Solution

Given
$A_1={3}$,
$A_2={5,7,9}$,
$A_3={11,13,15,17,19}$,
$A_4={21,23,25,27,29,31,33}$, …

Observation:

• $A_k$ contains $2k-1$ numbers
• All elements are consecutive odd numbers
• So the average of $A_k$ is simply its middle term

First term of $A_k$:
$3,5,11,21,\dots$

This follows
$a_k = 3 + 2(k-1)^2$

Middle term of $A_k$:
$\text{Average} = a_k + (2k-2)$

So,
$\text{Average} = 3 + 2(k-1)^2 + 2(k-1)$
$\text{Average} = 3 + 2k(k-1)$

For $k=20$:
$\text{Average} = 3 + 2\times20\times19$
$\text{Average} = 3 + 760$
$\boxed{763}$