If the (p + q)^{th} term of a G.P. is m and t | vedclass

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(B) Given that the $(p + q)^{th}$ term is $m = ar^{p+q-1}$ and the $(p - q)^{th}$ term is $n = ar^{p-q-1}$.Dividing the two equations: $\frac{m}{n} =

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$\sqrt{mn}$

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(B) Given that the $(p + q)^{th}$ term is $m = ar^{p+q-1}$ and the $(p - q)^{th}$ term is $n = ar^{p-q-1}$.Dividing the two equations: $\frac{m}{n} = \frac{ar^{p+q-1}}{ar^{p-q-1}} = r^{(p+q-1) - (p-q-1)} = r^{2q}$.Thus,$r^{2q} = \frac{m}{n}$,which implies $r = (\frac{m}{n})^{1/(2q)}$.The $p^{th}$ term is $T_p = ar^{p-1}$.We can express $m$ as $T_p \cdot r^q$ and $n$ as $T_p \cdot r^{-q}$.Multiplying $m$ and $n$: $m \cdot n = (T_p \cdot r^q) \cdot (T_p \cdot r^{-q}) = T_p^2 \cdot r^0 = T_p^2$.Therefore,$T_p = \sqrt{mn}$.Alternatively: In a $G.P.$,the $p^{th}$ term is the geometric mean of the terms equidistant from it. Since $(p+q)$ and $(p-q)$ are equidistant from $p$ (at a distance of $q$),the $p^{th}$ term is $\sqrt{mn}$.

If the $(p + q)^{th}$ term of a $G.P.$ is $m$ and the $(p - q)^{th}$ term is $n$,then the $p^{th}$ term will be

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