If 2^ (1)/(c) , 2^ (b)/(a c) , 2^ (1)/(a) are in GP, then which one of the following is correct?

Correct answer: A. a, b, c are in AP 2^ (1)/(c) , 2^ (b)/(a c) , 2^ (1)/(a) are in GP (2^ (b)/(a c) )^2=2^ (1)/(c) · 2^ (1)/(a) 2^ (2 b)/(a c) =2^ (1)/(c)+(1)/(a) (2 b)/(a c)=(1)/(c)+(1)/(a)=(a+c)/(a c) 2 b=a+c a, b, c are in AP.

NDA 2023

Exam: NDA
Year: 2023
Subject: Mathematics
Chapter: Sequences and Series

NDA 2023 · Mathematics · Sequences and Series

If 2^ (1)/(c) , 2^ (b)/(a c) , 2^ (1)/(a) are in GP, then which one...

2^{\frac{1}{c}}, 2^{\frac{b}{a c}}, 2^{\frac{1}{a}}

are in GP, then which one of the following is correct?

Options

A $a, b, c$ are in AP
B $a, b, c$ are in GP
C $a, b, c$ are in HP
D $a b, b c, c a$ are in AP

Correct answer

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

Step-by-step solution

2^{\frac{1}{c}}, 2^{\frac{b}{a c}}, 2^{\frac{1}{a}}

are in GP

\left(2^{\frac{b}{a c}}\right)^2=2^{\frac{1}{c}} \cdot 2^{\frac{1}{a}}2^{\frac{2 b}{a c}}=2^{\frac{1}{c}+\frac{1}{a}}\therefore \quad \frac{2 b}{a c}=\frac{1}{c}+\frac{1}{a}=\frac{a+c}{a c}\therefore \quad 2 b=a+c\therefore \quad a, b, c

are in AP.

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