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The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of

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 

  1. $\frac{4}{9}$

  2. $\frac{2}{9}$

  3. $\frac{2}{3}$

  4. $\frac{1}{3}$


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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