then show that a, b, c and d are in G.P.

Given:

$\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}$

Now, $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}$

Applying componendo and dividendo

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

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

$\Rightarrow \frac{a}{b}=\frac{b}{c}$

Similiarly, $\frac{(b+c x)+(b-c x)}{(b+c x)-(b-c x)}=\frac{(c+d x)+(c-d x)}{(c+d x)-(c-d x)}$

$\Rightarrow \frac{b}{c}=\frac{c}{d}$

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