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

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(D) The given series is $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \dots$ to $9$ terms.This is an arithmetic progression where the first term $a = \fr

Vedclass · Accepted Answer

$-\frac{3}{2}$

Vedclass · Answer

(D) The given series is $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \dots$ to $9$ terms.This is an arithmetic progression where the first term $a = \frac{1}{2}$.The common difference $d = \frac{1}{3} - \frac{1}{2} = \frac{2-3}{6} = -\frac{1}{6}$.The number of terms $n = 9$.The sum of an arithmetic progression is given by the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$.Substituting the values: $S_9 = \frac{9}{2} [2(\frac{1}{2}) + (9 - 1)(-\frac{1}{6})]$.$S_9 = \frac{9}{2} [1 + 8(-\frac{1}{6})] = \frac{9}{2} [1 - \frac{8}{6}] = \frac{9}{2} [1 - \frac{4}{3}]$.$S_9 = \frac{9}{2} [-\frac{1}{3}] = -\frac{9}{6} = -\frac{3}{2}$.

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

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