JEE Mathematics Sequence and Series MCQs Set E

Question. The first term of an A.P. of consecutive integer is \( p^2 + 1 \). The sum of \( (2p + 1) \) terms of this series can be expressed as
(a) \( (p + 1)^2 \)
(b) \( (2p + 1) (p + 1)^2 \)
(c) \( (p + 1)^3 \)
(d) \( p^3 + (p + 1)^3 \)
Answer: (d) \( p^3 + (p + 1)^3 \)
Solution:
a = \( P^2 + 1 \), d = 1, n = 2P + 1
\( S_{2P+1} = \frac{2P + 1}{2} [2(P^2 + 1) + (2P + 1 - 1) 1] \)
\( = \frac{(2P + 1)}{2} [2P^2 + 2 + 2P] = (2P + 1) (P^2 + P + 1) \)
\( = 2P^3 + 3P^2 + 3P + 1 = P^3 + (P + 1)^3 \)

Question. If \( a_1, a_2, a_3, \dots \) are in A.P. such that \( a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 225 \), then \( a_1 + a_2 + a_3 + \dots + a_{23} + a_{24} \) is equal to
(a) 909
(b) 75
(c) 750
(d) 900
Answer: (d) 900
Solution:
\( a_1, a_2, a_3, \dots \) in A.P.
\( a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 225 \)
\( 6a_1 + (74 - 5) d = 225 \)
\( 3(2a_1 + 23d) = 225 \Rightarrow (2a_1 + 23d) = \frac{225}{3} \)
\( a_1 + a_2 + \dots + a_{24} = \frac{24}{2} [2a_1 + 23d] \)
\( = 12 \times \frac{225}{3} = 900 \)

Question. The sum of integers from 1 to 100 that are divisible by 2 or 5 is
(a) 2550
(b) 1050
(c) 3050
(d) None of the options
Answer: (c) 3050
Solution:
Sum of divisible by 2
\( S_2 = \frac{50}{2} [2 + 100] = 25 \times 102 = 2550 \)
Sum of divisible by 5
\( S_5 = \frac{20}{2} [5 + 100] = 10 \times 105 = 1050 \)
Sum of divisible by 2 & 5
\( S_{2,5} = \frac{10}{2} [10 + 100] = 5 \times 110 = 550 \)
required sum = 2550 + 1050 - 550 = 3050

Question. Consider an A.P. with first term 'a' and the common difference 'd'. Let \( S_k \) denote the sum of its first K terms. If \( \frac{S_{kx}}{S_x} \) is independent of x, then
(a) a = d/2
(b) a = d
(c) a = 2d
(d) None of the options
Answer: (a) a = d/2
Solution:
\( S_k = \frac{k}{2} [2a + (k - 1) d] \), \( S_x = \frac{x}{2} [2a + (x - 1)d] \)
& \( S_{kx} = \frac{kx}{2} [2a + (kx - 1)d] \)
\( \frac{S_{kx}}{S_x} = \frac{k[2a + (kx - 1)d]}{[2a + (x - 1)d]} = \frac{k[(2a - d) + kxd]}{[(2a - d) + xd]} \)
= Independent of x if
2a - d = 0 \( \Rightarrow a = \frac{d}{2} \)

Question. If \( x \in R \), the numbers \( 5^{1 + x} + 5^{1 - x}, a/2, 25^x + 25^{-x} \) form an A.P. then 'a' must lie in the interval;
(a) [1, 5]
(b) [2, 5]
(c) [5, 12]
(d) [12, \( \infty \))
Answer: (d) [12, \( \infty \))
Solution:
\( x \in R, (5^{1+x} + 5^{1-x}), \frac{a}{2}, 25^x + 25^{-x} \)
\( 2 \cdot \frac{a}{2} = (5^{1+x} + 5^{1-x}) + (25^x + 25^{-x}) \)
\( \Rightarrow a = 5 (5^x + 5^{-x}) + (5^{2x} + 5^{-2x}) \)
\( \because 5^x + 5^{-x} \ge 2 \) & \( 5^{2x} + 5^{-2x} \ge 2 \)
\( \Rightarrow a \ge 5.2 + 2 \Rightarrow a \ge 12 \)

