In a GP. if $(m+n{)}^{\text{th }}$ terms is p and $(m-n{)}^{\text{th }}$ term is q, then $m^{th}$ term is:

For a G. P, $a_{m+n}=p$ and $a_{m-n}=q$

We know that $a_n=A{R}^{n-1}$ (in G.P.)

where A= first term and R = ratio 

$\therefore$$a_{m+n}=p$

$\Rightarrow {\mathrm{AR}}^{\mathrm{m}+\mathrm{n}-1}=\mathrm{p}\dots \text{ (i) }$

and $a_{m-n}=q$

$\Rightarrow {\mathrm{AR}}^{\mathrm{m}-\mathrm{n}-1}=\mathrm{q}\dots \text{ (ii) }$

On multiply equation (i) and (ii), we have 

$\left({\mathrm{AR}}^{\mathrm{m}+\mathrm{n}-1}\right)\cdot \left({\mathrm{AR}}^{\mathrm{m}-\mathrm{n}-1}\right)=\mathrm{pq}$

$\Rightarrow {\mathrm{A}}^2\cdot {\mathrm{R}}^{2(\mathrm{m}-1)}=\mathrm{pq}$

$\begin{array}{l}\Rightarrow {\left({\mathrm{AR}}^{\mathrm{m}-1}\right)}^2=\mathrm{pq}\\ \Rightarrow {\mathrm{AR}}^{\mathrm{m}-1}=\sqrt{\mathrm{pq}}\\ \Rightarrow {\mathrm{a}}_{\mathrm{m}}=\sqrt{\mathrm{pq}}\end{array}$