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

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

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

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

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

$\Rightarrow b^2=ac$

$\Rightarrow \frac{b}{a}=\frac{c}{b}$ ...... (1)

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

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

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

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

$\Rightarrow c^2=bd$

$\Rightarrow \frac{c}{b}=\frac{d}{c}$ ....... (2)

From (1) and (2), we get

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

which shows that a, b, c and d are in G.P.