Question. There are n A.M.'s between 3 and 54, such that the 8th mean : \( (n - 2)^{th} \) mean :: 3 : 5. The value of n is.
(a) 12
(b) 16
(c) 18
(d) 20
Answer: (b) 16
Solution:
3, \( A_1, A_2 = \dots, A_n \), 54, & \( \frac{A_8}{A_{n-2}} = \frac{3}{5} \)
\( d = \frac{54 - 3}{n + 1} = \frac{51}{n + 1} \)
\( A_8 = 3 + 8d \) & \( A_{n-2} = 3 + (n - 2) d \)
\( \frac{A_8}{A_{n-2}} = \frac{3 + 8d}{3 + (n - 2)d} = \frac{3}{5} \)
\( \Rightarrow \frac{3 + 8\left(\frac{51}{n + 1}\right)}{3 + (n - 2)\left(\frac{51}{n + 1}\right)} = \frac{3}{5} \)
\( \Rightarrow \frac{n + 1 + 136}{n + 1 + (n - 2)17} = \frac{3}{5} \)
\( \Rightarrow 5(n + 137) = 3 (18n - 33) \Rightarrow 784 = 49n \)
\( \Rightarrow n = \frac{784}{49} = \frac{112}{7} \Rightarrow n = 16 \)

Question. The third term of a G.P. is 4. The product of the first five terms is
(a) \( 4^3 \)
(b) \( 4^5 \)
(c) \( 4^4 \)
(d) None of the options
Answer: (b) \( 4^5 \)
Solution:
\( ar^2 = 4 \)
\( a \cdot ar \cdot ar^2 \cdot ar^3 \cdot ar^4 = a^5 r^{10} = (ar^2)^5 = 4^5 \)

Question. If S is the sum of infinity of a G.P. whose first term is 'a', then the sum of the first n terms is
(a) \( S \left(1 - \frac{a}{S}\right)^n \)
(b) \( S \left[1 - \left(1 - \frac{a}{S}\right)^n\right] \)
(c) \( a \left[1 - \left(1 - \frac{a}{S}\right)^n\right] \)
(d) None of the options
Answer: (b) \( S \left[1 - \left(1 - \frac{a}{S}\right)^n\right] \)
Solution:
Sum of G.P. is possible \( \Rightarrow |r| < 1 \)
\( S = \frac{a}{1 - r} \Rightarrow r = \left(1 - \frac{a}{S}\right) \)
\( S_n = \frac{a(1 - r^n)}{1 - r} = \frac{a \left[1 - \left(1 - \frac{a}{S}\right)^n\right]}{\frac{a}{S}} = S \left[1 - \left(1 - \frac{a}{S}\right)^n\right] \)

Question. The sum of the series \( \frac{1}{\log_2 4} + \frac{1}{\log_4 4} + \frac{1}{\log_8 4} + \dots + \frac{1}{\log_{2^n} 4} \) is
(a) \( \frac{1}{2} n (n + 1) \)
(b) \( \frac{1}{12} n (n + 1) (2n + 1) \)
(c) \( \frac{1}{n(n + 1)} \)
(d) \( \frac{1}{4} n (n + 1) \)
Answer: (d) \( \frac{1}{4} n (n + 1) \)
Solution:
\( \frac{1}{\log_2 4} + \frac{1}{\log_4 4} + \frac{1}{\log_8 4} + \dots + \frac{1}{\log_{2^n} 4} \)
\( = \log_4 2 + \log_4 2^2 + \log_4 2^3 + \dots + \log_4 2^n \)
\( = (\log_4 2) [1 + 2 + 3 + \dots + n] \)
\( = \frac{1}{2} \times \frac{n(n + 1)}{2} = \frac{n(n + 1)}{4} \)

Question. For a sequence \( \{a_n\} \), \( a_1 = 2 \) and \( \frac{a_{n+1}}{a_n} = \frac{1}{3} \). Then \( \sum_{r=1}^{20} a_r \) is
(a) \( \frac{20}{2} [4 + 19 \times 3] \)
(b) \( 3 \left(1 - \frac{1}{3^{20}}\right) \)
(c) \( 2 (1 - 3^{20}) \)
(d) None of the options
Answer: (b) \( 3 \left(1 - \frac{1}{3^{20}}\right) \)
Solution:
\( a_1 = 2 \), & \( \frac{a_{n+1}}{a_n} = \frac{1}{3} = r \)
\( \sum_{r=1}^{20} a_r = \frac{a_1 (1 - r^{20})}{1 - r} = \frac{2 \left[1 - \left(\frac{1}{3}\right)^{20}\right]}{\frac{2}{3}} = 3 \left(1 - \frac{1}{3^{20}}\right) \)

