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

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$ According to the given condition,

$a_{5}=a r^{5-1}=a r^{4}=p$          .........$(1)$

$a_{8}=a r^{8-1}=a r^{7}=q$          .........$(2)$

$a_{11}=a r^{11-1}=a r^{10}=s$          .........$(3)$

Dividing equation $(2)$ by $(1),$ we obtain

$\frac{a r^{7}}{a r^{4}}=\frac{q}{p}$

$r^{3}=\frac{q}{p}$       .........$(4)$

Dividing equation $(3)$ by $(2),$ we obtain

$\frac{a r^{10}}{a r^{7}}=\frac{s}{q}$

$\Rightarrow r^{3}=\frac{s}{q}$        .......$(5)$

Equating the values of $r^{3}$ obtained in $(4)$ and $(5),$ we obtain

$\frac{q}{p}=\frac{s}{q}$

$\Rightarrow q^{2}=p s$

Thus, the given result is proved.