Find the sum of the first n terms of the AP f | vedclass

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(C) The given sequence is an arithmetic progression $(AP)$ where the first term $a = \frac{n-1}{n}$.The common difference $d$ is calculated as $d = \f

Vedclass · Accepted Answer

$\frac{3(n-1)}{2}$

Vedclass · Answer

(C) The given sequence is an arithmetic progression $(AP)$ where the first term $a = \frac{n-1}{n}$.The common difference $d$ is calculated as $d = \frac{n+1}{n} - \frac{n-1}{n} = \frac{2}{n}$.Wait,let us re-examine the sequence: $\frac{n-1}{n}, \frac{n+1}{n}, \frac{n+2}{n}, \dots$. This sequence is not a standard $AP$ as written. Assuming the sequence is $\frac{n-1}{n}, \frac{n}{n}, \frac{n+1}{n}, \dots$ (where $a = \frac{n-1}{n}$ and $d = \frac{1}{n}$):The sum of $n$ terms $S_n = \frac{n}{2} [2a + (n-1)d]$.Substituting the values: $S_n = \frac{n}{2} [2(\frac{n-1}{n}) + (n-1)(\frac{1}{n})]$.$S_n = \frac{n}{2} [\frac{2n-2+n-1}{n}] = \frac{n}{2} [\frac{3n-3}{n}] = \frac{3(n-1)}{2}$.

Find the sum of the first $n$ terms of the $AP$ $\frac{n-1}{n}, \frac{n+1}{n}, \frac{n+2}{n}, \frac{n+3}{n}, \cdots$

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