Question

If $\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}(x\ne 0),$ then show that a, b, c and d are in G.P.

Answer 1

Given :

$\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$

Now, $\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}$

Applying componendo and dividendo

$\Rightarrow \frac{(a+bx)+(a-bx)}{(a+bx)-(a-bx)}=\frac{(b+cx)+(b-cx)}{(b+cx)-(b-cx)}$

$\Rightarrow \frac{2a}{2bx}=\frac{2b}{2cx}$

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

Similiarly,

$\left(\frac{(b+cx)+(b-cx)}{(b+cx)-(b-cx)}\right)=\left(\frac{(c+dx)+(c-dx)}{(c+dx)-(c-dx)}\right)$

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

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