SSC CGL 2022 PYQ — Question 72

The correct answer is Option 3. i.e. ${2n + 1} \over 3$.

We know that,

The sum of squares of the first 'n' natural numbers = ${n(n + 1)(2n + 1)}\over 6$

The sum of the first 'n' natural numbers = ${n(n + 1)}\over 2$

Now,

First 'n' natural numbers = 1, 2, 3, ......,n

Their corresponding weights = 1, 2, 3, ......,n

Now, weighted arithmetic mean = ${(1 \times 1 + 2 \times 2 + 3 \times 3 + .....+ n \times n)} \over {(1 + 2+ 3+ ....+ n)}$

⇒ ${1 ^2 + 2^2 + 3^2 + ....+ n^2} \over 1 + 2 +3+....+ n$ 

⇒ ${{n(n + 1)(2n + 1)}\over 6} \over {{n(n+1)}\over 2}$

⇒ ${2n + 1} \over 3$

Hence, the weighted arithmetic mean of the first ‘n’ natural numbers is ${2n + 1} \over 3$ .