The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is…
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:
$\frac{1}{3}$
$\frac{2}{3}$
$\frac{2}{9}$
$\frac{4}{9}$
Solution
Let the terms of infinite series are $a, a r, a r^{2}, a r^{3}, \ldots$
So, $\quad \frac{a}{1-r}=3$
Since, sum of cubes of its terms is $\frac{27}{19}$ that is sum of $a^{3}$,
$a^{3} r^{3}, \ldots \infty$ is $\frac{27}{19}$
So, $\frac{a^{3}}{1-r^{3}}=\frac{27}{19}$
$\Rightarrow \quad \frac{a}{1-r} \times \frac{a^{2}}{\left(1+r^{2}+r\right)}=\frac{27}{19}$
$\Rightarrow \frac{9\left(1+r^{2}-2 r\right) \times 3}{1+r^{2}+r}=\frac{27}{19}$
$\Rightarrow \quad 6 r^{2}-13 r+6=0$
$\Rightarrow(3 r-2)(2 r-3)=0$
$\Rightarrow r=\frac{2}{3},$ or $\frac{3}{2}$
As $|r| < 1$
So, $r=\frac{2}{3}$