If x = ∑ _ n = 0 ^ ∞ a ^ n , y = ∑ _ n = 0 ^ ∞ b ^ n , z = ∑ _ n = 0 ^ ∞ c ^ n , where a, b, c are in A.P. and |a| < 1, |b| < 1, |c| < 1 , abc ≠ 0 , then

Correct answer: C. (1)/(x), (1)/(y), (1)/(z) are in A.P. x = 1 + a + a ^ 2 = . . . . . . x = (1)/(1 - a) ⇒ a = 1 - (1)/(x) y = (1)/(1 - b) ⇒ b = 1 - (1)/(y) z = (1)/(1 - c) ⇒ c = 1 - (1)/(z) a, b, c are in A.P. ⇒ 1 - (1)/(x), 1 - (1)/(y), 1 - (1)/(z) are in A.P. ⇒ - (1)/(x), - (1)/(y), - (1)/(z) are in A.P. ⇒ (1)/(x), (1)/(y), (1)/(z) are in A.P.

JEE Main 202227 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 27 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2022 · 27 June · Morning Shift · Mathematics · Sequences and Series

If x = ∑ _ n = 0 ^ ∞ a ^ n , y = ∑ _ n = 0 ^ ∞ b ^ n , z = ∑ _ n =...

x = \mathop{∑}_{n = 0}^{\infty}{a}^{n}, y = \mathop{∑}_{n = 0}^{\infty}{b}^{n}, z = \mathop{∑}_{n = 0}^{\infty}{c}^{n}

, where

a, b, c

,

abc \ne 0$, then

Options

A $x, y, z$ are in A.P.
B $x, y, z$ are in G.P.
C $\dfrac{1}{x}, \dfrac{1}{y}, \dfrac{1}{z}$ are in A.P.
D $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1 - \left(a + b + c\right)$

Correct answer

C. $\dfrac{1}{x}, \dfrac{1}{y}, \dfrac{1}{z}$ are in A.P.

Step-by-step solution

$x = 1 + a + {a}^{2} = . . . . . .$

$x = \dfrac{1}{1 - a} ⇒ a = 1 - \dfrac{1}{x}$

$y = \dfrac{1}{1 - b} ⇒ b = 1 - \dfrac{1}{y}$

$z = \dfrac{1}{1 - c} ⇒ c = 1 - \dfrac{1}{z}$

$a, b, c$ are in A.P.

$⇒ 1 - \dfrac{1}{x}, 1 - \dfrac{1}{y}, 1 - \dfrac{1}{z}$ are in A.P.

$⇒ - \dfrac{1}{x}, - \dfrac{1}{y}, - \dfrac{1}{z}$ are in A.P.

$⇒ \dfrac{1}{x}, \dfrac{1}{y}, \dfrac{1}{z}$ are in A.P.

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