Question. \( \alpha, \beta \) be the roots of the equation \( x^2 - 3x + a = 0 \) and \( \gamma, \delta \) the roots of \( x^2 - 12x + b = 0 \) and numbers \( \alpha, \beta, \gamma, \delta \) (in this order) form an increasing G.P., then
(a) a = 3, b = 12
(b) a = 12, b = 3
(c) a = 2, b = 32
(d) a = 4, b = 16
Answer: (c) a = 2, b = 32
Solution:
\( x^2 - 3x + a = 0 \) roots \( \alpha, \beta \)
\( x^2 - 12x + b = 0 \) roots \( \gamma, \delta \)
\( \alpha, \beta, \gamma, \delta \) in increasing G.P.
Let \( \alpha, \alpha r, \alpha r^2, \alpha r^3 \)
\( \alpha + \alpha r = 3 \) & \( \alpha r^2 + \alpha r^3 = 12 \)
\( \alpha(1 + r) = 3 \)
\( \alpha r^2 (1 + r) = 12 \)
\( r^2 \cdot 3 = 12 \)
\( \therefore \alpha = \frac{3}{1 + r} \), \( r^2 = 4 \)
\( \alpha = \frac{3}{3} \Rightarrow \alpha = 1 \), \( r = 2 \), (\( r = -2 \) reject)
\( \because \) G.P. increasing
\( \alpha(\alpha r) = a \) & \( (\alpha r^2)(\alpha r^3) = b \)
a = \( 1 \cdot 2 \), b = \( \alpha^2 r^5 \)
a = 2, b = \( 1 \cdot 2^5 = 32 \)

Question. If \( 3 + \frac{1}{4}(3 + d) + \frac{1}{4^2}(3 + 2d) + \dots + \text{upto } \infty = 8 \), then the value of d is
(a) 9
(b) 5
(c) 1
(d) None of the options
Answer: (a) 9
Solution:
\( 3 + \frac{1}{4}(3 + d) + \frac{1}{4^2}(3 + 2d) + \dots + \infty = 8 \)
a = 3, \( r = \frac{1}{4} \)
(A.G.P.)
\( S_\infty = \frac{a}{1 - r} + \frac{dr}{(1 - r)^2} \)
\( \Rightarrow 8 = \frac{3}{(3 / 4)} + \frac{d\left(\frac{1}{4}\right)}{3^2 / 4^2} \Rightarrow 8 = 4 + \frac{4d}{3^2} \)
\( \Rightarrow 4 = \frac{4d}{3^2} \Rightarrow d = 3^2 \Rightarrow d = 9 \)

Question. If A, G & H are respectively te A.M., G.M. & H.M. of three positive numbers a, b, & c then the equation whose roots are a, b & c is given by
(a) \( x^3 - 3Ax^2 + 3G^3x - G^3 = 0 \)
(b) \( x^3 - 3Ax^2 + 3(G^3/H)x - G^3 = 0 \)
(c) \( x^3 + 3Ax^2 + 3(G^3/H)x - G^3 = 0 \)
(d) \( x^3 - 3Ax^2 - 3(G^3/H)x + G^3 = 0 \)
Answer: (b) \( x^3 - 3Ax^2 + 3(G^3/H)x - G^3 = 0 \)
Solution:
\( A = \frac{a + b + c}{3} \), \( G = (a b c)^{1/3} \), \( H = \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \)
\( H = \frac{3(abc)}{ab + bc + ca} \)
\( x^3 - (a + b + c) x^2 + (ab + bc + ca) x - abc = 0 \)
\( \Rightarrow x^3 - 3Ax^2 + (3G^3/H)x - G^3 = 0 \)

Question. If \( a^x = b^y = c^z = d^t \) and a, b, c, d are in G.P., then x, y, z, t are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) None of the options
Answer: (c) H.P.
Solution:
\( a^x = b^y = c^z = d^t = k \) (let)
\( x \log a = \log k, y \log b = \log k \)
\( z \log c = \log k, t \log d = \log k \)
a, b, c, d in G.P.
\( \Rightarrow \log a, \log b, \log c, \log d \) in A.P.
\( \Rightarrow \frac{\log k}{x}, \frac{\log k}{y}, \frac{\log k}{z}, \frac{\log k}{t} \) in A.P.
\( \Rightarrow \frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{t} \) in A.P. \( \Rightarrow \) x, y, z, t in H.P.

