Question

If $a,b,c$ are in GP and $x,y$ are the arithmetic means between $a,b$and $b,c$ respectively, then $\frac{a}{x}+\frac{c}{y}$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is C

$2$

Solve series in GP:

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

$$\begin{gathered} b^2=ac...\left(1\right) \\ x=\frac{(a+b)}{2},y=\frac{(b+c)}{2} \\ \frac{a}{x}+\frac{c}{y}=\frac{2a}{(a+b)}+\frac{2c}{(b+c)} \\ =\frac{\left[2a\right(b+c)+2c(a+b\left)\right]}{(a+b)(b+c)} \\ =\frac{(2ab+2ac+2ac+2bc)}{(ab+b^2+ac+bc)} \\ =2\frac{(ab+2ac+bc)}{(ab+2ac+bc)}(\sin ce{b}^2=ac) \\ =2 \\ \Rightarrow \frac{a}{x}+\frac{c}{y}=2 \end{gathered}$$

Hence, the correct option is (C).