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ધારો કે $a, a r, a r^2$, ......... એક સમગુણોતર શ્રેણી છે. જો $\sum_{n=0}^{\infty} a r^n=57$ અને $\sum_{n=0}^{\infty} a^3 r^{3 n}=9747$ હોય, તો $a+18 r=$ ..........
$27$
$46$
$38$
$31$
Solution
$ \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 $
$ \text { On solving, } \mathrm{r}=\frac{2}{3} \text { and } \mathrm{r}=\frac{3}{2}(\text { rejected) } $
$ \mathrm{a}=19 $
$ \therefore \mathrm{a}+18 \mathrm{r}=19+18 \times \frac{2}{3}=31$
