SPEC A24, A25 (H) • MODULE 2: ALGEBRA
Quadratic, Geometric & Special Sequences
Beyond linear: constant second differences, multiplying by a fixed ratio (even a surd), Fibonacci-type and number family sequences.
HIGHER TIER ONLY • LESSON 2.27
Four Types Beyond Linear
Linear sequences have a constant first difference. If they don't, the sequence is one of several other types — covered at Higher tier for deducing nth-term rules:
TypeSignatureTypical nth term rule
QuadraticConstant second differencean² + bn + c
GeometricConstant ratio r between termsa × r⁽ⁿ⁻¹⁾
Fibonacci-typeEach term = sum of previous twoDefined recursively — no closed form expected
Triangular / Square / CubeNamed number familiesn(n+1)/2 | n² | n³
First step: check first differences
Compute the differences between consecutive terms. If they're constant → linear (use lesson 2.26). If they grow by a constant amount → quadratic. If consecutive terms are in constant ratio → geometric. Otherwise look for Fibonacci behaviour or a known number family.
Spotting a Quadratic: Second Differences
Take the differences of the differences. For a quadratic sequence, the second differences are constant.
+11
↑ first differences (not constant — so not linear)
↑ second differences — constant at 2 → quadratic
Key link
second difference = 2aIf the nth term is an² + bn + c, then the constant second difference equals 2a. So the coefficient of n² is a = (second difference) ÷ 2.
Common slip
Second difference is 2, so a = 1 (not 2). Second difference of 6 → a = 3. Second difference of 10 → a = 5. Always halve.
Finding the nth Term: Four Steps
Every quadratic-sequence nth-term question follows the same four-step routine.
STEP 1 Compute first and second differences. Confirm second differences are constant.
STEP 2a = second difference ÷ 2. This is the coefficient of n².
STEP 3 Subtract an² from each term. The result is a linear sequence — find its nth term (bn + c).
STEP 4 Combine: nth term = an² + bn + c. Verify at n = 1.
The key insight
Subtracting an² from each term removes the quadratic part, leaving a linear sequence — which you already know how to handle from lesson 2.26. So the method just reduces a hard problem to an easier one.
Walkthrough Part 1: Finding a
Sequence: 5, 10, 17, 26, 37
Step 1 — Differences
Sequence: 5 10 17 26 37
First differences: +5 +7 +9 +11 (not constant)
Second differences: +2 +2 +2 (constant → quadratic) ✓
Step 2 — Find a
second difference = 2a = 2 → a = 1
So the coefficient of n² is 1. The nth term begins with n².
Step 3 — Subtract n² from each term
n12345Term510172637n²1491625Term − n²4681012The new sequence 4, 6, 8, 10, 12 is linear (constant difference 2) — now find its nth term to finish step 3.
Walkthrough Part 2: Combine
Step 3 (continued) — nth term of the linear sequence
Linear sequence: 4, 6, 8, 10, 12
d = 2, c = first − d = 4 − 2 = 2
Linear nth term = 2n + 2
Verify n = 1: 2(1) + 2 = 4 ✓
Step 4 — Combine
nth term = n² + 2n + 2= (the n² part) + (the linear nth term 2n + 2)
Final verification — always at n = 1
nn² + 2n + 2ActualCheck11 + 2 + 2 = 55✓24 + 4 + 2 = 1010✓39 + 6 + 2 = 1717✓
MCE Worked Example01
Find the nth term of the sequence 6, 11, 18, 27, 38, …[4 marks]
M
Method: compute first and second differences. If second differences constant → quadratic. a = second diff ÷ 2. Subtract an² to get linear sequence; find linear nth term; combine.
C
First differences: 5, 7, 9, 11. Second differences: 2, 2, 2. Constant → quadratic.
C
a = 2 ÷ 2 = 1. So coefficient of n² is 1.
C
Subtract n²: 6−1, 11−4, 18−9, 27−16, 38−25 = 5, 7, 9, 11, 13.
C
Linear: d = 2, c = 5 − 2 = 3. Linear nth term = 2n + 3.
C
Combine: nth term = n² + 2n + 3. Check n = 1: 1 + 2 + 3 = 6 ✓
E
nth term = n² + 2n + 3.
Geometric Sequences: Multiply by r
A geometric sequence multiplies by a fixed ratio r to get the next term. The differences grow, but the ratios stay the same.
General form
nth term = a × r⁽ⁿ⁻¹⁾where a is the first term and r is the common ratio (each term ÷ the previous).
Finding r
Divide any term by the previous. If the ratio is the same throughout, the sequence is geometric.
Example: 3, 6, 12, 24, 48. Ratios: 6/3 = 2, 12/6 = 2, 24/12 = 2, 48/24 = 2. r = 2.
