Question

The weighted mean of first n natural numbers whose weights are equal to the corresponding number is equal to

• A. $\frac{1}{3}(2n+1)$
• B. $\frac{1}{2}(n+4)$
• C. $\frac{1}{5}(2n+3)$
• D. None of these

Answer 1

The correct option is A $\frac{1}{3}(2n+1)$

$x_i\to 1,2,3,\mathrm{4...}n$
Similarily weight is
$f_i\to 1,2,3,4,...,n$ respectively
Weighted $AM$$=\frac{f_1{x}_1+f_2{x}_2+f_3{x}_3+.....+f_n{x}_n}{f_1+f_2+f_3+.....+f_n}=\frac{(1\times 1)+(2\times 2)+(3\times 3)+....+(n\times n)}{1+2+3+4+...+n}=\frac{\sum _{k=1}^n{k}^2}{\sum _{k=1}^{n}k}=\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}=\frac{2n+1}{3}$