The mean of the first n natural numbers is

\frac{n}{2}

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\frac{n + 1}{2}

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\frac{n}{2} + 1

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\frac{n^{2} + n + 1}{n}

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Solution

The correct option is B $\frac{n + 1}{2}$
We know that sum of first n natural numbers is $\frac{n (n + 1)}{2}$
So mean of the first n natural numbers = $\frac{s u m o f n n a t u r a l n u m b e r s}{n u m b e r} = \frac{\frac{n (n + 1)}{2}}{n} = \frac{(n + 1)}{2}$

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Prove that: 1 + 2 + 3 + ......... + n = $\frac{n (n + 1)}{2}$ i.e., the sum of the first n natural numbers is $\frac{n (n + 1)}{2}$ .

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