Given an arithmetic sequence $\{a_n\}$ with $a_1 = 3$ and $a_5 = 11$, find the sum of the first 7 terms, $S_7$.

1. Problem Description

Given an arithmetic sequence $\{a_n\}$ with $a_1 = 3$ and $a_5 = 11$, find the sum of the first 7 terms, $S_7$.

2. Solution Steps

First, we need to find the common difference $d$ of the arithmetic sequence.

We know that $a_n = a_1 + (n-1)d$.

Since $a_5 = 11$ and $a_1 = 3$, we have

$a_5 = a_1 + (5-1)d$

$11 = 3 + 4d$

$4d = 11 - 3$

$4d = 8$

$d = \frac{8}{4}$

$d = 2$

Now that we have $a_1 = 3$ and $d = 2$, we can find $a_7$:

$a_7 = a_1 + (7-1)d = a_1 + 6d = 3 + 6(2) = 3 + 12 = 15$

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

$S_n = \frac{n(a_1 + a_n)}{2}$

So, the sum of the first 7 terms is

$S_7 = \frac{7(a_1 + a_7)}{2} = \frac{7(3 + 15)}{2} = \frac{7(18)}{2} = 7(9) = 63$

Alternatively, we can use the formula $S_n = \frac{n}{2}[2a_1 + (n-1)d]$. Then,

$S_7 = \frac{7}{2}[2a_1 + (7-1)d] = \frac{7}{2}[2(3) + 6(2)] = \frac{7}{2}[6 + 12] = \frac{7}{2}(18) = 7(9) = 63$

3. Final Answer

$S_7 = 63$