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

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$\sqrt{mn}$

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(B) Given that the $(p + q)^{th}$ term $T_{p+q} = ar^{p+q-1} = m$ and the $(p - q)^{th}$ term $T_{p-q} = ar^{p-q-1} = n$.Dividing the two equations:$\frac{T_{p+q}}{T_{p-q}} = \frac{ar^{p+q-1}}{ar^{p-q-1}} = r^{(p+q-1) - (p-q-1)} = r^{2q} = \frac{m}{n}$.Thus,$r^{2q} = \frac{m}{n} \Rightarrow r = (\frac{m}{n})^{1/(2q)}$.The $p^{th}$ term is $T_p = ar^{p-1}$.Since $T_{p+q} = ar^{p-1} \cdot r^q = m$,we have $T_p \cdot r^q = m \Rightarrow T_p = \frac{m}{r^q}$.Since $T_{p-q} = ar^{p-1} \cdot r^{-q} = n$,we have $T_p \cdot r^{-q} = n \Rightarrow T_p = n \cdot r^q$.Multiplying these two expressions for $T_p$:$T_p^2 = (\frac{m}{r^q}) \cdot (n \cdot r^q) = mn$.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 $p$,the $p^{th}$ term is $\sqrt{T_{p+q} \cdot T_{p-q}} = \sqrt{mn}$.

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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