If the pth and qth terms of a G.P. are q and p, respectively,

As, $a_{p}=q$

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

Also, $a_{q}=p$

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

Dividing (i) by (ii), we get

$\frac{a r^{(p-1)}}{a r^{(q-1)}}=\frac{q}{p}$

$\Rightarrow r^{(p-1-q+1)}=\frac{q}{p}$

$\Rightarrow r^{(p-q)}=\frac{q}{p}$

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

Substituting the value of $r$ in (ii), we get

$a\left[\left(\frac{q}{p}\right)^{\frac{1}{(p-q)}}\right]^{(q-1)}=p$

$\Rightarrow a\left[\left(\frac{q}{p}\right)^{\frac{(q-1)}{(p-q)}}\right]=p$

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

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

Now,

$a_{(p+q)}=a r^{(p+q-1)}$

$=p\left(\frac{p}{q}\right)^{\frac{(q-1)}{(p-q)}} \times\left[\left(\frac{q}{p}\right)^{\frac{1}{(p-q)}}\right]^{(p+q-1)}$

$=p\left(\frac{p}{q}\right)^{\frac{(q-1)}{(p-q)}} \times\left(\frac{q}{p}\right)^{\frac{(p+q-1)}{(p-q)}}$$=p \times\left(\frac{q}{p}\right)^{\frac{-(q-1)}{(p-q)}+\frac{(p+q-1)}{(p-q)}}$

$=p\left(\frac{q}{p}\right)^{\frac{-(q-1)}{(p-q)}} \times\left(\frac{q}{p}\right)^{\frac{(p+q-1)}{(p-q)}}$

$=p \times\left(\frac{q}{p}\right)^{\frac{-(q-1)}{(p-q)}+\frac{(p+q-1)}{(p-q)}}$

$=p \times\left(\frac{q}{p}\right)^{\frac{-q+1+p+q-1}{(p-q)}}$

$=p \times\left(\frac{q}{p}\right)^{\frac{p}{(p-q)}}$

$=\frac{p \times q^{\frac{p}{(p-q)}}}{p \frac{p}{(p-q)}}$

$=\frac{q^{\frac{p}{(p-q)}}}{p \frac{p}{(p-q)}-1}$

$=\frac{q \frac{p}{(p-q)}}{p^{\frac{p-p+q}{(p-q)}}}$

$=\frac{q \frac{p}{(p-q)}}{p^{\frac{q}{(p-q)}}}$

$=\frac{q^{p \times \frac{1}{(p-q)}}}{p^{q \times \frac{1}{(p-q)}}}$

$=\left(\frac{q^{p}}{p^{q}}\right)^{\frac{1}{p-q}}$