JEE Main · 2005 · 08 May 2005 · Offline · Sequences and Series

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 A.P. and

∣ a ∣ < 1, ∣ b ∣ < 1, ∣ c ∣ < 1

, then

x, y, z

are in

Answer: (D)

Solution

Each quantity is the sum of an infinite geometric series, so

x = \frac{1}{1 - a}

y = \frac{1}{1 - b}

and

z = \frac{1}{1 - c}

, giving

a = 1 - \frac{1}{x}

b = 1 - \frac{1}{y}

c = 1 - \frac{1}{z}

. Since

a, b, c

are in A.P.,

2 b = a + c

, which becomes

2 (1 - \frac{1}{y}) = (1 - \frac{1}{x}) + (1 - \frac{1}{z})

, simplifying to

\frac{2}{y} = \frac{1}{x} + \frac{1}{z}

. This is exactly the condition for

x, y, z

to be in H.P.

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option