Given : $\large\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$
To prove $a,b,c\:and\:d$ are in G.P., we have to prove $\large\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$.
$\Rightarrow\:\large\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}$
Using componendo and dividendo we get
$\large\frac{(a+bx)+(a-bx)}{(a+bx)-(a-bx)}=\frac{(b+cx)+(b-cx)}{(b+cx)-(b-cx)}$
$\Rightarrow\:\large\frac{2a}{2bx}=\frac{2b}{2cx}$
$\Rightarrow\:\large\frac{a}{b}=\frac{b}{c}$...(i)
Step 2
Similarly by using componendo and dividento for
$\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$ we get
$\large\frac{b}{c}=\frac{c}{d}$.......(ii)
From (i) and (ii) we get
$\large\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$
$\Rightarrow\: a,b,c\:and\:d$ are in G.P.