Explanation

(b)
We have,
$\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$
Applying componendo and dividendo rule, we get
$\frac{2a}{2bx} = \frac{2b}{2cx} = \frac{2c}{2dx}$
$\Rightarrow \ \frac{a}{b} = \frac{b}{c} = \frac{c}{d}$
$\Rightarrow \ b^{2} = ac\ \ \ and\ c^{2} = bd$
$\Rightarrow a,\ b,\ c$ and $b,\ c,\ d$ are in GP, therefore $a,\ b,\ c,\ d$ are in GP.