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

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(b) Given that $m = a{r^{p + q - 1}}$ and $n = a{r^{p - q - 1}}$

${r^{p + q - 1 - p + q + 1}} = \frac{m}{n}$

$\Rightarrow r = {\left( {\frac{m}{

Vedclass · Accepted Answer

$\sqrt {mn} $

Vedclass · Answer

(b) Given that $m = a{r^{p + q - 1}}$ and $n = a{r^{p - q - 1}}$

${r^{p + q - 1 - p + q + 1}} = \frac{m}{n}$

$\Rightarrow r = {\left( {\frac{m}{n}} \right)^{1/(2q)}}$

and $a = \frac{m}{{{{\left( {\frac{m}{n}} \right)}^{(p + q - 1)/2q}}}}$

Now ${p^{th}}$ term $ = a{r^{p - 1}} = \frac{m}{{{{\left( {\frac{m}{n}} \right)}^{(p + q - 1)/2q}}}}{\left( {\frac{m}{n}} \right)^{(p - 1)/2q}}$

$ = m{\left( {\frac{m}{n}} \right)^{(p - 1)/2q - (p + q - 1)/(2q)}} = m{\left( {\frac{m}{n}} \right)^{ - 1/2}} = {m^{1 - 1/2}}{n^{1/2}}$

$ = {m^{1/2}}{n^{1/2}} = \sqrt {mn} $.

Aliter : As we know each term in a $G.P.$ is geometric mean of the terms equidistant from it.

Here ${(p + q)^{th}}$and ${(p - q)^{th}}$ terms are equidistant from ${p^{th}}$ term

$i.e.$ at a distance of $q$.

Therefore, ${p^{th}}$ term will be $G.M.$ of ${(p + q)^{th}}$ and ${(p - q)^{th}}$

$i.e.$ $\sqrt {mn} $.

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

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