What should I learn from this Further Maths lesson?

By the end of this lesson, you should be able to Find the nth term formula for a quadratic sequence by calculating first and second differences; Identify the coefficient of $n^2$ from the second difference of a quadratic sequence; Apply the method of comparing a quadratic sequence to multiples of $n^2$ to find the full nth term; Verify an nth term formula by substituting values of $n$ back into the sequence.

How should I revise Finding the nth Term of Quadratic Sequences?

Read each section, then close the notes and explain the idea from memory. Check your explanation against the page and finish with an exam question from Algebra.

Finding the nth Term of Quadratic Sequences | GCSE AQA Further Maths (H)

Finding the nth Term of Quadratic Sequences is part of nth terms of quadratic sequences in GCSE AQA Higher Further Maths.

What this lesson covers

Finding the nth Term of Quadratic Sequences: Learn how to determine the nth term rule for quadratic sequences.

Find the nth term formula for a quadratic sequence by calculating first and second differences
Identify the coefficient of $n^2$ from the second difference of a quadratic sequence
Apply the method of comparing a quadratic sequence to multiples of $n^2$ to find the full nth term
Verify an nth term formula by substituting values of $n$ back into the sequence

Key ideas to keep in view

Use this lesson to stay inside nth terms of quadratic sequences while connecting the detail back to the wider Algebra topic.

Quadratic sequence: A sequence whose second differences are constant and non-zero. The nth term contains an $n^2$ term as its highest power.
Second difference: The difference between consecutive first differences. A constant second difference confirms the sequence is quadratic and determines the coefficient of $n^2$.
nth term of a quadratic sequence: A formula of the form $an^2 + bn + c$ that generates every term in the sequence when $n = 1, 2, 3, \ldots$
First difference: The result of subtracting each term from the next term in a sequence. Used as the first step when identifying sequence type.

Revision notes

Read the core explanations from Finding the nth Term of Quadratic Sequences before testing yourself from memory.

1. Why Quadratic Sequences Matter

You already know how to find the nth term of a linear sequence — one where the differences between terms are constant. Quadratic sequences take that one step further: the differences between terms are not constant, but the differences of those differences are.

That second layer of differences is the key that unlocks the whole method. Once you spot it, finding the nth term becomes a reliable, repeatable process — and examiners test it regularly at GCSE.

2. Recognising a Quadratic Sequence

Start with a sequence: $3, 8, 15, 24, 35, \ldots$ The first differences are $5, 7, 9, 11, \ldots$ — these are not constant, so the sequence is not linear. Now take the differences of those first differences: $2, 2, 2, \ldots$ That constant second difference tells you the sequence is quadratic.

3. The Method: Finding the nth Term

The nth term of any quadratic sequence has the form $an^2 + bn + c$. The method works in three stages: find $a$ from the second difference, subtract the $an^2$ sequence from the original to leave a linear sequence, then find $b$ and $c$ from that linear remainder.

Here is the logic behind $a = \frac{\text{second difference}}{2}$: if you expand $an^2 + bn + c$ and compute its second differences, you always get $2a$. So dividing the constant second difference by $2$ gives you $a$ directly.

Worked example: Finding the nth Term of a Quadratic Sequence

Find the nth term of the sequence $5, 11, 21, 35, 53, \ldots$

$11 - 5 = 6$ $21 - 11 = 10$ $35 - 21 = 14$ $53 - 35 = 18$ First differences: $6, 10, 14, 18$ — Subtract each term from the next. These are not constant, so the sequence is not linear. [M1]
$10 - 6 = 4$ $14 - 10 = 4$ $18 - 14 = 4$ Second difference: $4$ (constant) — The constant second difference confirms a quadratic sequence. [M1]
$a = \frac{4}{2} = 2$ So the $n^2$ part is $2n^2$ — Divide the second difference by 2 to get $a$.
$n = 1$: $2(1)^2 = 2$, so $5 - 2 = 3$ $n = 2$: $2(2)^2 = 8$, so $11 - 8 = 3$ $n = 3$: $2(3)^2 = 18$, so $21 - 18 = 3$ $n = 4$: $2(4)^2 = 32$, so $35 - 32 = 3$ Remainder sequence: $3, 3, 3, 3, \ldots$ — The remainder is constant, so $b = 0$ and $c = 3$. The linear part is $3$. [M1]
$T_n = 2n^2 + 3$ — Combine the $an^2$ part with the linear remainder. [A1]
$n = 1$: $2(1 \times 1) + 3 = 2 + 3 = 5$ ✓ $n = 3$: $2(3 \times 3) + 3 = 2(9) + 3 = 18 + 3 = 21$ ✓ — Always substitute back to confirm your formula matches the original sequence.

Answer: $T_n = 2n^2 + 3$

4. When the Remainder is Linear (Not Constant)

The previous example gave a constant remainder, making $b = 0$. More often, the remainder sequence itself has a non-zero gradient — meaning you need to find the nth term of that linear remainder too.

