Question

The sum to infinite terms of the arithmetic-geometric progression $3,4,4,\frac{32}{9},....$ is equal to

• A. $27$
• B. $30$
• C. $24$
• D. $25$

Answer 1

Answer: (A)

Let the A.G.P. be $3,\left(3+d\right)r,\left(3+2d\right)r^2....$
So, $\left(3+d\right)r=4$ and $\left(3+2d\right)r^2=4$
$\frac{{\left(3+d\right)}^2{r}^2}{\left(3+2d\right)r^2}=\frac{16}{4}$
$\Rightarrow 9+d^2+6d=12+8d$
$\Rightarrow d^2-2d-3=0$
$\Rightarrow \left(d+1\right)\left(d-3\right)=0\Rightarrow d=-1,3$
$r=\frac{4}{3+d}\Rightarrow r=\frac{4}{3-1},\frac{4}{3+3}=2,\frac{2}{3}$
But $\left|r\right|<1\Rightarrow d=3$ and $r=\frac{2}{3}$
$S_{\in fty}=\frac{a}{1-r}+\frac{dr}{{\left(1-r\right)}^2}=\frac{3}{1-\frac{2}{3}}+\frac{3\times \frac{2}{3}}{{\left(1-\frac{2}{3}\right)}^2}$
$=9+18=27$