Mathematics - Sequence & Series Question with Solution | TestHub

Let $a,ar,a{r}^2,a{r}^3,a{r}^4,a{r}^5$ be six terms of a G.P. where ' $a$ ' is first term and $\mathrm{r}$ is common ratio. According to given conditions, we have $a{r}^3-a=5\Rightarrow a\left(r^3-1\right)=52$ and $a+ar+a{r}^2=26$ $\Rightarrow a\left(1+r+r^2\right)=26$ To find: $a\left(1+r+r^2+r^3+r^4+r^5\right)$ Consider $$\begin{array}{rl}& a\left[1+r+r^2+r^3+r^4+r^5\right]\\ & =a\left[1+r+r^2+r^3\left(1+r+r^2\right)\right]\\ & =a\left[1+r+r^2\right]\left[1+r^3\right]\end{array}$$ Divide (1) by (2), we get $$\frac{r^3-1}{1+r+r^2}=2$$ we know $r^3-1=(r-1)\left(1+r+r^2\right)$ $$\begin{array}{rl}& \therefore r-1=2\Rightarrow r=3\text{ and }a=2\\ & \therefore a\left(1+r+r^2+r^3+r^4+r^5\right)\\ & =a\left(1+r+r^2\right)\left(1+r^3\right)\\ & =2(1+3+9)(1+27)\\ & =26\times 28=728\end{array}$$