JEE Mathematics Sequence and Series MCQs Set 06

Question. Consider an infinite geometric series with first term a and common ratio r. If the sum is 4 and the second term is 3/4, then
(a) \( a = \frac{4}{7}, r = \frac{3}{7} \)
(b) \( a = 2, r = \frac{3}{8} \)
(c) \( a = \frac{3}{2}, r = \frac{1}{2} \)
(d) \( a = 3, r = \frac{1}{4} \)
Answer: (d) \( a = 3, r = \frac{1}{4} \)

Question. If a, b, c, d are positive real numbers such that \( a + b + c + d = 2 \), then \( M = (a + b) (c + d) \) satisfies the relation:
(a) \( 0 \le M \le 1 \)
(b) \( 1 \le M \le 2 \)
(c) \( 2 \le M \le 3 \)
(d) \( 3 \le M \le 4 \)
Answer: (a) \( 0 \le M \le 1 \)

Question. The fourth power of the common difference of an arithmetic progression with integer entries added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.
Answer: Let the four consecutive terms of the A.P. be \( (a - 3d), (a - d), (a + d), (a + 3d) \). The common difference is \( 2d \).
The given expression is \( (a - 3d)(a - d)(a + d)(a + 3d) + (2d)^4 \)
\( \implies (a^2 - 9d^2)(a^2 - d^2) + 16d^4 \)
\( \implies a^4 - 10a^2d^2 + 9d^4 + 16d^4 \)
\( \implies a^4 - 10a^2d^2 + 25d^4 \)
\( \implies (a^2 - 5d^2)^2 \)
Since \( a \) and \( d \) are related to integer entries and common difference, the result is the square of an integer.

Question. Given that \( \alpha, \gamma \) are roots of the equation, \( Ax^2 - 4x + 1 = 0 \) and \( \beta, \delta \) the roots of the equation, \( Bx^2 - 6x + 1 = 0 \), find values of A and B, such that \( \alpha, \beta, \gamma \text{ & } \delta \) are in H.P.
Answer: \( A = 3, B = 8 \)

Question. The sum of roots of the equation \( ax^2 + bx + c = 0 \) is equal to the sum of squares of their reciprocals. Find whether \( bc^2, ca^2 \text{ and } ab^2 \) in A.P., G.P. or H.P. ?
Answer: \( \alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2} \)
\( \implies (\alpha + \beta) = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} \)
\( \implies (\alpha + \beta) = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} \)
\( \implies -\frac{b}{a} = \frac{(-b/a)^2 - 2c/a}{(c/a)^2} \)
\( \implies -\frac{b}{a} = \frac{b^2 - 2ac}{c^2} \)
\( \implies -bc^2 = ab^2 - 2a^2c \)
\( \implies 2ca^2 = ab^2 + bc^2 \)
This indicates that \( bc^2, ca^2, ab^2 \) are in A.P.

Question. Solve the following equations for x and y:
\[ \log_2 x + \log_4 x + \log_{16} x + \dots = y \]
\[ \frac{5 + 9 + 13 + \dots + (4y + 1)}{1 + 3 + 5 + \dots + (2y - 1)} = 4 \log_4 x \]
Answer: From the first equation:
\( \log_2 x + \frac{1}{2} \log_2 x + \frac{1}{4} \log_2 x + \dots = y \)
\( \implies (\log_2 x) \left[ 1 + \frac{1}{2} + \frac{1}{4} + \dots \right] = y \)
\( \implies (\log_2 x) \left[ \frac{1}{1 - 1/2} \right] = y \)
\( \implies 2 \log_2 x = y \implies \log_2 x = \frac{y}{2} \)
Substituting into the second equation:
\( \frac{\frac{y}{2} [5 + 4y + 1]}{\frac{y}{2} [1 + 2y - 1]} = 2 \log_2 x \)
\( \implies \frac{4y + 6}{2y} = y \)
\( \implies 2y + 3 = y^2 \)
\( \implies y^2 - 2y - 3 = 0 \)
\( \implies (y - 3)(y + 1) = 0 \)
Since \( y \in N \), \( y = 3 \).
Then \( \log_2 x = 3/2 \implies x = 2^{3/2} = 2\sqrt{2} \).
Thus, \( x = 2\sqrt{2}, y = 3 \).

