We are given an arithmetic sequence $\{a_n\}$ and the sums of the first 4 terms $S_4 = 20$ and the first 2 terms $S_2 = 4$. We need to find the common difference $d$.

1. Problem Description

We are given an arithmetic sequence $\{a_n\}$ and the sums of the first 4 terms $S_4 = 20$ and the first 2 terms $S_2 = 4$. We need to find the common difference $d$.

2. Solution Steps

Let $a_1$ be the first term and $d$ be the common difference of the arithmetic sequence. The sum of the first $n$ terms of an arithmetic sequence is given by:

$S_n = \frac{n}{2}[2a_1 + (n-1)d]$

Using the given information, we have:

$S_4 = \frac{4}{2}[2a_1 + (4-1)d] = 2[2a_1 + 3d] = 4a_1 + 6d = 20$

$S_2 = \frac{2}{2}[2a_1 + (2-1)d] = 2a_1 + d = 4$

Dividing the first equation by 2, we get:

$2a_1 + 3d = 10$

We now have a system of two equations with two variables:

$2a_1 + d = 4$

$2a_1 + 3d = 10$

Subtracting the first equation from the second equation, we get:

$(2a_1 + 3d) - (2a_1 + d) = 10 - 4$

$2d = 6$

$d = 3$

3. Final Answer

The common difference $d = 3$.