Question

The sum to infinity of the progression $9-3+1-\frac{1}{3}+\dots$ is

• A. $9$
• B. $\frac{9}{2}$
• C. $\frac{27}{4}$
• D. $\frac{15}{2}$

Answer 1

The correct option is C

$\frac{27}{4}$

Explanation for the correct option:

Given series is, $9-3+1-\frac{1}{3}+\dots$.

We can observe that this is a geometric progression.

A geometric progression is a sequence of numbers where consecutive pairs of numbers have a constant ratio.

The general form is,

$a,ar,a{r}^2,a{r}^3,\dots$

Here, the first term $a=9$ and the common ratio $r=\frac{-3}{9}=-\frac{1}{3}$.

The sum of a geometric series is given as,

$S=\frac{a}{1-r},r<1$

Thus, the sum of the given geometric series is,

$\begin{array}{rcl}S& =& \frac{9}{1-\left(-{\displaystyle \frac{1}{3}}\right)}\\ & =& \frac{9}{1+{\displaystyle \frac{1}{3}}}\\ & =& \frac{9}{{\displaystyle \frac{4}{3}}}\\ & =& \frac{27}{4}\end{array}$

Hence, $\mathrm{Option}\left(\mathrm{C}\right)$ is correct.