Quadratic Sequences and nth Term: GCSE Maths

A quadratic sequence has a constant second difference — the differences between consecutive terms change at a steady rate. The nth term always includes an n² component. Spotting the second difference is the key first step: if it equals 2a, the coefficient of n² is a, giving you the framework to build the full formula.

What makes a sequence quadratic?

The quickest way to classify a sequence is to calculate the differences between consecutive terms, then calculate the differences of those differences. In a linear sequence the first differences are constant. In a quadratic sequence the first differences are not constant, but the second differences are.

Sequence | First differences | Second differences | Type
3, 5, 7, 9, 11 | 2, 2, 2, 2 | 0, 0, 0 | Linear
2, 5, 10, 17, 26 | 3, 5, 7, 9 | 2, 2, 2 | Quadratic
1, 4, 9, 16, 25 | 3, 5, 7, 9 | 2, 2, 2 | Quadratic (square numbers)

Notice that the square numbers also have a constant second difference of 2 — no coincidence. Their nth term is n², and the second difference is always twice the coefficient of n², in this case 2 × 1 = 2.

For a sequence with more than four terms you should have enough second differences to confirm they are all equal before concluding the sequence is quadratic. If the second differences are themselves changing, the sequence may be cubic, which is beyond GCSE.

How do you find the coefficient of n² from the second difference?

This is the gateway step, and the relationship is simple:

If the second difference = 2a, then the nth term contains an².

So you halve the second difference to get a:

Second difference = 2 → a = 1 → the nth term contains n²

Second difference = 4 → a = 2 → the nth term contains 2n²

Second difference = 6 → a = 3 → the nth term contains 3n²

Second difference = 10 → a = 5 → the nth term contains 5n²

This value of a is exact, not an estimate. Once you have it, the remaining work is to find what linear expression (bn + c) must be added to an² to reproduce the original sequence.

How do you find the full nth term formula?

The method has five clear steps. Find a from the second difference, generate the sequence an², subtract it term by term from the original sequence to reveal a "remainder" sequence, find the nth term of that linear remainder, and add everything together.

Worked example 1: Find the nth term of 5, 11, 21, 35, 53.

Calculate first differences: 11 − 5 = 6, 21 − 11 = 10, 35 − 21 = 14, 53 − 35 = 18.

Calculate second differences: 10 − 6 = 4, 14 − 10 = 4, 18 − 14 = 4. Second difference = 4, so a = 4 ÷ 2 = 2.

Write out 2n² for n = 1, 2, 3, 4, 5: 2, 8, 18, 32, 50.

Subtract 2n² from each term of the original sequence:

5 − 2 = 3, 11 − 8 = 3, 21 − 18 = 3, 35 − 32 = 3, 53 − 50 = 3.

Remainder sequence: 3, 3, 3, 3, 3 (constant, so the linear part is simply 3).

Full nth term = 2n² + 3.

Verification: n = 1: 2(1)² + 3 = 2 + 3 = 5 ✓; n = 3: 2(9) + 3 = 21 ✓; n = 5: 2(25) + 3 = 53 ✓.

Always verify with at least two values before committing to your answer. If a single value works but another does not, recheck your second difference or the remainder sequence.

How do you handle a non-constant remainder sequence?

When the remainder sequence is not constant, it is linear — find its nth term using the standard linear method, then add it to an².

Worked example 2: Find the nth term of 3, 8, 15, 24, 35.

First differences: 8 − 3 = 5, 15 − 8 = 7, 24 − 15 = 9, 35 − 24 = 11.

Second differences: 7 − 5 = 2, 9 − 7 = 2, 11 − 9 = 2. Second difference = 2, so a = 1.

Write out n² for n = 1, 2, 3, 4, 5: 1, 4, 9, 16, 25.

Subtract n² from each term of the original sequence:

3 − 1 = 2, 8 − 4 = 4, 15 − 9 = 6, 24 − 16 = 8, 35 − 25 = 10.

Remainder sequence: 2, 4, 6, 8, 10.

The remainder sequence is linear with a common difference of 2 and first term 2 (when n = 1). Its nth term = 2n.

Full nth term = n² + 2n.

Verification: n = 1: 1 + 2 = 3 ✓; n = 2: 4 + 4 = 8 ✓; n = 4: 16 + 8 = 24 ✓; n = 5: 25 + 10 = 35 ✓.

The remainder sequence will always be constant or linear — never itself quadratic — because subtracting an² removes all quadratic growth from the original sequence.

How do you find a specific term using the nth term formula?

Once you have the formula, substitute the term number directly — there is no need to write out every preceding term.

Example 1: Using nth term = 2n² + 3, find the 10th term.

Substitute n = 10: 2(10)² + 3 = 2(100) + 3 = 200 + 3 = 203.

Example 2: Using nth term = n² + 2n, find the 20th term.

Substitute n = 20: (20)² + 2(20) = 400 + 40 = 440.

Direct substitution is vastly faster than continuing the sequence term by term, making the nth term formula particularly valuable for large values of n.

How do you check whether a number is in a quadratic sequence?

Set the nth term formula equal to the number in question and solve the resulting equation for n. If n is a positive integer, the number is in the sequence. If n is not a positive integer, it is not.

Example 1: Is 203 in the sequence with nth term 2n² + 3?

Set 2n² + 3 = 203.

2n² = 200 → n² = 100 → n = 10.

n = 10 is a positive integer, so yes — 203 is the 10th term. ✓

Example 2: Is 50 in the same sequence?

Set 2n² + 3 = 50.

2n² = 47 → n² = 23.5 → n = √23.5 ≈ 4.85.

n is not an integer, so 50 is not in the sequence.

This technique is clean and general — it works for any algebraic nth term formula and gives a definitive yes-or-no answer in just a few lines.

Frequently asked questions

How do I know if a sequence is quadratic and not linear or cubic?

Calculate first differences, then second differences. Constant first differences → linear. Non-constant first differences but constant second differences → quadratic. If the second differences also vary, calculate third differences — if those are constant, the sequence is cubic. For GCSE you only need to distinguish linear from quadratic, but non-constant second differences definitively rule out a quadratic.

Why do you halve the second difference to get a?

For a sequence with nth term an² + bn + c, the first differences are a(2n − 1) + b — a linear sequence with common difference 2a. So the second differences all equal 2a. If the observed second difference is k, then 2a = k and a = k/2. Try deriving this yourself: write out an² + bn + c for n = 1, 2, 3, 4, compute both sets of differences, and confirm the second difference is always 2a. Understanding the derivation is far more reliable than memorising the rule alone.

What if the second differences are negative?

A negative constant second difference means a is negative — the sequence contains −|a|n² in its nth term and eventually falls. The method is identical: halve the second difference to get a (negative), subtract an² from each term (equivalently, add |a|n²), and find the linear nth term of the remainder. Take extra care with signs during the subtraction step, and always verify your formula with two or three values from the original sequence.

Can the nth term of a quadratic sequence have a negative n² coefficient?

Yes. The nth term −n² + 6n + 1 gives: n = 1: 6, n = 2: 9, n = 3: 10, n = 4: 9, n = 5: 6. It rises to a maximum then falls — the hallmark of a negative quadratic. The second differences are −2, −2, −2, giving a = −1 and confirming the −n² term. If your second differences are negative, the method works exactly the same way — just keep track of the signs throughout.

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