Question. The sum \( \sum_{r=2}^\infty \frac{1}{r^2 - 1} \) is equal to
(a) 1
(b) 3/4
(c) 4/3
(d) None of the options
Answer: (b) 3/4
Solution:
\( \sum_{r=2}^\infty \frac{1}{r^2 - 1} = \sum_{r=2}^\infty \frac{1}{(r - 1)(r + 1)} = \frac{1}{2} \sum_{r=2}^\infty \left( \frac{1}{r - 1} - \frac{1}{r + 1} \right) \)
\( = \frac{1}{2} \left[ 1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \dots + \frac{1}{n - 1} - \frac{1}{n + 1} + \dots \right] \)
\( n \to \infty \Rightarrow \frac{1}{n + 1} \to 0 \)
\( = \frac{1}{2} \left[ 1 + \frac{1}{2} \right] = \frac{3}{4} \)

Question. If \( x_i > 0, i = 1, 2, \dots, 50 \) ans \( x_1 + x_2 + \dots + x_{50} = 50 \), then the minimum value of \( \frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_{50}} \) equal to
(a) 50
(b) \( (50)^2 \)
(c) \( (50)^3 \)
(d) \( (50)^4 \)
Answer: (a) 50
Solution:
\( x_i > 0, i = 1, 2, \dots, 50 \)
\( x_1 + x_2 + x_3 + \dots + x_{50} = 50 \)
or \( \sum_1^{50} x_i = 50 \Rightarrow \frac{\Sigma x_i}{50} = 1 \)
A.M. \( \ge \) H.M.
\( \frac{\left( \sum_1^{50} x_i \right)}{50} \ge \frac{50}{\left( \sum_1^{50} \frac{1}{x_i} \right)} \Rightarrow 1 \ge \frac{50}{\left( \sum_1^{50} \frac{1}{x_i} \right)} \)
\( \Rightarrow \sum_1^{50} \frac{1}{x_i} \ge 50 \)
Min value of \( \Sigma \frac{1}{x_i} = 50 \)

Question. If a, \( a_1, a_2, a_3, \dots, a_{2n}, b \) are in A.P. and a, \( g_1, g_2, g_3, \dots, g_{2n}, b \) are in G.P. and h is the harmonic mean of a and b, then \( \frac{a_1 + a_{2n}}{g_1 g_{2n}} + \frac{a_2 + a_{2n-1}}{g_2 g_{2n-1}} + \dots + \frac{a_n + a_{n+1}}{g_n g_{n+1}} \), is equal to
(a) 2n/h
(b) 2nh
(c) nh
(d) n/h
Answer: (a) 2n/h
Solution:
a, \( a_1, a_2, a_3, \dots, a_{2n}, b \) in A.P.
& a, \( g_1, g_2, g_3, \dots, g_{2n}, b \) in G.P.
\( (a + b) = (a_1 + a_{2n}) = (a_2 + a_{2n-1}) = \dots \)
& \( ab = g_1 g_{2n} = g_2 g_{2n-1} = \dots \)
\( \therefore \frac{a_1 + a_{2n}}{g_1 g_{2n}} + \frac{a_2 + a_{2n-1}}{g_2 g_{2n-1}} + \dots + \frac{a_n + a_{n+1}}{g_n g_{n+1}} \)
\( = \left( \frac{a + b}{ab} \right) + \left( \frac{a + b}{ab} \right) + \dots + \left( \frac{a + b}{ab} \right) \) (n times)
\( \left\{ h = \frac{2ab}{a + b} \right\} \Rightarrow \frac{a + b}{ab} = \frac{2}{h} \)
\( = n \left( \frac{a + b}{ab} \right) = n \cdot \frac{2}{h} = \frac{2n}{h} \)

Question. One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points are in turn joined to form still another triangle. This process continues indefinitely. Then the sum of the perimeters of all the trianlges is
(a) 144 cm
(b) 212 cm
(c) 288 cm
(d) None of the options
Answer: (a) 144 cm
Solution:
Perimeter \( P_1 = 3 \times 24 \) cm
\( P_2 = 3 \times \frac{24}{2} \) cm
\( P_3 = 3 \times \frac{24}{2^2} \) cm
\( S_\infty = P_1 + P_2 + P_3 + \dots \infty \)
\( = 3 \times 24 \left[ 1 + \frac{1}{2} + \frac{1}{2^2} + \dots \right] \)
\( = 3 \times 24 \times 2 = 144 \) cm