Question. Let \( \alpha, \beta \) be the roots of \( x^2 - x + p = 0 \) and \( \gamma, \delta \) the roots of \( x^2 - 4x + q = 0 \). If \( \alpha, \beta, \gamma, \delta \) are in G.P., then the integral values of p and q respectively, are
(a) -2, -32
(b) -2, 3
(c) -6, 3
(d) -6, -32
Answer: (a) -2, -32

Question. If the sum of the first 2n terms of the A.P. 2, 5, 8, ... is equal to the sum of the first n terms of the A.P. 57, 59, 61, ..., then n equals
(a) 10
(b) 12
(c) 11
(d) 13
Answer: (c) 11

Question. Let the positive numbers a, b, c, d be in A.P. Then abc, abd, acd, bcd are
(a) NOT in A.P./G.P./ H.P.
(b) in A.P.
(c) in G.P.
(d) in H.P.
Answer: (d) in H.P.

Question. Let \( a_1, a_2, \dots \) be positive real numbers in G.P. For each n, let \( A_n, G_n, H_n \) be respectively, the arithmetic mean, geometric mean, and harmonic mean of \( a_1, a_2, \dots, a_n \). Find an expression for the G.M. of \( G_1, G_2, \dots, G_n \) in terms of \( A_1, A_2, \dots, A_n, H_1, H_2, \dots, H_n \).
Answer: \( \text{G.M. of } G_1, G_2, \dots, G_n = \left[ (A_1 A_2 \dots A_n)(H_1 H_2 \dots H_n) \right]^{1/2n} \)

Question. Suppose a, b, c are in A.P. and \( a^2, b^2, c^2 \) are in G.P. if \( a < b < c \) and \( a + b + c = 3/2 \), then the value of a is
(a) \( \frac{1}{2\sqrt{2}} \)
(b) \( \frac{1}{2\sqrt{3}} \)
(c) \( \frac{1}{2} - \frac{1}{\sqrt{3}} \)
(d) \( \frac{1}{2} - \frac{1}{\sqrt{2}} \)
Answer: (d) \( \frac{1}{2} - \frac{1}{\sqrt{2}} \)

Question. Let a, b be positive real numbers. If \( a, A_1, A_2, b \) are in A.P.; \( a, G_1, G_2, b \) are in G.P. and \( a, H_1, H_2, b \) are in H.P. show that \( \frac{G_1 G_2}{H_1 H_2} = \frac{A_1 + A_2}{H_1 + H_2} = \frac{(2a + b)(a + 2b)}{9ab} \).
Answer: By property of means:
\( A_1 + A_2 = a + b \)
\( G_1 G_2 = ab \)
For H.P.: \( \frac{1}{H_1} + \frac{1}{H_2} = \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab} \)
\( \implies \frac{H_1 + H_2}{H_1 H_2} = \frac{a + b}{ab} \)
\( \implies \frac{ab}{H_1 H_2} = \frac{a + b}{H_1 + H_2} \)
\( \implies \frac{G_1 G_2}{H_1 H_2} = \frac{A_1 + A_2}{H_1 + H_2} \)
Now, \( H_1 = \frac{3ab}{2b + a} \) and \( H_2 = \frac{3ab}{2a + b} \).
Substituting these values proves the second equality: \( \frac{(2a + b)(a + 2b)}{9ab} \).

Question. If a, b, c are in A.P., \( a^2, b^2, c^2 \) are in H.P., then prove that either \( a = b = c \) or \( a, b, -\frac{c}{2} \) form a G.P.
Answer: \( 2b = a + c \) and \( b^2 = \frac{2a^2c^2}{a^2 + c^2} \)
\( \implies (a^2 + c^2)b^2 = 2a^2c^2 \)
\( \implies (a^2 + c^2) \frac{(a + c)^2}{4} = 2a^2c^2 \)
\( \implies (a^2 + c^2)(a^2 + c^2 + 2ac) = 8a^2c^2 \)
Let \( a^2 + c^2 = x \) and \( ac = y \).
\( \implies x(x + 2y) = 8y^2 \)
\( \implies x^2 + 2xy - 8y^2 = 0 \)
\( \implies (x + 4y)(x - 2y) = 0 \)
Case 1: \( x - 2y = 0 \implies a^2 + c^2 - 2ac = 0 \implies a = c \). Since \( 2b = a + c \), \( a = b = c \).
Case 2: \( x + 4y = 0 \implies a^2 + c^2 + 4ac = 0 \). This lead to \( b^2 = a(-c/2) \), thus \( a, b, -c/2 \) are in G.P.

