Let S be the sum, P the product and R the sum of reciprocals of n…

Question

Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that $P^2R^n = S^n$.

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Answer

Let the G.P be $a, ar, ar^2, ar^3 ..........,ar^{n - 1}$
Here $\mathrm { S } = \frac { a \left( r ^ { n } - 1 \right) } { r - 1 }$
P = $a.ar.ar^2 ......... ar^{n-1}$
$= {a^n}.{r^{1 + 2 + 3 + ....... + (n - 1)}} = {a^n}.{r^{{{n(n - 1)} \over 2}}}$
and R $= \frac { 1 } { a } + \frac { 1 } { a r } + \frac { 1 } { a r ^ { 2 } } + \ldots \ldots \frac { 1 } { a r ^ { n - 1 } }$$= {{{r^{n - 1}} + {r^{n - 2}} + {r^{n - 3}} + .......... + 1} \over {a{r^{n - 1}}}}$
$= {{1({r^n} - 1)} \over {r - 1}}.{1 \over {a{r^{n - 1}}}}$$= {{{r^n} - 1} \over {a{r^{n - 1}}(r - 1)}}$
Now ${p^2}{R^n} = {{{a^{2n}}.{r^{n(n - 1)}}{{({r^n} - 1)}^n}} \over {{a^n}{r^{n(n - 1)}}{{(r - 1)}^n}}} = {{{a^n}{{({r^n} - 1)}^n}} \over {{{(r - 1)}^n}}} = {a^n}{\left( {{{{r^n} - 1} \over {r - 1}}} \right)^n} = {S^n}$
Hence proved.

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