Exponent is (n−1), not n
At n = 1 you want just the first term a. So the formula is a × r⁰ = a (since r⁰ = 1). Writing a × rⁿ instead of a × rⁿ⁻¹ gives the second term at n = 1 — off by one.
Geometric Example: Integer Ratio
Extracting a and r
First term a = 3
Ratio r = 6 ÷ 3 = 2 (same at every step: 12/6, 24/12, 48/24 = 2)
nth term = 3 × 2⁽ⁿ⁻¹⁾
Verify
n = 1: 3 × 2⁰ = 3 × 1 = 3 ✓
n = 2: 3 × 2¹ = 3 × 2 = 6 ✓
n = 5: 3 × 2⁴ = 3 × 16 = 48 ✓
Growth pattern
Geometric sequences grow (or shrink) far faster than linear/quadratic ones. Term 10 of this sequence is 3 × 2⁹ = 3 × 512 = 1536. Term 20 is over 1.5 million.
Higher-Only: Geometric with Surd Ratio
The Higher tier specifically includes geometric sequences where the ratio is a surd. The formula is the same — just handle the surd algebra carefully.
2√3
6√3
Extract a and r
a = 2
r = 2√3 ÷ 2 = √3
Check: (√3)² = 3, so T₃ = 2 × 3 = 6 ✓; (√3)³ = 3√3, so T₄ = 2 × 3√3 = 6√3 ✓; (√3)⁴ = 9, so T₅ = 18 ✓
nth term = 2 × (√3)⁽ⁿ⁻¹⁾
Why the terms "alternate" in form
Odd powers of √3 stay as multiples of √3 (like √3, 3√3, 9√3, …). Even powers become whole numbers (3, 9, 27, …). So the terms alternate between rational and irrational — but the rule a × r⁽ⁿ⁻¹⁾ generates them all.
Higher-only content
Foundation's geometric progressions are restricted to rational r > 0. Surds in r are Higher-tier only, so if you see a term like 2√3 in a geometric sequence, it's a Higher-specific variation and the formula a × r⁽ⁿ⁻¹⁾ handles it the same way.
Fibonacci-Type Sequences
Each term is the sum of the previous two. You're not expected to write a closed-form nth-term formula at GCSE — these are handled recursively. The exam usually asks you to find missing terms given enough information.
Definition
Tₙ = Tₙ₋₁ + Tₙ₋₂You need two starting terms to generate the sequence. The classic Fibonacci starts 1, 1 (or 0, 1), but GCSE questions often use other starting values.
Example: A Fibonacci-type sequence starts 4, 3. Find the next three terms.
T₃ = T₂ + T₁ = 3 + 4 = 7
T₄ = T₃ + T₂ = 7 + 3 = 10
T₅ = T₄ + T₃ = 10 + 7 = 17
Sequence: 4, 3, 7, 10, 17, …
Other recursive sequences
The spec note for Higher says "other recursive sequences will be defined in the question". The approach is the same — the question gives you the rule (e.g. uₙ₊₁ = 2uₙ + 1 with u₁ = 1, giving 1, 3, 7, 15, 31, …) and you apply it to generate each term.
Working backwards
If you're given two consecutive terms in the middle (e.g. T₃ = 11 and T₄ = 18), you can find earlier terms too: T₂ = T₄ − T₃ = 7, then T₁ = T₃ − T₂ = 4.
Recognise Special Number Sequences
Four sequences you should be able to recognise on sight from their first few terms.
NameSequencenth termHow it's built
Triangular1, 3, 6, 10, 15, 21, …n(n + 1)/2Add 1, 2, 3, 4, … (row counts in a triangle)
Square1, 4, 9, 16, 25, 36, …n²Each term is n × n
Cube1, 8, 27, 64, 125, …n³Each term is n × n × n
Fibonacci1, 1, 2, 3, 5, 8, 13, …(recursive)Each term = sum of previous two
The test
If a question shows a sequence starting 1, 3, 6, 10 — it's triangular. 1, 4, 9, 16 — square. 1, 8, 27 — cube. Recognise the pattern and state the nth term directly; there's no need to run the full four-step quadratic method.
Triangular numbers ARE quadratic
Tₙ = n(n+1)/2 = ½n² + ½n — so the four-step method would also give this. But recognition is faster.
Examiner Alert: Halve the Second Difference
Weak Student
Second differences are all 2, so:
"coefficient of n² is 2"
Proceeds with rule starting 2n² — every subsequent line of working is wrong.
Strong Student
Second differences are 2, so a = 2 ÷ 2 = 1:
"coefficient of n² is 1"
Proceeds with rule starting n². Full marks.
