Question:
The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $\frac{27}{19}$. Then the common ratio of this
series is
Correct Option: , 2
Solution:
$\frac{a}{1-r}=3$ ........(1)
$\frac{\mathrm{a}^{3}}{1-\mathrm{r}^{3}}=\frac{27}{19} \Rightarrow \frac{27(1-\mathrm{r})^{3}}{1-\mathrm{r}^{3}}=\frac{27}{19}$
$\Rightarrow 6 r^{2}-13 r+6=0$
$\Rightarrow r=\frac{2}{3}$ as $|r|<1$
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