Question

The sum up to $9$ terms of the series $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+...$ is

• A. $-\frac{5}{6}$
• B. $-\frac{1}{2}$
• C. $1$
• D. $-\frac{3}{2}$

Answer 1

Answer: (D)

$-\frac{3}{2}$

Given that series is an A.P with first term, $a=\frac{1}{2}$ and
Common difference, $d=\frac{1}{3}-\frac{1}{2}=\frac{1}{6}-\frac{1}{3}=\frac{-1}{6}$
So using sum of A.P formula $S_n=\frac{n}{2}(2a+(n-1\left)d\right)$
Here $n=9\Rightarrow S_9=\frac{9}{2}(2.\frac{1}{2}+(9-1).\frac{-1}{6})=\frac{-3}{2}$