Given an arithmetic sequence $\{a_n\}$, we are given that $a_1 = 2$ and $a_2 + a_3 = 13$. We are asked to find $S_6$, which is the sum of the first 6 terms of the sequence.

1. Problem Description

Given an arithmetic sequence $\{a_n\}$, we are given that $a_1 = 2$ and $a_2 + a_3 = 13$. We are asked to find $S_6$, which is the sum of the first 6 terms of the sequence.

2. Solution Steps

Since $\{a_n\}$ is an arithmetic sequence, we can write $a_n = a_1 + (n-1)d$, where $d$ is the common difference.

We are given $a_1 = 2$.

We are also given $a_2 + a_3 = 13$.

Expressing $a_2$ and $a_3$ in terms of $a_1$ and $d$:

$a_2 = a_1 + d = 2 + d$

$a_3 = a_1 + 2d = 2 + 2d$

Therefore, $a_2 + a_3 = (2+d) + (2+2d) = 4 + 3d = 13$.

Solving for $d$:

$3d = 13 - 4 = 9$

$d = 3$

Now we know $a_1 = 2$ and $d = 3$.

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

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

We want to find $S_6$, so we plug in $n=6$, $a_1=2$, and $d=3$:

$S_6 = \frac{6}{2} [2(2) + (6-1)(3)] = 3 [4 + 5(3)] = 3 [4 + 15] = 3(19) = 57$

3. Final Answer

57