Question

If $a, b, c$ are in $G.P.$ and $ x,y $ are arithmetic mean of $a, b$ and $b, c$ respectively, then $ \frac{1}{x}+\frac{1}{y} $ is equal to

• A. $ \frac{2}{b} $
• B. $ \frac{3}{b} $
• C. $ \frac{b}{3} $
• D. $ \frac{b}{2} $
• E. $ \frac{1}{b} $

Answer 1

Answer: (A)

Given, $ {{b}^{2}}=ac,x=\frac{a+b}{2} $ and $ y=\frac{b+c}{2} $
$ \therefore $ $ \frac{1}{x}+\frac{1}{y}=\frac{2}{a+b}+\frac{2}{b+c} $
$=\frac{2(2b+a+c)}{ab+{{b}^{2}}+bc+ac} $
$=\frac{2(2b+a+c)}{ab+2{{b}^{2}}+bc} $
$=\frac{2(2b+a+c)}{b(2b+a+c)} $
$=\frac{2}{b} $