Question. In a G.P. of positive terms, any term is equal to the sum of the next two terms. The common ratio of the G.P. is
(a) \( 2 \cos 18^\circ \)
(b) \( \sin 18^\circ \)
(c) \( \cos 18^\circ \)
(d) \( 2 \sin 18^\circ \)
Answer: (d) \( 2 \sin 18^\circ \)
Solution:
\( a_k = a_{k+1} + a_{k+2} \quad \forall a_k > 0 \)
\( \Rightarrow a r^{k-1} = a r^k + a r^{k+1} \quad \Rightarrow r > 0 \)
\( \Rightarrow 1 = r + r^2 \quad \Rightarrow r^2 + r - 1 = 0 \)
\( \Rightarrow r = \frac{-1 \pm \sqrt{1 + 4}}{2} \quad \) {r = -ve reject}
\( \Rightarrow r = \frac{\sqrt{5} - 1}{2} = 2 \left( \frac{\sqrt{5} - 1}{4} \right) = 2 \sin 18^\circ \)

Question. If \( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots \text{upto } \infty = \frac{\pi^2}{6} \), then \( \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \dots = \)
(a) \( \pi^2 / 12 \)
(b) \( \pi^2 / 24 \)
(c) \( \pi^2 / 8 \)
(d) None of the options
Answer: (c) \( \pi^2 / 8 \)
Solution:
\( S = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots \infty = \frac{\pi^2}{6} \)
Now \( S_{even} = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots \infty \)
\( = \frac{1}{2^2} \left[ \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots \infty \right] = \frac{1}{2^2} \frac{\pi^2}{6} = \frac{\pi^2}{24} \)
\( S_{odd} = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \dots \infty \)
\( = S - S_{even} \)
\( = \frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{3\pi^2}{24} = \frac{\pi^2}{8} \)

Question. If \( a_1, a_2 \dots a_n \) are in A.P. with common difference \( d \neq 0 \), then the sum of the series
\( (\sin d) [\csc a_1 \csc a_2 + \csc a_2 \csc a_3 + \dots + \csc a_{n-1} \csc a_n] \)
(a) \( \sec a_1 - \sec a_n \)
(b) \( \csc a_1 - \csc a_n \)
(c) \( \cot a_1 - \cot a_n \)
(d) \( \tan a_1 - \tan a_n \)
Answer: (c) \( \cot a_1 - \cot a_n \)
Solution:
d = \( (a_2 - a_1) = (a_3 - a_2) = \dots = (a_n - a_{n-1} \neq 0) \)
\( \sin d [\csc a_1 \csc a_2 + \csc a_2 \csc a_3 + \dots + \csc a_{n-1} \csc a_n] \)
\( = \frac{\sin d}{\sin a_1 \sin a_2} + \frac{\sin d}{\sin a_2 \sin a_3} + \dots + \frac{\sin d}{\sin a_{n-1} \sin a_n} \)
\( = \sum_{r=2}^n \frac{\sin (a_r - a_{r-1})}{\sin a_{r-1} \sin a_r} \)
\( = \sum_{r=2}^n \left( \frac{\sin a_r \cos a_{r-1}}{\sin a_{r-1} \sin a_r} - \frac{\cos a_r \sin a_{r-1}}{\sin a_{r-1} \sin a_r} \right) \)
\( = \sum_{r=2}^n (\cot a_{r-1} - \cot a_r) = \cot a_1 - \cot a_n \)

Question. Sum of the series
\( S = 1^2 - 2^2 + 3^2 - 4^2 + \dots - 2002^2 + 2003^2 \) is
(a) 2007006
(b) 1005004
(c) 2000506
(d) None of the options
Answer: (a) 2007006
Solution:
\( S = 1^2 - 2^2 + 3^2 - 4^2 + \dots - 2002^2 + 2003^2 \)
\( = 1 + (3^2 - 2^2) + (5^2 - 4^2) + \dots + (2003^2 - 2002^2) \)
\( = 1 + 2 + 3 + 4 + 5 + \dots + 2002 + 2003 \)
\( = \frac{2003}{2} [1 + 2003] = 2003 (1002) \)
\( = (2000 + 3) (1000 + 2) = 2007006 \)

