If, in a G.P., the (p + q)th term is a and the (p – q)th term is b, then pth term is

Let the G.P be $\mathrm{x},\mathrm{xy},{\mathrm{xy}}^2,\dots$

then ${\mathrm{t}}_{\mathrm{p}+\mathrm{q}}={\mathrm{xy}}^{\mathrm{p}+\mathrm{q}-1}=\mathrm{a}$                 (1)

and  ${\mathrm{t}}_{\mathrm{p}-\mathrm{q}}={\mathrm{xy}}^{\mathrm{p}-\mathrm{q}-1}=\mathrm{b}$                 (2)

Dividing Eqs (1) by (2),

${\mathrm{y}}^{2\mathrm{q}}=\frac{\mathrm{a}}{\mathrm{b}}$

$\therefore \mathrm{y}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\frac{1}{2\mathrm{q}}}$

From Eq. (1),

$\mathrm{x}=\mathrm{a}{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^{\frac{\mathrm{p}+\mathrm{q}-1}{2\mathrm{q}}}$

$\begin{array}{l}\therefore {\mathrm{t}}_{\mathrm{p}}={\mathrm{xy}}^{\mathrm{p}-1}=\mathrm{a}\cdot {\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^{\frac{\mathrm{p}+\mathrm{q}-1}{2\mathrm{q}}}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\frac{\mathrm{p}-1}{2\mathrm{q}}}\\ =1-\frac{\mathrm{p}+\mathrm{q}-1}{2\mathrm{q}}+\frac{\mathrm{p}-1}{2\mathrm{q}}\times {\mathrm{b}}^{\frac{\mathrm{p}+\mathrm{q}-1}{2\mathrm{q}}-\frac{\mathrm{p}-1}{2\mathrm{q}}}\\ ={\mathrm{a}}^{\frac{1}{2}}{\mathrm{b}}^{\frac{1}{2}}=\sqrt{\mathrm{ab}}\end{array}$