Question

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

Answer 1

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

Applying Componendo Dividendo

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

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

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

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

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

Thus, $a,b,c,d$ are in G.P as their common ratio is same