Question

The sum of the series $5-\frac{10}{3}+\frac{20}{9}-\frac{40}{27}+....$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $None\, of\, these$

Answer 1

Answer: (C)

$3$

Answer 2

Let $S=5-\frac{10}{3}+\frac{20}{9}-\frac{40}{27}+...$ ..(i)
$\frac{1}{3}S=\frac{5}{3}-\frac{10}{9}+\frac{20}{27}-.....$ ..(ii)
On adding Eqs. (i) and (ii), we get
$\frac{4}{3}S=5-\frac{5}{3}+\frac{10}{9}-\frac{20}{27}$
$=5-\frac{5}{3}\left( 1-\frac{2}{3}+\frac{4}{9}-.... \right)$
$=5-\frac{5}{3}\left( \frac{1}{1+\frac{2}{3}} \right)$
$=5-\frac{5}{3}\left( \frac{3}{5} \right)$
$\Rightarrow$ $\frac{4}{3}S=4$
$\Rightarrow$ $S=3$