Find the average of cubes of first 49 positive integers.

Answer: Option A

Sum of cubes of first $$n$$ positive consecutive numbers is
$$\begin{array}{rl}& =\frac{{\left\{n\left(n+1\right)\right\}}^2}{4}\\ & \therefore \text{Average}=\frac{n{\left(n+1\right)}^2}{4}\\ & \Rightarrow n=49\\ & \Rightarrow \frac{49{\left(50\right)}^2}{4}=30625\end{array}$$