Examiner alert
The relationship is second difference = 2a, so a is HALF the second difference. Second diff = 2 → a = 1. Second diff = 6 → a = 3. Second diff = 10 → a = 5. Always divide by 2.
Why 2a?
If the nth term is an² + bn + c, look at consecutive terms. First differences grow linearly (2an + constant), and the change in first differences — the second difference — is exactly 2a. Proofs for this come in A-level; at GCSE, just remember: halve it.
Examiner Alert: Show the Difference Tables
Even if you spot the nth term instantly, write out the difference tables. Method marks are awarded for showing the process — not just for the final answer.
Weak Student
Spots "that's n² + 2n + 2" immediately and writes just:
"nth term = n² + 2n + 2"
Correct answer but zero working. If there's any slip in the quick mental computation, the whole mark is lost.
Strong Student
Writes differences table, subtracts n², finds linear rule, combines:
"1st diffs: 5, 7, 9, 11
2nd diffs: 2, 2, 2 → a = 1
Term − n²: 4, 6, 8, 10, 12 → 2n + 2
Full: n² + 2n + 2"
Secures every method mark even if they slip at the last step.
4-mark quadratic rule
A typical "find the nth term of this quadratic sequence" question is worth 3–4 marks: M1 for correct second difference, M1 for correct a, M1 for correct linear part, A1 for the combined rule. Show every step to secure all the method marks.
Your Turn
Find the nth term of the sequence:
5, 11, 19, 29, 41, …
[4 marks]
M
Check differences. If second differences constant → quadratic. Apply four-step method.
C
First differences: 11 − 5 = 6, 19 − 11 = 8, 29 − 19 = 10, 41 − 29 = 12. So 6, 8, 10, 12.
C
Second differences: 8 − 6 = 2, 10 − 8 = 2, 12 − 10 = 2. Constant → quadratic.
C
a = 2 ÷ 2 = 1. Coefficient of n² is 1.
C
Subtract n²: 5−1, 11−4, 19−9, 29−16, 41−25 = 4, 7, 10, 13, 16.
C
Linear: d = 3, c = 4 − 3 = 1. Linear nth term = 3n + 1.
C
Combine: nth term = n² + 3n + 1. Check n = 1: 1 + 3 + 1 = 5 ✓
E
nth term = n² + 3n + 1.
Common Errors To Avoid
Students often writeCorrect versionWhy it matters
Second difference 2 → coefficient of n² is 2a = 2 ÷ 2 = 1 → coefficient is 1Halve the second difference
Geometric: nth term = a × rⁿnth term = a × r⁽ⁿ⁻¹⁾Exponent is (n − 1), not n
Only write final nth term, no workingWrite out difference tables before concludingSecure method marks (M1 each stage)
After subtracting an² get a constant — stopThat's a valid linear sequence (d = 0); include bn + cLinear can be trivial (b = 0)
Try to write a formula for a Fibonacci-type sequenceState the recursion Tₙ = Tₙ₋₁ + Tₙ₋₂; generate terms as neededNo closed form expected at GCSE
Confuse first and second differencesFirst = change in terms; second = change in first differencesEssential distinction for quadratics
Quick Self-Check
Before moving on: Did I halve the second difference to get a? Did I verify at n = 1? Did I show my difference tables? For geometric — is the exponent (n − 1), not n?
Summary
Quadratic Sequence — Four Steps
Differences : find first, then second. Constant second difference → quadratic.
a = (second difference) ÷ 2 : the coefficient of n².
Subtract an² from each term; the result is linear.
Find linear nth term (bn + c) and combine: nth term = an² + bn + c. Verify at n = 1.
Other Sequence Types
Geometric : nth term = a × r⁽ⁿ⁻¹⁾. Higher tier includes surd ratios (e.g. r = √3).
Fibonacci-type : Tₙ = Tₙ₋₁ + Tₙ₋₂. No closed form expected — generate terms from two starting values.
Triangular : n(n + 1)/2 | Square : n² | Cube : n³. Recognise on sight.
Spec Coverage
Spec refContent
A24 (H)Recognise & use sequences — quadratic, Fibonacci-type, geometric (including r as a surd), and other sequences defined in the question.
A25 (H)Deduce nth term expressions for linear and quadratic sequences.
Exam Tip
Higher-only topic, typically sitting mid-to-late in a paper. Quadratic nth-term is usually 3–4 marks; geometric with surd r is commonly a 2–3 mark extension. Show working at every stage — the method marks (M1 for each difference, M1 for the linear sub-rule, A1 for the combined answer) are where the marks live.
ExamRizz GCSE Maths (AQA 8300) • Lesson 2.27 • Module 2: Algebra • Spec A24 (H), A25 (H) • Higher tier only
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