If (p + q)th and (p − q)th terms of a G.P. are m and n respectively,

Here, $(p+q)^{\text {th }}$ term $=\mathrm{m}$

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

And, $(\mathrm{p}-\mathrm{q})^{\mathrm{th}}$ term $=\mathrm{n}$

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

Dividing (i) by (ii):

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

$\Rightarrow r^{2 q}=\frac{m}{n}$

$\Rightarrow r^{q}=\sqrt{\frac{m}{n}}$

Now, from (i):

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

$\Rightarrow a r^{p-1} \times \sqrt{\frac{m}{n}}=m$

$\Rightarrow a r^{p-1}=m \times \frac{\sqrt{n}}{\sqrt{m}}$

$\Rightarrow a r^{p-1}=\frac{m \sqrt{n}}{\sqrt{m}}$

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