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

Solve for prove

$\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}(x\ne 0)$

By using $\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}$

On cross multiplying,

$(a+bx)(b-cx)=(b+cx)(a-bx)$

$\Rightarrow ab-acx+b^{2}x-b{x}^{2}c$$=ab-b^{2}x+acx-bc{x}^2$

$\Rightarrow 2b^{2}x=2acx$

$\Rightarrow b^2=ac$

$\Rightarrow \frac{b}{a}=\frac{c}{b}.....\left(i\right)$

Also, given $\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$

On cross multiplying,

$(b+cx)(c-dx)=(b-cx)(c+dx)$

$\Rightarrow bc-bdx+c^{2}x-cd{x}^2$$=bc+bdx-c^{2}x-cd{x}^2$

$\Rightarrow 2c^{2}x=2bdx$

$\Rightarrow c^2=bd$

$\Rightarrow \frac{c}{b}=\frac{d}{c}$ $....\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$, we get

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

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