Let a, a r, a r^2, ldots . . .. be an infinit | vedclass

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Question

$ \sum_{\mathrm{n}=0}^{\infty} \mathrm{ar}^{\mathrm{n}}=57 $

$ \mathrm{a}+\mathrm{ar}+\mathrm{ar}^2+\infty=57 $

$ \frac{\mathrm{a}}{1-\mathrm{r}

Vedclass · Accepted Answer

$31$

Vedclass · Answer

$ \sum_{\mathrm{n}=0}^{\infty} \mathrm{ar}^{\mathrm{n}}=57 $

$ \mathrm{a}+\mathrm{ar}+\mathrm{ar}^2+\infty=57 $

$ \frac{\mathrm{a}}{1-\mathrm{r}}=57 \ldots \ldots \ldots \ldots .(\mathrm{I}) $

$ \sum_{\mathrm{n}=0}^{\infty} \mathrm{a}^3 \mathrm{r}^{3 \mathrm{n}}=9747 $

$ \mathrm{a}^3+\mathrm{a}^3 \cdot \mathrm{r}^3+\mathrm{a}^3 \cdot \mathrm{r}^6+\ldots \ldots \ldots \infty=9746 $

$ \frac{\mathrm{a}^3}{1-\mathrm{r}^3}=9746 \ldots \ldots \ldots \ldots(\mathrm{II}) $

$ \frac{(\mathrm{I})^3}{(\mathrm{II})} \Rightarrow \frac{\mathrm{a}^3}{\frac{(1-\mathrm{r})^3}{\mathrm{a}^3}}=\frac{57^3}{9717}=19 $

$ 	ext { On solving, } \mathrm{r}=\frac{2}{3} 	ext { and } \mathrm{r}=\frac{3}{2}(	ext { rejected) } $

$ \mathrm{a}=19 $

$ 	herefore \mathrm{a}+18 \mathrm{r}=19+18 	imes \frac{2}{3}=31$

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

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