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

Hint: Sum of an infinite geometric series is $\dfrac{a}{1-r}$ where $a$ is the first term and $r$ is the common difference.

Complete step-by-step answer:

We know that the sum of an infinite series of a geometric progression is given by the general formula $\dfrac{a}{1-r}$ where a is the first term of series, and r is the common ratio.

So, let the first term of series be a and the common ratio of series as r.

Then the sum of an infinite series of geometric progression $=\dfrac{a}{1-r}$.

We know that it is given in the question that the sum of an infinite geometric series is 162.

So,

\[\dfrac{a}{1-r}=162....\left( i \right)\]

Now it is given in the question that the sum of its first n terms is 160.

Now we know that the sum of n terms of a geometric progression is given by general formula i.e. $\dfrac{a(1-{{r}^{n}})}{1-r}$ .

So, we can say that

Sum of first $n$ terms of G.P $=160$

$\dfrac{a(1-{{r}^{n}})}{1-r}=160....\left( ii \right)$

Now we can substitute the value from equation (i) in equation (ii).

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

Now ,

$1-{{r}^{n}}=\dfrac{160}{162}$

$\begin{align}

  & {{r}^{n}}=1-\dfrac{160}{162} \\

 & {{r}^{n}}=\dfrac{162-160}{162} \\

 & {{r}^{n}}=\dfrac{2}{162} \\

 & {{r}^{n}}=\dfrac{1}{81} \\

\end{align}$

Now it is given in the question that $\dfrac{1}{r}$ is an integer.

So now we will reciprocal both the left-hand side of the equation and the right-hand side of the equation.

${{\left( \dfrac{1}{r} \right)}^{n}}=81$

We will find the values of n such that $\dfrac{1}{r}$ is an integer. We will find the value of $n$ using the hit and trial method.

If $n=1$ then $\dfrac{1}{r}=81$ which is an integer.

And if $n=2$ then$\dfrac{1}{r}=9$ which is an integer.

And if $n=3$ then $\dfrac{1}{r}$ is not an integer so we cannot take $n=3$ .

And if $n=4$ then $\dfrac{1}{r}=3$ which is an integer.

And so on we will have $\dfrac{1}{r}$ as an integer only at $n=1,2,4$

Possible value of $n=1,2,4$

And if $n=1$ then $r=\dfrac{1}{81}$ .

If $n=2$ then $r=\dfrac{1}{9}$ .

If$n=4$ then$r=\dfrac{1}{3}$ .

So possible value of $r=\dfrac{1}{81},\dfrac{1}{9},\dfrac{1}{3}$ .

Now, we need to calculate the value of the first term $a$ .

We use equation (i),

$\dfrac{a}{1-r}=162$

Substitute the value of $r$ in it.

When $r=\dfrac{1}{81}$ then $a=160$ .

When $r=\dfrac{1}{9}$ then $a=144$ .

When $r=\dfrac{1}{3}$ then $a=108$

So, possible values of $a=160,144,108$ .


Note: The formula for the sum of infinite series of a geometric progression $\dfrac{a}{1-r}$ is applicable only when $r$ is less than 1. And here we can observe that the values of $r$ we get are less than 1.