Short answer
A quadratic sequence has an nth term of the form an² + bn + c, where the second differences between consecutive terms are constant. Recognising this pattern is the first skill; finding a, b and c from it is the second. This guide shows you both, step by step, using a single worked example.
At a glance
- Key stage
- GCSE
- Subject
- Algebra
- Type
- How-to guide
- For
- Students
- Read time
- 5 min
- Last updated
- 14 October 2026
Where this fits
- Key Stage 3Years 7–9
- GCSEYears 10–11This article
Method at a glance
- Write out the sequence and find the first differences
- Find the second differences
- Find a
- Subtract an² from each term
- Find the nth term of the linear sequence
- Combine the parts
What makes a sequence quadratic?
In a linear sequence the differences between consecutive terms are constant. In a quadratic sequence the first differences change, but the second differences — the differences between the first differences — are constant. That constant second difference is always equal to 2a, where a is the coefficient of n² in the nth term formula.
For example, if the second differences are all 4, then 2a = 4, so a = 2 and your nth term starts with 2n².
What is the nth term formula for a quadratic sequence?
Every quadratic sequence has an nth term of the form:
T(n) = an² + bn + c
Your job is to find the three constants a, b and c. The second-differences method gives you a directly, and then you use simultaneous-style reasoning to find b and c.
How do you find the nth term of a quadratic sequence step by step?
Work through these steps in order:
- Write out the sequence and find the first differences. Subtract each term from the next: T(2) − T(1), T(3) − T(2), and so on.
- Find the second differences. Subtract each first difference from the next. Check they are all equal — if they are not, the sequence is not quadratic.
- Find a. Divide the constant second difference by 2: a = second difference ÷ 2.
- Subtract an² from each term. Create a new sequence by computing T(n) − an² for n = 1, 2, 3, 4, 5. This leaves a linear sequence.
- Find the nth term of the linear sequence. It has the form bn + c; use the standard method (first term minus common difference to find c, common difference to find b).
- Combine the parts. Write the full nth term as an² + bn + c and verify by substituting n = 1 and n = 5 back in.
Worked example: finding the nth term
Find the nth term of the sequence: 5, 14, 27, 44, 65
Step 1 — First differences:
14 − 5 = 9, 27 − 14 = 13, 44 − 27 = 17, 65 − 44 = 21
Step 2 — Second differences:
13 − 9 = 4, 17 − 13 = 4, 21 − 17 = 4. The second difference is 4.
Step 3 — Find a:
a = 4 ÷ 2 = 2
Step 4 — Subtract 2n² from each term:
| n | T(n) | 2n² | T(n) − 2n² |
|---|---|---|---|
| 1 | 5 | 2 | 3 |
| 2 | 14 | 8 | 6 |
| 3 | 27 | 18 | 9 |
| 4 | 44 | 32 | 12 |
| 5 | 65 | 50 | 15 |
The residual sequence is 3, 6, 9, 12, 15 — a linear sequence with common difference 3 and first term 3.
Step 5 — Find the linear nth term:
Residual nth term = 3n + 0, so b = 3 and c = 0.
Step 6 — Write the full nth term:
T(n) = 2n² + 3n
Verification: T(1) = 2 + 3 = 5 ✓ T(5) = 50 + 15 = 65 ✓
What if c is not zero?
Most GCSE questions produce a non-zero value of c. In the example above, if the residual sequence had been 5, 8, 11, 14, 17, the nth term of that residual would be 3n + 2, giving a final answer of 2n² + 3n + 2. Always read off the linear nth term carefully rather than assuming c = 0.
How do you verify your nth term formula?
Always check at least two values — preferably n = 1 and the last term given. Substitute each into your formula and confirm the outputs match the sequence. If n = 1 works but n = 5 does not, you have likely made an error in finding b or c rather than a.
A quick sanity check: the second differences of the sequence produced by your formula should equal the original second difference. For 2n² + 3n, the second differences are always 4 — matching what you found in step 2.
What types of quadratic sequence questions appear in GCSE exams?
Typical exam questions ask you to:
- Find the nth term, given the first five terms.
- Find a specific term (e.g. the 20th term) once you have the nth term.
- Decide whether a given number (e.g. 200) is in the sequence, by setting T(n) = 200 and solving the resulting quadratic equation.
- Show that a sequence is quadratic, by demonstrating that second differences are constant.
For the "is 200 in the sequence?" type, set your nth term equal to 200 and solve. If the solution is a positive integer, the number is in the sequence; if it is a fraction or negative, it is not.
Frequently asked questions
Q: What if the second differences are not constant?
If the second differences are not equal, the sequence is not quadratic. Check your arithmetic first. If the second differences are themselves constant — i.e. the third differences are constant — the sequence is cubic and beyond standard GCSE scope.
Q: Can the coefficient a be a fraction?
Yes. If the second difference is an odd number such as 3, then a = 3/2. This is more common on Higher tier papers. The method is identical; you simply carry the fraction through the calculation.
Q: How is a quadratic sequence different from a quadratic equation?
A quadratic sequence is a list of numbers whose nth term follows a quadratic rule. A quadratic equation (e.g. x² + 3x − 10 = 0) is something you solve to find values of x. The link is that you often set a sequence's nth term equal to a number and solve a quadratic equation to answer questions about membership.
Q: Is this topic on Foundation tier?
Identifying and working with quadratic sequences, including finding the nth term, is assessed at Higher tier. Foundation tier focuses on linear sequences. However, Foundation students may be asked to recognise that a sequence is not linear and describe a pattern in words.
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Key terms
- second differences
- Find the second differences
- Find a
- Combine the parts
- Verification