Let \(a,ar,a{r}^{2},\ldots ..\). be an infinite G.P. If \(\sum _{n=0}^{\infty }a{r}^{n}=57\) and \(\sum _{n=0}^{\infty }…

\(\sum_{n=0}^{\infty} a r^n=57\)

\(\frac{a}{1-r}=57\) \(\ldots(i)\)

\(\sum_{n=0}^{\infty} a^3 r^{3 n}=9747\)

\(\frac{\mathrm{a}^3}{1-\mathrm{r}^3}=9747\) \(\ldots (ii)\)

from (i) and (ii)

\(\frac{\frac{a^3}{(1-r)^3}}{\frac{a^3}{1-r^3}}=\frac{57^3}{9747}=19\)

On solving, \(r=\frac{2}{3}\) and \(r=\frac{3}{2}\) (rejected)

\(a=19\)

\(\therefore a+18 r=19+18 \times \frac{2}{3}=31\)