Quadratic Sequences and Nth Term

Quadratic sequences take the pattern-spotting skills you learned with linear sequences and push them one level further. Instead of a constant first difference, quadratic sequences have a constant second difference. This topic is exclusive to the Higher tier and is tested regularly by AQA, Edexcel, and OCR. On this page you will learn how to recognise a quadratic sequence, find its nth term formula in the form an² + bn + c, and apply the method to exam-style questions. The process has more steps than for linear sequences, but each step is straightforward once you know the routine.

What Is a Quadratic Sequence?

A quadratic sequence is a sequence whose nth term is a quadratic expression — that is, it contains an n² term. The defining feature is that the first differences change, but the second differences are constant.

Example: 3, 8, 17, 30, 47, ...

First differences: 5, 9, 13, 17

Second differences: 4, 4, 4 — constant, so this is quadratic.

nth term = an² + bn + c

a = (second difference) ÷ 2

Once you know a, you subtract an² from each term, leaving a linear sequence from which you find bn + c using the linear nth term method.

Step-by-Step Method

Write out the sequence and label the terms T₁, T₂, T₃, etc.
Calculate the first differences (gaps between consecutive terms).
Calculate the second differences (gaps between the first differences). If these are constant, the sequence is quadratic.
Find a. Divide the constant second difference by 2. This is the coefficient of n².
Subtract an² from each original term. Write out the values of an² for n = 1, 2, 3, 4, ... and subtract them from the corresponding terms.
Find the nth term of the remaining linear sequence. The result is bn + c.
Combine to get the full formula: an² + bn + c.
Check by substituting values of n back into the formula.

Worked Example 1 — Foundation-style approach at Higher

Question: Find the nth term of 5, 12, 23, 38, 57, ...

Working:

Step 1: First differences: 7, 11, 15, 19.

Step 2: Second differences: 4, 4, 4. Constant ✓

Step 3: a = 4 ÷ 2 = 2. So the n² part is 2n².

Step 4: Subtract 2n² from each term.

n	Term	2n²	Term − 2n²
1	5	2	3
2	12	8	4
3	23	18	5
4	38	32	6
5	57	50	7

Step 5: The remaining sequence is 3, 4, 5, 6, 7, ... which has nth term = n + 2.

Step 6: Full nth term = 2n² + n + 2.

Check: n = 3 → 2(9) + 3 + 2 = 23 ✓; n = 5 → 2(25) + 5 + 2 = 57 ✓

Answer: nth term = 2n² + n + 2

Worked Example 2 — Higher Level

Question: Find the nth term of 0, 5, 14, 27, 44, ...

Working:

Step 1: First differences: 5, 9, 13, 17.

Step 2: Second differences: 4, 4, 4. Constant ✓

Step 3: a = 4 ÷ 2 = 2. n² part is 2n².

Step 4: Subtract 2n².

n	Term	2n²	Term − 2n²
1	0	2	−2
2	5	8	−3
3	14	18	−4
4	27	32	−5

Step 5: Remaining sequence: −2, −3, −4, −5, ... Common difference = −1. When n = 1, value = −2. nth term of linear part: −n − 1.

Step 6: Full nth term = 2n² − n − 1.

Check: n = 1 → 2 − 1 − 1 = 0 ✓; n = 4 → 32 − 4 − 1 = 27 ✓

Answer: nth term = 2n² − n − 1

Common Mistakes

Forgetting to divide the second difference by 2. The second difference is 2a, not a. If the second difference is 6, then a = 3.
Errors in the subtraction table. Calculate each value of an² carefully. A single arithmetic slip will throw off the entire linear part.
Not recognising a quadratic sequence. If the first differences are not constant, always check the second differences before assuming the sequence is something more exotic.
Mixing up first and second differences. First differences are the gaps between terms; second differences are the gaps between first differences. Label your rows clearly.
Skipping the check step. Always verify your formula against at least two original terms. This catches sign and arithmetic errors.

Exam Tips

Set out your working in a table. Examiners find this much easier to follow and it reduces your own errors. Columns for n, term, an², and the remainder work well.
This question is typically worth 3-4 marks. You usually get one mark for finding the second difference, one for the an² term, and one or two for the complete formula. Show each step to maximise marks.
Some questions give you the nth term and ask for a specific term. Simply substitute n into the formula. These are free marks if you can handle substitution into quadratics.
Occasionally, the sequence might start with n = 0 or use different notation. Read the question carefully to see which term corresponds to which value of n.

Practice Questions

Q1: Find the nth term of 4, 13, 26, 43, 64, ...

Answer: 2n² + 3n − 1 (second difference = 4, a = 2; subtract 2n² to get 2, 5, 8, 11, 14 which is 3n − 1)

Q2: Find the nth term of 1, 8, 19, 34, 53, ...

Answer: 2n² + n − 2 (second difference = 4, a = 2; subtract 2n² to get −1, 0, 1, 2, 3 which is n − 2)

Q3: The nth term of a sequence is 3n² − 2n + 1. Find the 10th term and the difference between the 10th and 9th terms.

Answer: 10th term = 3(100) − 20 + 1 = 281; 9th term = 3(81) − 18 + 1 = 226; difference = 55

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Summary

A quadratic sequence has a constant second difference.
The nth term takes the form an² + bn + c.
Find a by halving the second difference.
Subtract an² from every term to reveal a linear sequence, then find bn + c.
Always present your working in a clear table.
Check your final formula by substituting at least two values of n.
This topic builds directly on linear sequences and connects to solving and graphing quadratics.

Quadratic Sequences & Nth Term –