Explanation

(a)
Let $r$ be the common ratio of the GP. Since $\alpha,\ \beta\ ,\gamma,\delta$ are in GP, then $\beta = \alpha r,\ \gamma = \alpha r^{2}\ and\ \delta = \alpha r^{3}$.
For equations, $x^{2} - x + p = 0$
$\therefore\ \alpha + \beta = 1$
$\Rightarrow \ \alpha + \alpha r = 1$
$\Rightarrow \ \alpha(1 + r) = 1$ ….(i)
and $\alpha\beta = p\ \Rightarrow \ \alpha(\alpha r) = p\$
$\Rightarrow \ \alpha^{2}r = p$ ...(ii)
For equation, $x^{2} - x + q = 0$
$\gamma + \delta = 4$
$\Rightarrow \ \alpha r^{2} + \alpha r^{3} = 4$
$\Rightarrow \ \alpha r^{2}(1 + r) = 4$ ...(iii)
and $\gamma\delta = q \Rightarrow \ \alpha r^{3} \bullet \ \alpha r^{2} = q$
$\Rightarrow \ \alpha^{2}r^{5} = q$ ...(iv)
On dividing Eq. (iii) by Eq. (i),we get
$r^{2} = 4\ \Rightarrow r = \pm 2$
If we take $r = 2$, then $\alpha$ is not integral, so we take $r = - 2.$
Substituting $r = - 2$ in Eq. (i), we get
$\alpha = - 1$
Now, from Eq. (ii), we have
$p = \alpha^{2}r = ( - 1)^{2}( - 2) = - 2$
and from Eq. (iv), we have
$q = \alpha^{2}r^{5} = ( - 1)^{2}( - 2)^{5} = - 32$
$\Rightarrow (p,q) = ( - 2, - 32)$