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, find all possible values of the common ratio, $n$ and the first term of the series.

Answer 1

Given that sum of an infinite geometric series is $162$

We know that sum of an infinite geometric series is $\frac{a}{1-r}$

where $a=$ first term and $r=$ common ratio

Therefore, $\frac{a}{1-r}=162$ ------(1)

Given that sum of first $n$ terms is $160$

We know that sum of first $n$ terms is $\frac{a(1-r^n)}{1-r}$

Therefore, $\frac{a(1-r^n)}{1-r}=160$ ------(2)

Dividing (2) by (1) we get

$1-r^n=\frac{160}{162}=\frac{80}{81}$

$⟹r^n=1-\frac{80}{81}$

$⟹r^n=\frac{1}{81}$

$⟹(\frac{1}{r}{)}^n=81$

Given that $\frac{1}{r}$ is an integer and $n$ is also an integer.

Therefore, $\frac{1}{r}=81,9$ or $3$ for $n=1,2$ or $4$

substituting $r$ values in (1) we get

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

$⟹a=160,144$ or $108$