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 $a+a r+a r^2+\ldots \ldots+a r^{n-1}+\ldots \ldots . . \infty$
Now, $\frac{a}{1-r}=162$ (given)$\ldots .$. (1)
Also, $\frac{a\left(1-r^n\right)}{1-r}=160$ ....(2)
$\therefore(2) \div(1)$
$\Rightarrow 1- r ^{ n }=\frac{160}{162}=\frac{80}{81} \Rightarrow r ^{ n }=\frac{1}{81} \Rightarrow n =1$ 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 $a=216$.