I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.
First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If [tex]a_n[/tex] is the [tex]n[/tex]th term in the sequence, then the next term [tex]a_{n+1}[/tex] is a fixed constant (the common difference [tex]d[/tex]) added to the previous term. As a recursive formula, that's
[tex]a_{n+1}=a_n+d[/tex]
This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since [tex]a_{n+1}=a_n+d[/tex], this means that [tex]a_n=a_{n-1}+d[/tex], so you plug this into the recursive formula and end up with
[tex]a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d[/tex]
You can continue in this pattern, since every term in the sequence follows this rule:
[tex]a_{n+1}=a_{n-1}+2d[/tex]
[tex]a_{n+1}=(a_{n-2}+d)+2d[/tex]
[tex]a_{n+1}=a_{n-2}+3d[/tex]
[tex]a_{n+1}=(a_{n-3}+d)+3d[/tex]
[tex]a_{n+1}=a_{n-3}+4d[/tex]
and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to [tex]n+1[/tex]. You have, for example, [tex](n-2)+3=n+1[/tex] in the third equation above.
Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one [tex]a_1[/tex]. In order for the pattern mentioned above to hold, you would end up with
[tex]a_{n+1}=a_1+nd[/tex]
or, shifting the index by one so that the formula gives the [tex]n[/tex]th term explicitly,
[tex]a_n=a_1+(n-1)d[/tex]
Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio [tex]r[/tex] between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,
[tex]a_{n+1}=ra_n[/tex]
Well, since [tex]a_n[/tex] is just the term after [tex]a_{n-1}[/tex] scaled by [tex]r[/tex], you can write
[tex]a_{n+1}=r(ra_{n-1})=r^2a_{n-1}[/tex]
Doing this again and again, you'll see a similar pattern emerge:
[tex]a_{n+1}=r^2a_{n-1}[/tex]
[tex]a_{n+1}=r^2(ra_{n-2})[/tex]
[tex]a_{n+1}=r^3a_{n-2}[/tex]
[tex]a_{n+1}=r^3(ra_{n-3})[/tex]
[tex]a_{n+1}=r^4a_{n-3}[/tex]
and so on. Notice that the subscript and the exponent of the common ratio both add up to [tex]n+1[/tex]. For instance, in the third equation, [tex]3+(n-2)=n+1[/tex]. Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:
[tex]a_{n+1}=r^na_1[/tex]
or, to give the formula for [tex]a_n[/tex] explicitly,
[tex]a_n=r^{n-1}a_1[/tex]
