Quadratic sequences come up on every GCSE Higher paper — AQA 8300, Edexcel 1MA1, and OCR J560 all include them, typically on Paper 2 or Paper 3. They often appear as a multi-part question worth 4–6 marks, so getting this method locked in is one of the most efficient things you can do before your exam.
What You Need to Know First
Before tackling quadratic sequences, make sure you're comfortable with these:
- Finding the nth term of a linear sequence (e.g. 3n + 2) — you'll need this as part of the method
- What first differences and second differences mean — the gap between consecutive terms, and the gap between those gaps
- Substituting values into an expression — you'll be checking your answer by putting n = 1, 2, 3 back in
- Squaring whole numbers up to at least 10 — you'll be working with n² repeatedly
- The standard form of a quadratic nth term: an² + bn + c, where a, b, and c are numbers to find
How to Spot a Quadratic Sequence
The key test is the second differences.
Take the sequence 3, 8, 15, 24, 35.
Write out the first differences — the gaps between terms:
- 8 − 3 = 5
- 15 − 8 = 7
- 24 − 15 = 9
- 35 − 24 = 11
First differences: 5, 7, 9, 11 — not constant, so this isn't a linear sequence.
Now write out the second differences — the gaps between the first differences:
- 7 − 5 = 2
- 9 − 7 = 2
- 11 − 9 = 2
Second differences: 2, 2, 2 — constant. That tells you the sequence is quadratic.
The value of a in your formula an² + bn + c is always:
a = second difference ÷ 2
So here, a = 2 ÷ 2 = 1, and the formula starts with n².
Finding the Nth Term: The Full Method
Let's find the nth term of 3, 8, 15, 24, 35 step by step.
Step 1: Confirm it's quadratic and find a. Second differences are all 2, so a = 2 ÷ 2 = 1.
Step 2: Write out the values of an². Since a = 1, we write out n²:
| n | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| n² | 1 | 4 | 9 | 16 | 25 |
Step 3: Subtract an² from the original sequence.
| Position (n) | Sequence | n² | Remainder |
|---|---|---|---|
| 1 | 3 | 1 | 2 |
| 2 | 8 | 4 | 4 |
| 3 | 15 | 9 | 6 |
| 4 | 24 | 16 | 8 |
| 5 | 35 | 25 | 10 |
The remainders are 2, 4, 6, 8, 10 — a linear sequence.
Step 4: Find the nth term of the remainder sequence. The remainder sequence 2, 4, 6, 8, 10 has a common difference of 2 and starts at 2. Nth term = 2n.
Step 5: Add everything together. Nth term = n² + 2n
Check: n = 1 → 1 + 2 = 3 ✓. n = 4 → 16 + 8 = 24 ✓. n = 5 → 25 + 10 = 35 ✓
A Harder Example: When a Is Not 1
When the second difference is bigger than 2, a will be greater than 1. The method is identical — you just subtract an² instead of n².
Find the nth term of 5, 14, 29, 50, 77.
Step 1: Find the differences.
- First differences: 9, 15, 21, 27
- Second differences: 6, 6, 6 → constant ✓
Step 2: Find a. a = 6 ÷ 2 = 3. The formula starts with 3n².
Step 3: Write out 3n².
| n | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| 3n² | 3 | 12 | 27 | 48 | 75 |
Step 4: Subtract 3n² from the original sequence.
| n | Sequence | 3n² | Remainder |
|---|---|---|---|
| 1 | 5 | 3 | 2 |
| 2 | 14 | 12 | 2 |
| 3 | 29 | 27 | 2 |
| 4 | 50 | 48 | 2 |
| 5 | 77 | 75 | 2 |
The remainders are all 2 — a constant. A constant remainder means b = 0 and c = 2.
The nth term is 3n² + 2.
Check: n = 2 → 3(4) + 2 = 12 + 2 = 14 ✓. n = 5 → 3(25) + 2 = 75 + 2 = 77 ✓
Using the Formula Once You Have It
Examiners also ask you to find a specific term, or to decide whether a given number is in the sequence.
Example using 3n² + 2:
"Is 110 a term in the sequence?"
Set 3n² + 2 = 110, then solve:
- 3n² = 108
- n² = 36
- n = 6
n = 6 is a positive whole number, so yes, 110 is the 6th term.
"Is 80 a term in the sequence?"
- 3n² = 78
- n² = 26
- n = √26 ≈ 5.1
Not a whole number, so 80 is not in the sequence.
This type of question appears frequently on Edexcel 1MA1 Higher and AQA 8300 — it's worth 1–2 marks and only takes a minute if you know the formula.
Common Mistakes Students Make
- Dividing by 2 is essential — and often forgotten. The second difference equals 2a, not a. If the second difference is 6, then a = 3, not 6. This is the single most common error on quadratic sequence questions.
- Subtracting in the wrong direction. In Step 4, the remainder = (original sequence) − (an²). Getting this the wrong way round gives you the wrong linear sequence.
- Not checking by substituting back. Once you have your formula, always substitute n = 1 and n = 2 to check. This takes 30 seconds and will catch sign errors every time.
- Stopping at the remainder. If the remainder sequence in Step 4 isn't a constant, students sometimes panic and stop. It's just a linear sequence — find its nth term and add it to an².
- Confusing the n value. n is the position of the term, starting at 1 — not the term itself. The 4th term uses n = 4.
Exam Tips for Quadratic Sequences
- Show both rows of differences. AQA 8300 typically awards a method mark for finding the second differences, even if the final nth term is wrong. Never skip this working.
- State the value of a explicitly. Writing "a = 3" on your working line makes it clear to the examiner exactly where your 3n² comes from. It earns method marks even if arithmetic errors creep in later.
- Learn to recognise the three possible outcomes in Step 4. The remainder is always either (1) a constant, (2) a linear sequence, or (3) something that doesn't simplify — if it's the third, you've made an error somewhere.
- On OCR J560, the sequence may be presented as a diagram. Count the objects in each pattern to build your number sequence, then apply the method above. The maths is identical.
- Practise working backwards. Given the nth term, you should also be able to write down the first 5 terms. Edexcel sometimes tests this in one-mark starter questions.
Practise Now
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