If the sum of the n terms of G.P. is S produc | vedclass

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(d) Given that sum $S = \frac{{a({r^n} - 1)}}{{r - 1}} = \frac{{a\,(1 - {r^n})}}{{1 - r}}$......(i)

$P = a(ar)(a{r^2})..........(a{r^{n - 1}}) = {a

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${\left( {\frac{S}{R}} \right)^n}$

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(d) Given that sum $S = \frac{{a({r^n} - 1)}}{{r - 1}} = \frac{{a\,(1 - {r^n})}}{{1 - r}}$......(i)

$P = a(ar)(a{r^2})..........(a{r^{n - 1}}) = {a^n}{r^{1 + 2 + ......... + (n - 1)}}$

$ = {a^n}{r^{(n - 1)n/2}}\;i.e.,\;{P^2} = {a^{2n}}{r^{n(n - 1)}}$ ......(ii)

and $R = \frac{1}{a} + \frac{1}{{ar}} + \frac{1}{{a{r^2}}} + ..........$ upto $\frac{1}{{49}}$ terms

$ = \frac{1}{a}\left( {1 + \frac{1}{r} + \frac{1}{{{r^2}}} + .........{m{upto}}\;n\;{m{terms}}} ight)$

$ = \frac{{\frac{1}{a}\left[ {{{\left( {\frac{1}{r}} ight)}^n} - 1} ight]}}{{\left( {\frac{1}{r} - 1} ight)}}\left( {\;\frac{1}{r} > 1} ight)$if $r < 1$

$ = \frac{{(1 - {r^n})}}{{a{r^{n - 1}}(1 - r)}}$....... (iii)

Therefore , $\frac{S}{R} = \frac{{a(1 - {r^n})}}{{1 - r}} 	imes \frac{{a{r^{n - 1}}(1 - r)}}{{(1 - {r^n})}} = {a^2}{r^{n - 1}}$

or ${\left( {\frac{S}{R}} ight)^n} = {({a^2}{r^{n - 1}})^n} = {a^{2n}}{r^{n(n - 1)}} = {P^2}$.

If the sum of the $n$ terms of $G.P.$ is $S$ product is $P$ and sum of their inverse is $R$, than ${P^2}$ is equal to

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