Question:
The $5^{\text {th }}, 8^{\text {th }}$ and $11^{\text {th }}$ terms of a G.P. are $p, q$ and $s$, respectively. Show that $q^{2}=p s$.
Solution:
Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
$a_{5}=a r^{5-1}=a r^{4}=p \ldots(1)$
$a_{8}=a r^{8-1}=a r^{7}=q \ldots(2)$
$a_{11}=a r^{11-1}=a r^{10}=s \ldots(3)$
Dividing equation (2) by (1), we obtain
$\frac{a r^{7}}{a r^{4}}=\frac{q}{p}$
$r^{3}=\frac{q}{p}$ $\ldots(4)$
Dividing equation (3) by (2), we obtain
$\frac{a r^{10}}{a r^{7}}=\frac{s}{q}$
$\Rightarrow r^{3}=\frac{s}{q}$ $\ldots(5)$
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.