Question. The first term of an infinite geometric progression is x and its sum is 5. Then
(a) \( 0 \le x \le 10 \)
(b) \( 0 < x < 10 \)
(c) \( -10 < x < 0 \)
(d) \( x > 10 \)
Answer: (b) \( 0 < x < 10 \)

Question. If a, b, c are positive real numbers. then prove that \( [(1 + a) (1 + b) (1 + c)]^7 > 7^7 a^4 b^4 c^4 \).
Answer: Using the inequality \( \frac{1+a+b+c+ab+bc+ca+abc}{7} > (a \cdot b \cdot c \cdot ab \cdot bc \cdot ca \cdot abc)^{1/7} \)
\( \implies \frac{(1+a)(1+b)(1+c) - 1}{7} > (a^4 b^4 c^4)^{1/7} \)
\( \implies (1+a)(1+b)(1+c) > 7(a^4 b^4 c^4)^{1/7} \)
\( \implies [(1+a)(1+b)(1+c)]^7 > 7^7 a^4 b^4 c^4 \).

Question. In the quadratic equation \( ax^2 + bx + c = 0 \), If \( \Delta = b^2 - 4ac \) and \( \alpha + \beta, \alpha^2 + \beta^2, \alpha^3 + \beta^3 \), are in G.P. where \( \alpha, \beta \) are the roots of \( ax^2 + bx + c = 0 \), then
(a) \( \Delta \ne 0 \)
(b) \( b\Delta = 0 \)
(c) \( c\Delta = 0 \)
(d) \( \Delta = 0 \)
Answer: (c) \( c\Delta = 0 \)

Question. If total number of runs scored in n matches is \( \left( \frac{n + 1}{4} \right) (2^{n+1} - n - 2) \) where \( n > 1 \), and the runs scored in the \( k^{th} \) match are given by \( k \cdot 2^{n+1-k} \), where \( 1 \le k \le n \). Find n.
Answer: Total runs \( S_n = \sum_{k=1}^n k \cdot 2^{n+1-k} = 2^{n+1} \sum_{k=1}^n \frac{k}{2^k} \)
Using the sum of an A.G.P. series:
\( \sum_{k=1}^n \frac{k}{2^k} = 2 - \frac{n+2}{2^n} \)
\( \implies S_n = 2^{n+1} \left[ 2 - \frac{n+2}{2^n} \right] = 2^{n+2} - 2(n+2) = 4 \cdot 2^n - 2n - 4 \)
Equating to the given expression \( \left( \frac{n + 1}{4} \right) (2^{n+1} - n - 2) \):
\( \implies n = 7 \)

Question. If \( A_n = \frac{3}{4} - \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 + \dots + (-1)^{n - 1} \left( \frac{3}{4} \right)^n \) and \( B_n = 1 - A_n \), then find the minimum natural number \( n_0 \) such that \( B_n > A_n \forall n > n_0 \).
Answer: \( A_n \) is a G.P. sum: \( A_n = \frac{3/4 [1 - (-3/4)^n]}{1 - (-3/4)} = \frac{3}{7} [1 - (-3/4)^n] \)
Condition \( B_n > A_n \implies 1 - A_n > A_n \implies 2A_n < 1 \)
\( \implies \frac{6}{7} [1 - (-3/4)^n] < 1 \)
\( \implies 1 - (-3/4)^n < 7/6 \)
\( \implies -1/6 < (-3/4)^n \)
This is satisfied for \( n \ge 6, 7, 8, \dots \).
Thus, the minimum natural number \( n_0 = 5 \).

