Question

Let the sum of an infinite geometric progression is 162 and the sum of its first n terms is 160. Also the reciprocal of its common ratio is an integer
The possible value of the first term of the geometric progression can be

• A. 108
• B. 144
• C. 180
• D. 216

Answer 1

Answer: (A), (B), (C), (D)

We have $a+ar+a{r}^2+\dots \dots +a{r}^{n-1}+\dots \dots ..\infty$
Now, $\frac{a}{1-r}=162$ (given)$\dots .$. (1)
Also, $\frac{a\left(1-r^n\right)}{1-r}=160$ ....(2)
$\therefore (2)÷(1)$
$\Rightarrow 1-r^n=\frac{160}{162}=\frac{80}{81}\Rightarrow r^n=\frac{1}{81}\Rightarrow n=1$ or 2 or 4 so that reciprocal of $r$ is an integer.

If $r=\frac{1}{81}$, then $a=160;$ If $r=\frac{1}{9}$, then $a=144$.
If $r=\frac{-1}{9}$, then $a=180;$ If $r=\frac{1}{3}$, then $a=108$.
If $r=\frac{-1}{3}$, then $a=216$.