Question

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

Answer 1

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

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

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

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

$\Rightarrow b^2=ac$

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

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

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

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

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

$\Rightarrow c^2=bd$

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

From $\left(1\right)$ and $\left(2\right)$, we obtain

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

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