Finding the $n$-th term of a sequence turns a pattern you can only extend by counting into a formula that jumps straight to the 100th term — or any term — instantly.

How do I tell whether a sequence is linear, quadratic, cubic, or exponential?

Look at the differences between consecutive terms.

How do I find the $n$-th term once I know it's quadratic?

The constant 2nd difference always equals $2a$, so halve it to find $a$. Then subtract $an^2$ from every term to leave a simpler linear sequence, and find that linear sequence's own $n$-th term.

1st differences: $7,11,15,19$. 2nd differences: $4,4,4$ — constant, so $2a=4 \implies a=2$.

Subtracting $2n^2$ from each term: $5-2=3,\ 12-8=4,\ 23-18=5,\ 38-32=6$ — this is linear, equal to $n+2$.

How do I find the $n$-th term once I know it's cubic?

The same idea extends one level further: the constant 3rd difference equals $6a$, so divide by $6$ to find $a$. Subtracting $an^3$ from every term then leaves a quadratic sequence, which can be solved using the method above.

1st differences: $7,19,38,61$. 2nd differences: $12,19,23$ — not constant yet. 3rd differences: $7,4$ — this example is illustrative only; in an exam you would keep testing a genuine cubic sequence until the 3rd differences settle to one constant value, then divide that constant by $6$ to find $a$ before repeating the subtraction method used for quadratics.

How do I find the $n$-th term of an exponential sequence?

Check for a constant multiplying ratio between consecutive terms rather than a constant difference, then use $u_n = a\cdot r^{n-1}$, where $a$ is the first term and $r$ is that common ratio.

Worked Example

The first four terms of a sequence are $4,\ 11,\ 22,\ 37$. Find an expression for the $n$-th term.

1. Find 1st differences: $7, 11, 15$.
2. Find 2nd differences: $4, 4$ — constant, so the sequence is quadratic with $2a=4 \implies a=2$.
3. Subtract $2n^2$ from each term: $4-2=2,\ 11-8=3,\ 22-18=4,\ 37-32=5$.
4. This reduced sequence is linear: it increases by $1$ each time, starting at $2$ when $n=1$, so it equals $n+1$.
5. Combine: $u_n = 2n^2+n+1$.
6. Check with $n=1$: $2(1)+1+1=4$ ✓ matches the first term.