Explanation

(a)
Since, $\alpha,\ \beta$ and $\gamma,\delta$ are the roots of the equation $ax^{2} + 2bx + c = 0$ and $px^{2} + 2qx + r = 0$ respectively, then
$\alpha + \beta = - \frac{2b}{a},\ \alpha\beta = \frac{c}{a},\ \gamma + \delta = - \frac{2q}{p},\ \gamma\delta = \frac{r}{p}$
As given $\alpha,\ \beta,\ \gamma\ and\ \delta$ are in GP, therefore
$\frac{\alpha}{\gamma} = \frac{\beta}{\delta}$ ...(i)
But $\frac{\alpha\beta}{\gamma\delta} = \frac{pc}{ar}\ \ \Rightarrow \left( \ \frac{\beta}{\delta} \right)^{2} = \frac{pc}{ar}$ [from Eq. (i)]
Also, $\frac{\alpha}{\beta} = \frac{\gamma}{\delta}$ $\Rightarrow \ \frac{\alpha + \beta}{\beta} = \frac{\gamma + \delta}{\delta}\ \Rightarrow \ \frac{\alpha + \beta}{\gamma + \delta} = \frac{\beta}{\delta}$
$\Rightarrow \ \frac{bp}{aq} = \sqrt{\frac{pc}{ar}}\ \Rightarrow \ \frac{b^{2}p^{2}}{a^{2}q^{2}} = \frac{pc}{ar}\ \Rightarrow q^{2}ac = b^{2}pr$