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

(A) Given that $\frac{a+bx}{a-bx} = \frac{b+cx}{b-cx}$.
Applying componendo and dividendo or cross-multiplying:
$(a+bx)(b-cx) = (a-bx)(b+cx)$
$ab - acx + b^2x - bcx^2 = ab + acx - b^2x - bcx^2$
$2b^2x = 2acx$
Since $x \neq 0$,we get $b^2 = ac$,which implies $\frac{b}{a} = \frac{c}{b} \dots (1)$.
Similarly,given $\frac{b+cx}{b-cx} = \frac{c+dx}{c-dx}$.
$(b+cx)(c-dx) = (b-cx)(c+dx)$
$bc - bdx + c^2x - cdx^2 = bc + bdx - c^2x - cdx^2$
$2c^2x = 2bdx$
Since $x \neq 0$,we get $c^2 = bd$,which implies $\frac{c}{b} = \frac{d}{c} \dots (2)$.
From $(1)$ and $(2)$,we have $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.
Therefore,$a, b, c,$ and $d$ are in $G.P.$