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