In a G.P. if the

(c) $\sqrt{p q}$

Here, $a_{(m+n)}=p$

$\Rightarrow a r^{(m+n-1)}=p \quad \ldots \ldots(\mathrm{i})$

Also, $\mathbf{a}_{(m-n)}=q$

$\Rightarrow a r^{(m-n-1)}=q \quad \ldots \ldots$ (ii)

Mutliplying (i) and (ii):

$\Rightarrow a r^{(m+n-1)} a r^{(m-n-1)}=p q$

$\Rightarrow a^{2} r^{(2 m-2)}=p q$

$\Rightarrow\left(a r^{(m-1)}\right)^{2}=p q$

$\Rightarrow a r^{(m-1)}=\sqrt{p q}$

$\Rightarrow a_{m}=\sqrt{p q}$

Thus, the $m^{\text {th }}$ term is $\sqrt{p q}$.