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

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Question

(D) The given series is $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}, \dots$ Here,the first term $a = \frac{1}{2}$. The common difference $d = \frac{1}{3} -

Vedclass · Accepted Answer

$-\frac{3}{2}$

Vedclass · Answer

(D) The given series is $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}, \dots$ Here,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 $S_n = \frac{n}{2} [2a + (n-1)d]$. Substituting the values: $S_9 = \frac{9}{2} [2(\frac{1}{2}) + (9-1)(-\frac{1}{6})]$ $S_9 = \frac{9}{2} [1 + 8(-\frac{1}{6})]$ $S_9 = \frac{9}{2} [1 - \frac{4}{3}]$ $S_9 = \frac{9}{2} [-\frac{1}{3}] = -\frac{3}{2}$.

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

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