The sum of the series frac{1}{2} + frac{1}{3} | vedclass

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Question

(d) Given series $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + ........$

Here $a = \frac{1}{2}$, common difference $d = - \frac{1}{6}$ and $n = 9$.

Vedclass · Accepted Answer

$ - \frac{3}{2}$

Vedclass · Answer

(d) Given series $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + ........$

Here $a = \frac{1}{2}$, common difference $d = - \frac{1}{6}$ and $n = 9$.

$ \Rightarrow $ ${S_9} = \frac{9}{2}\left[ {2 	imes \frac{1}{2} + (9 - 1)\left( { - \frac{1}{6}} ight)} ight] = - \frac{3}{2}$.

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

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