Question. Let \( V_r \) denote the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r - 1). Let \( T_r = V_{r + 1} - V_r - 2 \) and \( Q_r = T_{r + 1} - T_r \) for \( r = 1, 2, \dots \).
(a) The sum \( V_1 + V_2 + \dots + V_n \) is
(a) \( \frac{1}{12} n(n+1)(3n^2 - n + 1) \)
(b) \( \frac{1}{12} n(n+1)(3n^2 + n + 2) \)
(c) \( \frac{1}{2} n(2n^2 - n + 1) \)
(d) \( \frac{1}{3} (2n^3 - 2n + 3) \)
Answer: (b) \( \frac{1}{12} n(n+1)(3n^2 + n + 2) \)

Question. [Based on the above passage] \( T_r \) is always
(a) an odd number
(b) an even number
(c) a prime number
(d) a composite number
Answer: (d) a composite number

Question. [Based on the above passage] Which one of the following is a correct statement?
(a) \( Q_1, Q_2, Q_3, \dots \) are in A.P., with common difference 5
(b) \( Q_1, Q_2, Q_3, \dots \) are in A.P., with common difference 6
(c) \( Q_1, Q_2, Q_3, \dots \) are in A.P., with common difference 11
(d) \( Q_1 = Q_2 = Q_3 = \dots \)
Answer: (b) \( Q_1, Q_2, Q_3, \dots \) are in A.P., with common difference 6

Question. Let \( A_1, G_1, H_1 \) denote the arithmetic, geometric and harmonic means, respectively, of two distinct positive numbers. For \( n \ge 2 \), Let \( A_{n - 1} \) and \( H_{n - 1} \) have arithmetic, geometric and harmonic means as \( A_n, G_n, H_n \) respectively.
(a) Which one of the following statements is correct?
(a) \( G_1 > G_2 > G_3 > \dots \)
(b) \( G_1 < G_2 < G_3 < \dots \)
(c) \( G_1 = G_2 = G_3 = \dots \)
(d) \( G_1 < G_3 < G_5 < \dots \text{ and } G_2 > G_4 > G_6 > \dots \)
Answer: (c) \( G_1 = G_2 = G_3 = \dots \)

Question. [Based on the above passage] Which one of the following statement is correct?
(a) \( A_1 > A_2 > A_3 > \dots \)
(b) \( A_1 < A_2 < A_3 < \dots \)
(c) \( A_1 > A_3 > A_5 > \dots \text{ and } A_2 < A_4 < A_6 < \dots \)
(d) \( A_1 < A_3 < A_5 < \dots \text{ and } A_2 > A_4 > A_6 > \dots \)
Answer: (a) \( A_1 > A_2 > A_3 > \dots \)

Question. [Based on the above passage] Which one of the following statement is correct?
(a) \( H_1 > H_2 > H_3 > \dots \)
(b) \( H_1 < H_2 < H_3 < \dots \)
(c) \( H_1 > H_3 > H_5 > \dots \text{ and } H_2 < H_4 < H_6 < \dots \)
(d) \( H_1 < H_3 < H_5 < \dots \text{ and } H_2 > H_4 > H_6 > \dots \)
Answer: (b) \( H_1 < H_2 < H_3 < \dots \)

Question. A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle, then
(a) \( \frac{1}{PS} + \frac{1}{ST} < \frac{2}{\sqrt{QS \times SR}} \)
(b) \( \frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{QS \times SR}} \)
(c) \( \frac{1}{PS} + \frac{1}{ST} < \frac{4}{QR} \)
(d) \( \frac{1}{PS} + \frac{1}{ST} > \frac{4}{QR} \)
Answer: (b) \( \frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{QS \times SR}} \)

Question. Suppose four distinct positive numbers \( a_1, a_2, a_3, a_4 \) are in G.P. Let \( b_1 = a_1, b_2 = b_1 + a_2, b_3 = b_2 + a_3 \) and \( b_4 = b_3 + a_4 \).
Statement-1: The numbers \( b_1, b_2, b_3, b_4 \) are neither in A.P. nor in G.P. and
Statement-2: The numbers \( b_1, b_2, b_3, b_4 \) are in H.P.
(a) Statement (1) is true and statement (2) is true and statement (2) is correct explanation for (1)
(b) Statement (1) is true and statement (2) is true and statement (2) is NOT correct explanation for (1)
(c) Statement (1) is true but (2) is false
(d) Statement (1) is false but (2) is true
Answer: (c) Statement (1) is true but (2) is false

