Question

The weighted mean of first n natural numbers when their weights are equal to corresponding natural number, is

• A. $\frac{n+1}{2}$
• B. $\frac{2n+1}{3}$
• C. $\frac{(n+1)(2n+1)}{6}$
• D. None of these

Answer 1

Answer: (B)

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

Given, $x_i=1,2,3,..............,n$
and $w_i=1,2,3,..............n$

Thus required mean $=\frac{x_1{w}_1+x_2{w}_2+..............+x_n{w}_n}{w_1+w_2+...........+w_n}$
$=\frac{1^2+2^2+3^2+.............n^2}{1+2+3+.........+n}=\frac{\frac{1}{6}n(n+1)(2n+1)}{\frac{1}{2}n(n+1)}=\frac{1}{3}(2n+1)$