Question

Prove "the mean of first n natural numbers is $\frac{n+2}{2}$".

Answer 1

We need to find mean of first $n$ natural numbers.

$meanofnumbers=\frac{sumofnnumbers}{n}-----\left(2\right)$

Sum of first $n$ natural numbers $=1+2+3+-----n$

the series is an $AP$ with $a=1$ and $d=1$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{n}{2}\left[2\right(1)+(n-1\left)1\right]=\frac{n}{2}[2+n-1]=\frac{n}{2}[n+1]$

$\therefore S_n=\frac{n(n+1)}{2}-----\left(1\right)$

From $\left(1\right)$ and $\left(2\right)$

mean of first $n$ natural numbers $=\frac{S_n}{n}$

$=\frac{n(n+1)}{2n}\Rightarrow \left[\frac{n+1}{2}\right]$

$\therefore$ $meanoffirstnnaturalnumbers=\left(\frac{n+1}{2}\right)$