Question

The sum of infinity of the progression $9-3+1-\frac{1}{3}+...$ 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:

The given series is $9-3+1-\frac{1}{3}+..$

Therefore, the first term is $a=9$

The second term is $ar=-3$

Therefore the common ratio is $r=-\frac{1}{3}$

We know that the sum of infinite GP series is $=\frac{a}{1-r}$

Therefore the sum of the given series is

$$\begin{gathered} =\frac{9}{1-\left(-{\displaystyle \frac{1}{3}}\right)} \\ =\frac{27}{4} \end{gathered}$$

Hence, option C is correct.