Find the average of squares of first 11 consecutive even numbers.

Average of their squares = \(\frac{ (2^2 + 4^2 + 6^2 + ...... 22^2)}{11}\)

Average of their squares = \(\frac{ (2^2 ) ( 1^2 + 2^2 + 3^2 + ...... 11^2)}{11}\)

Sum of squares of first n natural Number = \(\frac{ n (n + 1) (2n + 1) }{6}\)

Average of their squares = \(\frac{4 × 11 × 12 × 23}{11 × 6}\) =  184