Question

The sum of the series $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+.......$ to 9 terms be $k$. Find $-2k$.

Answer 1

Given series :-

$\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+--------9terms$

Now,

$\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=\frac{-1}{6}$

again

$\frac{1}{6}-\frac{1}{3}=\frac{1}{6}-\frac{2}{6}=\frac{-1}{6}$

$\therefore \frac{1}{3}-\frac{1}{2}=\frac{1}{6}-\frac{1}{3}=\frac{-1}{6}=d$

So, the given series is a A.P.

$a=1stterm=\frac{1}{2}$

$\therefore S_9=\frac{9}{2}[2\times \frac{1}{2}+(9-1)\times \frac{-1}{6}]$

$=\frac{9}{2}[1++8\times \frac{-1}{6}]$

$=\frac{9}{2}[1-\frac{4}{3}]=\frac{9}{2}\left[\frac{3-4}{3}\right]=\frac{9}{2}\times \frac{-1}{3}=\frac{-3}{2}$

$\therefore S_9=\frac{-3}{2}$

$A.T.Q.S_9=K=\frac{-3}{2}$

$\therefore -2k=-2\times \frac{-3}{2}=-1\times -3=3$

$\therefore -2K=3$