Question

What is the weighted arithmetic mean of the first $n$ natural numbers whose weights are equal to the corresponding numbers?

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

Answer 1

Answer: (D)

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

Weight Arithmetic mean $=\frac{1\ast 1+2\ast 2+3\ast 3+....n^2}{1+2+3+....n}$

$=\frac{n(2n+1)(n+1)}{6\ast \frac{n(n+1)}{2}}=\frac{2n+1}{3}=\frac{1}{3}(2n+1)$.