Worked example: nth Term with a Linear Remainder

Find the nth term of the sequence $4, 10, 20, 34, 52, \ldots$

$10 - 4 = 6$ $20 - 10 = 10$ $34 - 20 = 14$ $52 - 34 = 18$ First differences: $6, 10, 14, 18$ — Not constant — confirms the sequence is not linear.
$10 - 6 = 4$ $14 - 10 = 4$ $18 - 14 = 4$ Second difference: $4$ — Constant second difference confirms quadratic.
$a = \frac{4}{2} = 2$ — The $n^2$ coefficient is $2$.
$n = 1$: $2(1 \times 1) = 2$, so $4 - 2 = 2$ $n = 2$: $2(2 \times 2) = 8$, so $10 - 8 = 2$ $n = 3$: $2(3 \times 3) = 18$, so $20 - 18 = 2$ $n = 4$: $2(4 \times 4) = 32$, so $34 - 32 = 2$ Remainder sequence: $2, 2, 2, 2, \ldots$ — The remainder is constant at $2$, so $b = 0$ and $c = 2$.
$T_n = 2n^2 + 2$ — Combine the quadratic and constant parts. [A1]
$n = 2$: $2(2 \times 2) + 2 = 8 + 2 = 10$ ✓ $n = 5$: $2(5 \times 5) + 2 = 2(25) + 2 = 50 + 2 = 52$ ✓ — Both match the original sequence — the formula is correct.

Answer: $T_n = 2n^2 + 2$

Now consider a sequence where the remainder is genuinely linear — not constant. The sequence $2, 7, 16, 29, 46, \ldots$ has second difference $4$, giving $a = 2$. Subtracting $2n^2$ gives the remainder $0, 3, 6, 9, 12, \ldots$, which is a linear sequence with nth term $3n - 3$. The full nth term is therefore $2n^2 + 3n - 3$.

5. A Full Worked Example with a Linear Remainder

This is the version that catches most students out. When the remainder after subtracting $an^2$ is not constant, you must find the nth term of that linear remainder separately and add it on.

Worked example: nth Term with a Non-Constant Linear Remainder

Find the nth term of the sequence $1, 6, 15, 28, 45, \ldots$

$6 - 1 = 5$ $15 - 6 = 9$ $28 - 15 = 13$ $45 - 28 = 17$ First differences: $5, 9, 13, 17$ — Not constant — the sequence is not linear.
$9 - 5 = 4$ $13 - 9 = 4$ $17 - 13 = 4$ Second difference: $4$ — Constant second difference of $4$ confirms a quadratic sequence. [M1]
$a = \frac{4}{2} = 2$ — Halve the second difference to get the coefficient of $n^2$.
$n=1$: $2(1 \times 1)=2$, remainder $1-2=-1$ $n=2$: $2(2 \times 2)=8$, remainder $6-8=-2$ $n=3$: $2(3 \times 3)=18$, remainder $15-18=-3$ $n=4$: $2(4 \times 4)=32$, remainder $28-32=-4$ Remainder sequence: $-1, -2, -3, -4, \ldots$ — The remainder is linear (constant difference of $-1$). [M1]
Remainder sequence: $-1, -2, -3, -4, \ldots$ First term $= -1$, common difference $= -1$ nth term of remainder $= -1 + (n-1)(-1) = -1 - n + 1 = -n$ — Use the linear nth term method: $a + (n-1)d$ where $a = -1$ and $d = -1$.
$T_n = 2n^2 + (-n) = 2n^2 - n$ — Add the quadratic part and the linear remainder. [A1]
$n=1$: $2(1 \times 1) - 1 = 2 - 1 = 1$ ✓ $n=3$: $2(3 \times 3) - 3 = 18 - 3 = 15$ ✓ $n=5$: $2(5 \times 5) - 5 = 50 - 5 = 45$ ✓ — All three match — the formula is verified.

Answer: $T_n = 2n^2 - n$

6. Practice: Try These Yourself

The method is consistent across every quadratic sequence. Work through these problems using the same four-stage approach: find the second difference, halve it to get $a$, subtract $an^2$, then deal with the remainder.

7. Exam Technique and Common Mistakes

Examiners award method marks at each stage of this process, so showing your differences table is worth marks even if your final formula has a slip. Never skip straight to an answer — write out the first differences, second differences, and the subtracted sequence explicitly.

The most reliable way to pick up full marks is to verify your formula at the end. Substitute $n = 1$ and $n = 2$ (or any two terms) back into your formula and confirm they match the original sequence. If they do not match, you have time to spot where the error crept in.

Mastering the nth term of quadratic sequences means you can generate any term in the sequence instantly — and that skill connects directly to more advanced work on sequences, series, and algebraic proof at higher GCSE levels.

Sources and content standards

Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-14.

AQA Level 2 Certificate in Further Mathematics (8365) specification How we check content Report a correction — include this lesson URL

Finding the nth Term of Quadratic Sequences Revision Notes | GCSE AQA Higher Further Maths