33}{20} \)
Answer: (d) \( \frac{33}{20} \)

Question. The number of common terms in two A.P's 2, 7, 12, 17, ........... 500 terms and 1, 8, 15, 22, ......... 300 terms is
(a) 58
(b) 60
(c) 61
(d) 63
Answer: (b) 60

Question. In G.P. \( (p+q)^{\text{th}} \) term is \( m \), \( (p-q)^{\text{th}} \) term is \( n \), then \( p^{\text{th}} \) term is
(a) \( nm \)
(b) \( \sqrt{nm} \)
(c) \( m/n \)
(d) \( \sqrt{m/n} \)
Answer: (b) \( \sqrt{nm} \)

Question. If \( a_1, a_2, a_3 \) are three positive consecutive terms of a GP with common ratio K. then all values of K for which the in equality \( a_3 > 4a_2 - 3a_1 \), is satisfies
(a) \( (1, 3) \)
(b) \( (-\infty, 1) \cup (3, \infty) \)
(c) \( (-\infty, \infty) \)
(d) \( (0, \infty) \)
Answer: (b) \( (-\infty, 1) \cup (3, \infty) \)

Question. The series \( \frac{2x}{x+3} + \left(\frac{2x}{x+3}\right)^2 + \left(\frac{2x}{x+3}\right)^3 + \dots \text{to } \infty \) will have a definite sum when
(a) \( -1 < x < 3 \)
(b) \( 0 < x < 1 \)
(c) \( x = 0 \)
(d) \( x > 3 \)
Answer: (a) \( -1 < x < 3 \)

Question. If a,b,c,d,x are real and the roots of equation \( (a^2+b^2+c^2)x^2 - 2(ab+bc+cd)x + (b^2+c^2+d^2) = 0 \) real and equal, then a,b,c,d are in
(a) A.P
(b) G.P
(c) H.P
(d) None of the options
Answer: (b) G.P

Question. (666.... ndigits)² + (888.....n digits) =
(a) \( \frac{4}{9}(10^n - 1) \)
(b) \( \frac{4}{9}(10^{2n} - 1) \)
(c) \( \frac{4}{9}(10^n - 1)^2 \)
(d) \( \frac{4}{9}(10^n - 1)^2 \)
Answer: (b) \( \frac{4}{9}(10^{2n} - 1) \)

Question. Let a = 111....1(55 digits), \( b = 1 + 10 + 10^2 + \dots + 10^4 \), \( c = 1 + 10^5 + 10^{10} + 10^{15} + \dots + 10^{50} \), then
(a) a = b+c
(b) a = bc
(c) b = ac
(d) c = ab
Answer: (b) a = bc

Question. The sum to infinity of \( \frac{1}{7} + \frac{2}{7^2} + \frac{1}{7^3} + \frac{2}{7^4} + \dots \) is
(a) 1/5
(b) 7/24
(c) 5/48
(d) 3/16
Answer: (d) 3/16

Question. If each term of an infinite G.P is twice the sum of the terms following it , then the common ratio of G.P is
(a) 1/2
(b) 2/3
(c) 1/3
(d) 3/2
Answer: (c) 1/3

Question. Sum of infinite No.of terms in G.P is 20 and sum of their squares is 100 , then the common ratio of G.P.is
(a) 1/5
(b) 4/5
(c) 2/5
(d) 3/5
Answer: (d) 3/5

Question. If 's' is the sum to infinite terms of a G.P. whose first term is 1, then the sum of n terms is
(a) \( s\left(1 - \left(1 - \frac{1}{s}\right)^n\right) \)
(b) \( \frac{1}{s}\left(1 - \left(1 - \frac{1}{s}\right)^n\right) \)
(c) \( 1 - \left(1 - \frac{1}{s}\right)^n \)
(d) \( 1 + \left(1 - \frac{1}{s}\right)^n \)
Answer: (a) \( s\left(1 - \left(1 - \frac{1}{s}\right)^n\right) \)

Question. If \( r > 1 \) and \( x = a + a/r + a/r^2 + \dots \), \( y = b + b/r + b/r^2 + \dots \), And \( z = c + c/r + c/r^2 + \dots \), Then value of \( xy/z^2 \) is
(a) \( ab/c^2 \)
(b) \( abr/c \)
(c) \( ab/c^2r \)
(d) \( ab/c \)
Answer: (a) \( ab/c^2 \)

Question. If the A.M. and G.M. of two numbers are 13 and 12 respectively then the two numbers are
(a) 8, 12
(b) 8, 18
(c) 10, 18
(d) 12, 18
Answer: (b) 8, 18

Question. If n!, 3(n!) and (n+1)! are in G.P., then n!, 5(n!) and (n+1)! are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) None of the options
Answer: (a) A.P.