Question. If \( H_n = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \), then value of
\( 1 + \frac{3}{2} + \frac{5}{3} + \dots + \frac{2n - 1}{n} \) is
(a) \( 2n - H_n \)
(b) \( 2n + H_n \)
(c) \( H_n - 2n \)
(d) \( H_n + n \)
Answer: (a) \( 2n - H_n \)
Solution:
\( H_n = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \)
\( 1 + \frac{3}{2} + \frac{5}{3} + \dots + \left( \frac{2n - 1}{n} \right) \)
\( = \sum_{n=1}^n \left( \frac{2n - 1}{n} \right) = \sum_{n=1}^n \left( 2 - \frac{1}{n} \right) = \sum_{n=1}^n 2 - \sum_{n=1}^n \frac{1}{n} \)
\( = 2n - \left( 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \right) = 2n - H_n \)

Question. If \( S_1, S_2, S_3 \) are the sums of first n natural numbers, their squares, their cubes respectively, then
\( \frac{S_3(1 + 8S_1)}{S_2^2} \) is equal to
(a) 1
(b) 3
(c) 9
(d) 10
Answer: (c) 9
Solution:
\( S_1 = \sum_{1}^n n \), \( S_2 = \sum_{1}^n n^2 \), \( S_3 = \sum_{1}^n n^3 \)
\( S_1 = \frac{n(n + 1)}{2} \), \( S_2 = \frac{n(n + 1)(2n + 1)}{6} \), \( S_3 = \frac{n^2(n + 1)^2}{4} \)
\( \therefore \frac{S_3(1 + 8S_1)}{S_2^2} = \frac{\frac{n^2(n + 1)^2}{4} (1 + 4(n + 1)n)}{\left( \frac{n(n + 1)(2n + 1)}{6} \right)^2} \)
\( = \frac{9(1 + 4n^2 + 4n)}{(2n + 1)^2} = \frac{9(2n + 1)^2}{(2n + 1)^2} = 9 \)

Question. If \( a_1, a_2, \dots, a_n \) are in HP, then the expression \( a_1a_2 + a_2a_3 + \dots + a_{n-1} a_n \) is equal to
(a) \( (n - 1) (a_1 - a_n) \)
(b) \( n a_1 a_n \)
(c) \( (n - 1) a_1 a_n \)
(d) \( n (a_1 - a_n) \)
Answer: (c) \( (n - 1) a_1 a_n \)
Solution:
\( a_1, a_2, \dots, a_n \) in H.P.
\( \frac{1}{a_1}, \frac{1}{a_2}, \dots, \frac{1}{a_n} \) in A.P.
\( \therefore d = \frac{a_1 - a_2}{a_1 a_2} = \frac{a_2 - a_3}{a_2 a_3} = \dots = \frac{a_{n-1} - a_n}{a_{n-1} a_n} \)
\( a_1a_2 + a_2a_3 + \dots + a_{n-1} a_n \)
\( = \frac{1}{d} [ (a_1 - a_2) + (a_2 - a_3) + \dots + (a_{n-1} - a_n) ] \)
\( = \frac{a_1 - a_n}{d} \)
\( \left\{ \left( \frac{1}{a_n} \right) = \left( \frac{1}{a_1} \right) + (n - 1)d \right\} \)
\( \Rightarrow \frac{a_1 - a_n}{a_1 a_n} = (n - 1)d \Rightarrow \frac{(a_1 - a_n)}{d} = (n - 1) a_1 a_n \)

Question. If \( x = \sum_{n=0}^\infty a^n \), \( y = \sum_{n=0}^\infty b^n \), \( z = \sum_{n=0}^\infty c^n \) where a, b, c are in AP and \( |a| < 1, |b| < 1, |c| < 1 \), then x, y, z are in
(a) HP
(b) Arithmetic-Geometric Progression
(c) AP
(d) GP
Answer: (a) HP
Solution:
\( x = \sum_{n=0}^\infty a^n \), \( y = \sum_{n=0}^\infty b^n \), \( z = \sum_{n=0}^\infty c^n \) \( \left\{ \begin{array}{l} |a| < 1 \\ |b| < 1 \\ |c| < 1 \end{array} \right\} \)
\( x = \frac{1}{1 - a} \), \( y = \frac{1}{1 - b} \), \( z = \frac{1}{1 - c} \)
a, b, c in A.P.
1 - a, 1 - b, 1 - c in A.P.
\( \frac{1}{1 - a}, \frac{1}{1 - b}, \frac{1}{1 - c} \) in H.P. \( \Rightarrow \) x, y, z in H.P.