Question. If the sum of first n terms of an A.P. is \( cn^2 \), then the sum of squares of these n terms is
(a) \( \frac{n(4n^2 - 1)c^2}{6} \)
(b) \( \frac{n(4n^2 + 1)c^2}{3} \)
(c) \( \frac{n(4n^2 - 1)c^2}{3} \)
(d) \( \frac{n(4n^2 + 1)c^2}{6} \)
Answer: (c) \( \frac{n(4n^2 - 1)c^2}{3} \)

Question. Let \( S_k, K = 1, 2, \dots, 100 \) denote the sum of the infinite geometric series whose first term is \( \frac{k - 1}{k!} \) and the common ratio is 1/k. Then the value of \( \frac{100^2}{100!} + \sum_{k=1}^{100} (k^2 - 3k + 1) S_k \) is
Answer: \( S_k = \frac{(k-1)/k!}{1 - 1/k} = \frac{1}{(k-1)!} \)
The sum is \( \sum_{k=1}^{100} (k^2 - 3k + 1) \frac{1}{(k-1)!} \)
\( \implies \sum \frac{(k-1)^2 - k}{(k-1)!} = \sum \left[ \frac{k-1}{(k-2)!} - \frac{k}{(k-1)!} \right] \)
This is a telescoping sum. Sum \( = 3 - \frac{100}{99!} \).
Thus the expression \( \frac{100^2}{100!} + 3 - \frac{100}{99!} = \frac{100^2}{100!} + 3 - \frac{100^2}{100!} = 3 \).

Question. Let \( a_1, a_2, a_3, \dots, a_{11} \) be real numbers satisfying \( a_1 = 15, 27 - 2a_2 > 0 \) and \( a_k = 2a_{k-1} - a_{k-2} \) for \( k = 3, 4, \dots, 11 \). If \( \frac{a_1^2 + a_2^2 + \dots + a_{11}^2}{11} = 90 \), then the value of \( \frac{a_1 + a_2 + \dots + a_{11}}{11} \) is equal to
Answer: \( a_k \) terms are in A.P. Let common difference be d.
Mean square \( \frac{1}{11} \sum_{i=0}^{10} (15 + id)^2 = 90 \)
\( \implies \frac{1}{11} [11(15^2) + 2(15)d \frac{10 \times 11}{2} + d^2 \frac{10 \times 11 \times 21}{6}] = 90 \)
\( \implies 225 + 150d + 35d^2 = 90 \)
\( \implies 35d^2 + 150d + 135 = 0 \)
\( \implies 7d^2 + 30d + 27 = 0 \)
\( \implies (7d + 9)(d + 3) = 0 \)
Since \( a_2 < 13.5 \), \( d = -3 \).
AM of terms \( = \frac{11/2 [2(15) + 10(-3)]}{11} = 0 \).

Question. The minimum value of the sum of real numbers \( a^{-5}, a^{-4}, 3a^{-3}, 1, a^8 \text{ and } a^{10} \) with \( a > 0 \) is
Answer: Using AM \( \ge \) GM for terms \( \{ a^{-5}, a^{-4}, a^{-3}, a^{-3}, a^{-3}, 1, a^8, a^{10} \} \)
\( \frac{\text{Sum}}{8} \ge (a^{-5-4-3-3-3+0+8+10})^{1/8} = (a^0)^{1/8} = 1 \)
\( \implies \text{Sum} \ge 8 \).

Question. Let \( a_1, a_2, a_3, \dots, a_{100} \) be an arithmetic progression with \( a_1 = 3 \) and \( S_p = \sum_{i = 1}^p a_i, 1 \le p \le 100 \). For any integer n with \( 1 \le n \le 20 \). let m=5n. If \( \frac{S_m}{S_n} \) does not depend on n, then \( a_2 \) is
Answer: \( \frac{S_{5n}}{S_n} = \frac{5n/2 [2(3) + (5n-1)d]}{n/2 [2(3) + (n-1)d]} = 5 \frac{6-d+5nd}{6-d+nd} \)
For this to be independent of n, \( 6 - d = 0 \implies d = 6 \).
Then \( a_2 = a_1 + d = 3 + 6 = 9 \).

Question. Let \( a_1, a_2, a_3, \dots \) be in harmonic progression with \( a_1 = 5 \) and \( a_{20} = 25 \). The least positive integer n for which \( a_n < 0 \) is
(a) 22
(b) 23
(c) 24
(d) 25
Answer: (d) 25