Question

If $a,b,c$are in GP 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. $\frac{b}{2}$

Answer 1

The correct option is A

$\frac{2}{b}$

Explanation for the correct option:

Solve the series in GP:

Given $a,b,c$are in GP and $x,y$ are arithmetic mean of $a,b$and $b,c$

$b^2=ac..\left(i\right)$

$x=\frac{\left(a+b\right)}{2}$

$y=\frac{\left(b+c\right)}{2}$

$\begin{array}{rcl}\therefore \frac{1}{x}+\frac{1}{y}& =& \frac{2}{(a+b)}+\frac{2}{(b+c)}\\ & =& \frac{2(b+c)+2(a+b)}{(a+b)(b+c)}\\ & =& \frac{(2b+2c+2a+2b)}{(ab+b^2+ac+bc)}\\ & =& \frac{2(a+c+2b)}{(ab+b^2+b^2+bc)}\\ & =& \frac{2(a+c+2b)}{(ab+2b^2+bc)}\\ & =& \frac{2(a+c+2b)}{b(a+2b+c)}\\ & =& \frac{\mathbf{2}}{\mathbf{b}}\end{array}$ $\begin{array}{rcl}& & \left[\because b^2=ac\right]\end{array}$

Hence, correct option is (A).