Question

What is the average of the squares of the first $19$ natural numbers?

• A. $124$
• B. $127.5$
• C. $130$
• D. $133.5$

Answer 1

Answer: (C)

$130$

$1^2+2^2+3^2+4^2+...............+n^2=\frac{n(n+1)(2n+1)}{6}{1}^2+2^2+3^2+4^2+...............+{19}^2=\frac{19\times 20\times 39}{6}=2470$

average$=\frac{2470}{19}=130$

so option C is correct