Determine the common difference, the general formula, and the rule for the given arithmetic sequence: 5, 9, 13, 17, 21.

The terms are 5, 9, 13, 17, 21.

The difference between the consecutive terms is 4.

The general form for an arithmetic sequence is $T_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference. Here, $a=5$ and $d=4$.
The 1st term = 5
The 2nd term = $5 + (2-1) \times 4 = 5 + 1 \times 4$
The 3rd term = $5 + (3-1) \times 4 = 5 + 2 \times 4$
The 4th term = $5 + (4-1) \times 4 = 5 + 3 \times 4$
The 5th term = $5 + (5-1) \times 4 = 5 + 4 \times 4$
The 6th term = $5 + (6-1) \times 4 = 5 + 5 \times 4$

The completed expressions are:
The 1st term = 5
The 2nd term = $5 + 1 \times \boxed{4}$
The 3rd term = $5 + \boxed{2} \times \boxed{4}$
The 4th term = $5 + \boxed{3} \times \boxed{4}$
The 5th term = $5 + \boxed{4} \times \boxed{4}$
The 6th term = $5 + \boxed{5} \times \boxed{4}$

The rule for finding a term in any position in this sequence is to start with the first term (5), then add the product of the common difference (4) and one less than the term's position.
This can be written as $T_n = 5 + (n-1) \times 4$, where $T_n$ is the term at position $n$.