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