Question

Sol. [Key Concept: Take the four terms in G.P. as $a,ar,{ar}^{2},{ar}^{3}$; use $a + ar = 1$ and ${ar}^{2} + {ar}^{3} = 4$ to find $r,a$, then $p = \alpha\beta$, $q = \gamma\delta$.]

Let $\alpha = a,\beta = ar,\gamma = {ar}^{2},\delta = {ar}^{3}$

$$\begin{matrix} & a + ar = 1 \\ & ar^{2} + ar^{3} = 4 \\ & \ \Rightarrow ar^{2}(1 + r) = 4 \\ & \ \Rightarrow r = 2,a = \frac{1}{3}(\text{~}\text{reject as}\text{~}p \in z) \\ & \text{~}\text{or}\text{~}r = - 2,a = - 1 \\ & \text{~}\text{Now}\text{~}|P + q| = \left| a(ar) + ar^{2} - ar^{3} \right| \\ & \ = \left| a^{2}r\left( 1 + r^{4} \right) \right| \\ & \ |1 \times - 2 \times 17| = 34 \end{matrix}$$