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  4. The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is (27/19). Then the common ratio of this series is :

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Q. 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 :

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Solution:

$\frac{a}{1-r} = 3$
$ \frac{a^{3}}{1-r^{3}} = \frac{27}{19} \Rightarrow \frac{27\left(1-r\right)^{3}}{1-r^{3}} = \frac{27}{19} $
$ \Rightarrow 6r^{2} -13r+6 =0 $
$\Rightarrow r = \frac{2}{3}$ as $\left|r\right|<1$