Question. If G₁ and G₂ are two geometric means and A is the arithmetic mean inserted between two positive numbers then the value of \( \frac{G_1^2}{G_2} + \frac{G_2^2}{G_1} \) is
(a) A/2
(b) A
(c) 2A
(d) 3A
Answer: (c) 2A

Question. If \( x_i > 0, i = 1, 2, 3, \dots 50 \) and \( x_1 + x_2 + x_3 + \dots + x_{50} = 50 \) and minimum value of \( \frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \dots + \frac{1}{x_{50}} \) is \( \lambda \) then \( \lambda = \)
(a) 50
(b) 60
(c) 40
(d) 202
Answer: (a) 50

Question. If \( A_1, A_2, A_3, \dots \) belongs to A.P such that \( A_1 + A_4 + A_7 + \dots + A_{28} = 140 \) then maximum value of \( A_1 . A_2 \dots A_{28} \) is
(a) \( 2^{28} \)
(b) \( 7^{28} \)
(c) \( (14)^{28} \)
(d) \( (28)^{28} \)
Answer: (c) \( (14)^{28} \)

Question. Let a,b and c be the real numbers such that \( a + b + c = 6 \) then, the range of \( ab^2c^3 \) is
(a) \( (0, \infty) \)
(b) \( (0, 1) \)
(c) \( (0, 108] \)
(d) \( (6, 108] \)
Answer: (c) \( (0, 108] \)

Question. If none of \( b_1, b_2, \dots b_n \) is zero then \( \left(\frac{a_1}{b_1} + \frac{a_2}{b_2} + \dots + \frac{a_n}{b_n}\right)^2 \) is
(a) \( \ge (a_1^2 + a_2^2 + \dots + a_n^2)(b_1^{-2} + b_2^{-2} + \dots + b_n^{-2}) \)
(b) \( \le (a_1^2 + a_2^2 + \dots + a_n^2)(b_1^{-2} + b_2^{-2} + \dots + b_n^{-2}) \)
(c) \( > (a_1^2 + a_2^2 + \dots + a_n^2)(b_1^{-2} + b_2^{-2} + \dots + b_n^{-2}) \)
(d) \( < (a_1^2 + a_2^2 + \dots + a_n^2)(b_1^{-2} + b_2^{-2} + \dots + b_n^{-2}) \)
Answer: (b) \( \le (a_1^2 + a_2^2 + \dots + a_n^2)(b_1^{-2} + b_2^{-2} + \dots + b_n^{-2}) \)

Question. If a,b,c be the \( p^{\text{th}} \), \( q^{\text{th}} \) and \( r^{\text{th}} \) terms respectively of a G.P., then the equation \( a^q b^r c^p x^2 + pqrx + a^r b^p c^q = 0 \) has
(a) both roots zero
(b) at least one root zero
(c) no root zero
(d) both roots unity
Answer: (c) no root zero

Question. If –1 < a, b, c < 1 and a, b, c are in A.P. and \( x = \sum_{n=0}^{\infty} a^n, y = \sum_{n=0}^{\infty} b^n, z = \sum_{n=0}^{\infty} c^n \) then x, y, z are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) A.G.P
Answer: (c) H.P.

Question. If \( a_1, a_2, a_3, \dots, a_n \) are in H.P then \( \frac{a_1}{a_2 + a_3 + \dots a_n}, \frac{a_2}{a_1 + a_3 + \dots a_n}, \frac{a_3}{a_1 + a_2 + \dots a_n}, \dots, \frac{a_n}{a_1 + a_2 + \dots a_{n-1}} \)
(a) A.P.
(b) G.P.
(c) H.P.
(d) A.G.P
Answer: (c) H.P.

Question. If a, 8, b are in A.P; a, 4, b are in G.P; a, x, b are in H.P then x =
(a) 2
(b) 1
(c) 4
(d) 16
Answer: (a) 2

Question. Number of positive integral ordered pairs of \( (a,b) \) such that 6,a,b are in H.P is
(a) 5
(b) 6
(c) 7
(d) 8
Answer: (c) 7

Question. If a, b, c are in H.P, then the value of \( \frac{a+c}{a-c} \) is
(a) \( \frac{a}{a-b} \)
(b) \( \frac{a-b}{a} \)
(c) \( \frac{b}{a} \)
(d) \( \frac{a}{a+b} \)
Answer: (a) \( \frac{a}{a-b} \)

Question. If \( x > 1, y > 1, z > 1 \) are in G.P then \( \frac{1}{1 + \log x}, \frac{1}{1 + \log y}, \frac{1}{1 + \log z} \) are in
(a) AP
(b) GP
(c) HP
(d) AGP
Answer: (c) HP

Question. If \( a = \sum_{r=1}^{\infty} \frac{1}{r^2}, b = \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} \), then \( \frac{a}{b} = \)
(a) 5/4
(b) 4/3
(c) 3/4
(d) 4/5
Answer: (b) 4/3