The problem is about an arithmetic sequence $\{a_n\}$. Given that the sum of the first 3 terms is $S_3 = 6$ and the first term is $a_1 = 4$, we need to find the common difference $d$.

1. Problem Description

The problem is about an arithmetic sequence $\{a_n\}$. Given that the sum of the first 3 terms is $S_3 = 6$ and the first term is $a_1 = 4$, we need to find the common difference $d$.

2. Solution Steps

The sum of the first $n$ terms of an arithmetic sequence is given by the formula:

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

In this case, we have $n=3$, $S_3 = 6$, and $a_1 = 4$. Plugging these values into the formula, we get:

$6 = \frac{3}{2}(2(4) + (3-1)d)$

$6 = \frac{3}{2}(8 + 2d)$

Multiply both sides by 2/3:

$6 \cdot \frac{2}{3} = 8 + 2d$

$4 = 8 + 2d$

$2d = 4 - 8$

$2d = -4$

$d = -2$

3. Final Answer

The common difference $d$ is $-2$.