Explanation

(a)
Since, $a,\ b$ and $c$ are in GP
$\therefore\ \ b^{2} = ac$
Given, equation $ax^{2} + 2bx + c = 0$ becomes
$ax^{2} + 2\sqrt{acx} + c = 0$
$\Rightarrow \ \ \left( ax + \sqrt{c} \right)^{2} = 0$
$\Rightarrow \ \ x = - \sqrt{\frac{c}{a}}$ (respected roots)
Since, this root satisfy the second equation
$dx^{2} + 2cx + f = 0$
$\therefore\ \ \ d\frac{c}{a} - 2e\sqrt{\frac{c}{a}} + f = 0$
$\Rightarrow \ \ \ \frac{d}{a} + \frac{f}{c} = \frac{2e}{c}\sqrt{\frac{c}{a}} = \frac{2e}{b}$ ($\because b = \sqrt{ac}$)
$\Rightarrow \ \ \ \ \frac{d}{a},\ \frac{e}{b},\ \frac{f}{c}$ are in GP