Let a, a r, a r^2, ............ be an infinite G.P. If \sum\limits_{n=0}^{\infty} a r^n=57 and \sum\limits_{n=0}^{\infty} a^3 r^{3 n}=9747, then a+18 r is equal to

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Answer: C. 31 \begin{array}{ll} \sum_{n=0}^{\infty} a r^n=57 & \Rightarrow \frac{a}{1-r}=57 \quad 	ext{.... (i)}\ \sum_{n=0}^{\infty} a^3 r^{3 n}=9747 & \Rightarrow \frac{a^3}{1-r^3}=9747 \quad 	ext{.... (ii)} \end{array} \begin{aligned} & \frac{\left(1-r^3ight)}{(1-r)^3}=\frac{(57)^3}{9747}=19 \ & \Rightarrow \quad \frac{(1-r)\left(1+r+r^2ight)}{(1-r)^3}=19 \ & \Rightarrow \quad 18 r^2-39 r+18=0 \ & \Rightarrow \quad r=\frac{2}{3}, \frac{3}{2} 	ext { (rejected) } \ & 	herefore \quad a=19 \ & \quad a+18 r \ & \quad=19+12=31 \end{aligned}

Let $$a, a r, a r^2$$, ............ be an infinite G.P. If $$\sum\limits_{n=0}^{\infty} a r^n=57$$ and $$\sum\limits_{n=0}^{\infty} a^3 r^{3 n}=9747$$, then $$a+18 r$$ is equal to

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