Question

The weighted arithmetic mean ofthe first $n$ natural numbers whose weights are equal to the corresponding numbers is given by,

• A. $\frac{1}{2}(n+1)$
• B. $\frac{1}{2}(n+2)$
• C. $\frac{1}{3}(2n+1)$
• D. $\frac{1}{3}n(2n+1)$

Answer 1

Answer: (C)

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

$x_i$	$x$	Weight $\left(w\right)$
$x_1$	$1$	$1$
$x_2$	$2$	$2$
$x_3$ .... ....	$3$	$3$
$x_n$	$n$	$n$

First $n$ natural numbers are written in the table along with their respective weight.

Weighted arithmetic mean $=\frac{{\sum }_n^{r-1}{w}_r{x}_r}{{\sum }_n^{r-1}{w}_r}$

$=\frac{1\times 1+2\times 2+3\times 3+...+n\times n}{1+2+3+4+...+n}$

$=\frac{1^2+2^2+3^2+...+n^2}{1+2+3+4+...+n}$

Using formula of sum of series, we get

$=\frac{n(n+1)(2n+1)}{6}\times \frac{2}{n(n+1)}$

$=\frac{(2n+1)}{3}$

Hence, this is the required answer.