Each term of the sequence $5, 12, 19, 26, \cdots$ is $7$ more than the term that precedes it. What is the first term of the sequence that is greater than $2017$?

$\text{(A) }2018\qquad\text{(B) }2019\qquad\text{(C) }2020\qquad\text{(D) }2021\qquad\text{(E) }2022$

1. **Identify the sequence and its properties:**
   The given sequence is \(5, 12, 19, 26, \ldots\). This is an arithmetic sequence where the first term \(a_1 = 5\) and the common difference \(d = 7\).

2. **General formula for the \(n\)-th term of an arithmetic sequence:**
   The \(n\)-th term \(a_n\) of an arithmetic sequence can be expressed as:
   \[
   a_n = a_1 + (n-1)d
   \]
   Substituting the given values \(a_1 = 5\) and \(d = 7\), we get:
   \[
   a_n = 5 + (n-1) \cdot 7
   \]
   Simplifying, we have:
   \[
   a_n = 5 + 7n - 7 = 7n - 2
   \]

3. **Set up the inequality to find the smallest \(n\) such that \(a_n > 2017\):**
   \[
   7n - 2 > 2017
   \]
   Solving for \(n\):
   \[
   7n > 2019
   \]
   \[
   n > \frac{2019}{7}
   \]
   \[
   n > 288.428571
   \]
   Since \(n\) must be an integer, we take the smallest integer greater than \(288.428571\), which is \(n = 289\).

4. **Calculate the term \(a_{289}\):**
   \[
   a_{289} = 7 \cdot 289 - 2
   \]
   \[
   a_{289} = 2023 - 2
   \]
   \[
   a_{289} = 2021
   \]

5. **Verify that \(a_{289} > 2017\):**
   \[
   2021 > 2017
   \]
   This confirms that \(a_{289} = 2021\) is indeed the first term greater than 2017.

The final answer is \(\boxed{2021}\).