Question

The sum of an infinite geometric series is $162$ and the sum of its first $n$ terms is $160.$ If the inverse of its common ratio is an integer, then which of the following is not a possible first term?

• A. $108$
• B. $144$
• C. $160$
• D. none of these

Answer 1

The correct option is D none of these

Given,

$S_{\infty }=\frac{a}{1-r}=162...\left(i\right)S_n=\frac{a(1-r^n)}{1-r}=160...\left(ii\right)$

Dividing equation $\left(ii\right)$ by $\left(i\right)$,

$1-r^n=\frac{160}{162}=\frac{80}{81}\Rightarrow 1-\frac{80}{81}=r^n\Rightarrow r^n=\frac{1}{81}\Rightarrow {\left(\frac{1}{r}\right)}^n=81$

Now, it is given that $\frac{1}{r}$ is an integer and $n$ is also an integer.

Hence, $\frac{1}{r}=3,9,81$ so that $n=4,2,1$.

$\therefore$ $a=162(1-\frac{1}{3})$ or $162(1-\frac{1}{9})$ or $162(1-\frac{1}{81})$

$=108$ or $144$ or $160$

Hence, the option (D) is correct.