Find Second Term of Arithmetic Progression Independent of Ratio

Let the given arithmetic progression have the first term $a_1 = 3$ and common difference $d$.

The sum of the first $p$ terms of the arithmetic progression is given by:
$S_p = \sum_{i=1}^p a_i = \frac{p}{2} \left[ 2a_1 + (p-1)d \right] = \frac{p}{2} \left[ 6 + (p-1)d \right]$

For $m = 5n$, we have:
$S_m = S_{5n} = \frac{5n}{2} \left[ 6 + (5n-1)d \right]$
$S_n = \frac{n}{2} \left[ 6 + (n-1)d \right]$

Now, consider the ratio $\frac{S_m}{S_n}$:
$\frac{S_m}{S_n} = \frac{\frac{5n}{2} \left[ (6-d) + 5nd \right]}{\frac{n}{2} \left[ (6-d) + nd \right]} = 5 \cdot \frac{(6-d) + 5nd}{(6-d) + nd}$

For this ratio to be independent of $n$, the expression $\frac{(6-d) + 5nd}{(6-d) + nd}$ must be a constant for all $1 \le n \le 20$.

This occurs under two possible conditions:

1. Case 1: $6 - d = 0$
   $d = 6$
   Then,
   $\frac{S_m}{S_n} = 5 \cdot \frac{5nd}{nd} = 25 \quad (\text{independent of } n)$
   The second term is:
   $a_2 = a_1 + d = 3 + 6 = 9$

2. Case 2: $d = 0$
   Then,
   $\frac{S_m}{S_n} = 5 \cdot \frac{6 - 0 + 0}{6 - 0 + 0} = 5 \quad (\text{independent of } n)$
   The second term is:
   $a_2 = a_1 + d = 3 + 0 = 3$

Thus, the possible values for $a_2$ are $3$ or $9$.