Question

If a, b and c be in G .P. and x, y be the arithmetic means between a, b and b, c respectively then $\frac{a}{x}+\frac{c}{y}$ is-

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is A $2$

Given $a,b,c$ are in G.P.

Therefore, $b^2=ac$ $...\left(1\right)$

$\Rightarrow a=\frac{b^2}{c}$ or $c=\frac{b^2}{a}$ $...\left(2\right)$

Also, $x$ and $y$ are the arithmetic mean between $a,b$ and $b,c$ respectively.

Therefore, $2x=a+b$ and $2y=b+c$

$\Rightarrow x=\frac{a+b}{2}$ and $\Rightarrow y=\frac{b+c}{2}$ $...\left(3\right)$

Consider, $\frac{a}{x}$ and substitute the values of $a$ and $x$ from the equation $\left(2\right)$ and $\left(3\right)$.

$\Rightarrow \frac{a}{x}=\frac{\frac{b^2}{c}}{\frac{a+b}{2}}$

$\Rightarrow \frac{a}{x}=\frac{2b^2}{c(a+b)}$

Similarly, substitute the values of $c$ and $y$ from the equation

$\left(2\right)$ and $\left(3\right)$ in $\frac{c}{y}$.

$\Rightarrow \frac{c}{y}=\frac{\frac{b^2}{a}}{\frac{b+c}{2}}$

$\Rightarrow \frac{c}{y}=\frac{2b^2}{a(b+c)}$

Consider, $\frac{a}{x}+\frac{c}{y}=\frac{2b^2}{c(a+b)}+\frac{2b^2}{a(b+c)}$

$=2b^2[\frac{1}{c(a+b)}+\frac{1}{a(b+c)}]$

$=2b^2\left[\frac{ab+2ac+bc}{(ac+bc)(ab+ac)}\right]$

Substitute the value of $ac$ from equation $\left(1\right)$; we get

$\frac{a}{x}+\frac{c}{y}=2b^2\left[\frac{ab+2b^2+bc}{(b^2+bc)(ab+b^2)}\right]$

$=2b^2\left[\frac{ab+2b^2+bc}{b^2(b+c)(a+b)}\right]$

$=2\left[\frac{ab+2b^2+bc}{(b+c)(a+b)}\right]$

$=2\left[\frac{ab+2b^2+bc}{ab+2b^2+bc}\right]$

$=2$

Thus, $\frac{a}{x}+\frac{c}{y}=2$.