Mathematics - Sequence & Series Question with Solution | TestHub

Let the terms of infinite series are $a,ar,a{r}^2,a{r}^3,\dots$ So, $\frac{a}{1-r}=3$ Since, sum of cubes of its terms is $\frac{27}{19}$ that is sum of $a^3$, $a^3{r}^3,\dots \infty$ is $\frac{27}{19}$ So, $\frac{a^3}{1-r^3}=\frac{27}{19}$ $\Rightarrow \frac{a}{1-r}\times \frac{a^2}{\left(1+r^2+r\right)}=\frac{27}{19}$ $\Rightarrow \frac{9\left(1+r^2-2r\right)\times 3}{1+r^2+r}=\frac{27}{19}$ $\Rightarrow 6r^2-13r+6=0$ $\Rightarrow (3r-2)(2r-3)=0$ $\Rightarrow r=\frac{2}{3},$ or $\frac{3}{2}$ As $|r|<1$ So, $r=\frac{2}{3}$