If $\left(p+q\right)^{th}$ term and $\left(p-q\right)^{th}$ terms of a G.P are a and b respectively, prove that the $p^{th}$ term is $\sqrt{ab}$.
Let the first term of G.P $= A$ and common ratio $= R$
$n^{th}$ term of G.P $= t_n = AR^{n-1}$
$\therefore$ $\left(p+q\right)^{th}$ term $= t_{p+q} = AR^{p+q-1} = a$ (given) $\cdots$ (1)
and $\left(p-q\right)^{th}$ term $= t_{p-q} = AR^{p-q-1} = b$ (given) $\cdots$ (2)
$\therefore$ $\dfrac{t_{p+q}}{t_{p-q}} = \dfrac{AR^{p+q-1}}{AR^{p-q-1}} = \dfrac{a}{b}$
i.e. $R^{p+q-1-p+q+1} = \dfrac{a}{b}$
i.e. $R^{2q} = \dfrac{a}{b}$ $\implies$ $R = \sqrt[2q]{\dfrac{a}{b}} = \left(\dfrac{a}{b}\right)^{\frac{1}{2q}}$ $\cdots$ (3)
$\therefore$ From equations (1) and (3) we have
$A\left[\left(\dfrac{a}{b}\right)^{\frac{1}{2q}}\right]^{p+q-1} = a$
i.e. $A \left(\dfrac{a}{b}\right)^{\frac{p+q-1}{2q}} = a$
$\implies$ $A = a \times \left(\dfrac{b}{a}\right)^{\frac{p+q-1}{2q}}$
i.e. $A = a^{1-\frac{\left(p+q-1\right)}{2q}} b^{\frac{p+q-1}{2q}} = a^{\frac{1-p+q}{2q}} b^{\frac{p+q-1}{2q}}$
$
\begin{aligned}
\therefore p^{th} \text{term} = t_p & = AR^{p-1} \\
& = a^{\frac{1-p+q}{2q}} b^{\frac{p+q-1}{2q}} \times \left(\dfrac{a}{b}\right)^{\frac{p-1}{2q}} \\
& = a^{\frac{1-p+q+p-1}{2q}} b^{\frac{p+q-1}{2q} - \frac{p-1}{2q}} \\
& = a^{\frac{1}{2}} b^{\frac{1}{2}} = \sqrt{ab}
\end{aligned}
$