(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}}} + ………{\rm{upto}}\;n\;{\rm{terms}}} \right)$

$ = \frac{{\frac{1}{a}\left[ {{{\left( {\frac{1}{r}} \right)}^n} – 1} \right]}}{{\left( {\frac{1}{r} – 1} \right)}}\left( {\;\frac{1}{r} > 1} \right)$if $r < 1$

$ = \frac{{(1 – {r^n})}}{{a{r^{n – 1}}(1 – r)}}$……. (iii)

Therefore , $\frac{S}{R} = \frac{{a(1 – {r^n})}}{{1 – r}} \times \frac{{a{r^{n – 1}}(1 – r)}}{{(1 – {r^n})}} = {a^2}{r^{n – 1}}$

or ${\left( {\frac{S}{R}} \right)^n} = {({a^2}{r^{n – 1}})^n} = {a^{2n}}{r^{n(n – 1)}} = {P^2}$.