The $5^{\text{th}}$,$8^{\text{th}}$,and $11^{\text{th}}$ terms of a $G.P.$ are $p, q$,and $s$,respectively. Show that $q^{2} = ps$.

(N/A) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
According to the given condition:
$a_{5} = ar^{5-1} = ar^{4} = p$ .........$(1)$
$a_{8} = ar^{8-1} = ar^{7} = q$ .........$(2)$
$a_{11} = ar^{11-1} = ar^{10} = s$ .........$(3)$
Dividing equation $(2)$ by $(1)$,we obtain:
$\frac{ar^{7}}{ar^{4}} = \frac{q}{p} \Rightarrow r^{3} = \frac{q}{p}$ .........$(4)$
Dividing equation $(3)$ by $(2)$,we obtain:
$\frac{ar^{10}}{ar^{7}} = \frac{s}{q} \Rightarrow r^{3} = \frac{s}{q}$ .........$(5)$
Equating the values of $r^{3}$ from $(4)$ and $(5)$:
$\frac{q}{p} = \frac{s}{q}$
$\Rightarrow q^{2} = ps$
Thus,the result is proved.