If a, b, c are in G.P. and x, y are AM's between a, b and b,c respectively,

Solution:

(d) $\frac{1}{x}+\frac{1}{y}=\frac{2}{b}$

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

$\therefore b^{2}=a c \quad \ldots \ldots$ (i)

$a, x$ and $b$ are in A.P.

$\therefore 2 x=a+b \quad \ldots \ldots \ldots($ ii $)$

Also, $b, y$ and $c$ are in A.P.

$\therefore 2 y=b+c$

$\Rightarrow 2 y=b+\frac{b^{2}}{a} \quad[$ Using (i) $]$

$\Rightarrow 2 y=b+\frac{b^{2}}{(2 x-b)} \quad[$ Using (ii) $]$

$\Rightarrow 2 y=\frac{b(2 x-b)+b^{2}}{(2 x-b)}$

$\Rightarrow 2 y=\frac{2 b x-b^{2}+b^{2}}{(2 x-b)}$

$\Rightarrow 2 y=\frac{2 b x}{(2 x-b)}$

$\Rightarrow y=\frac{b x}{(2 x-b)}$

$\Rightarrow y(2 x-b)=b x$

$\Rightarrow 2 x y-b y=b x$

$\Rightarrow b x+b y=2 x y$

Dividing both the sides by $x y$ :

$\Rightarrow \frac{1}{y}+\frac{1}{x}=\frac{2}{b}$