Question

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

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

Answer 1

The correct option is D $-\frac{3}{2}$

This is an A.P. of common difference $-\frac{1}{6}$.
$\therefore Sum=\frac{9}{2}[2\times \frac{1}{2}+(9-1\left)(-\frac{1}{6})\right]=-\frac{3}{2}$.