A24: Types of Sequences
Recognise and use sequences: triangular, square, cube, arithmetic, geometric, Fibonacci-type, quadratic (Higher)
📋 Key Concepts
| Type | Description | Example |
|---|---|---|
| Arithmetic | Add/subtract same amount | 3, 7, 11, 15, 19, ... |
| Geometric | Multiply by same amount | 2, 6, 18, 54, ... |
| Square numbers | n² | 1, 4, 9, 16, 25, ... |
| Cube numbers | n³ | 1, 8, 27, 64, 125, ... |
| Triangular numbers | Sum of first n integers | 1, 3, 6, 10, 15, ... |
| Fibonacci | Add previous two terms | 1, 1, 2, 3, 5, 8, 13, ... |
📝 Arithmetic Sequences
Definition: A sequence where the difference between consecutive terms is constant.
nth term: aₙ = a + (n-1)d
where a = first term, d = common difference
where a = first term, d = common difference
Example 1
Find the nth term of: 5, 8, 11, 14, 17, ...
Solution:
First term a = 5
Common difference d = 8 - 5 = 3
nth term: 5 + (n-1) × 3 = 5 + 3n - 3 = 3n + 2
Example 2
Find the nth term of: 20, 17, 14, 11, ...
Solution:
a = 20, d = -3
nth term: 20 + (n-1)(-3) = 20 - 3n + 3 = 23 - 3n
📝 Geometric Sequences
Definition: A sequence where each term is found by multiplying the previous term by a constant (the common ratio).
nth term: aₙ = a × rn-1
where a = first term, r = common ratio
where a = first term, r = common ratio
Example 3
Find the nth term of: 3, 6, 12, 24, 48, ...
Solution:
a = 3, r = 2 (each term doubles)
nth term: 3 × 2n-1
Example 4
Find the 6th term of: 100, 50, 25, ...
Solution:
a = 100, r = ½
6th term: 100 × (½)5 = 100 × 1⁄32 = 3.125
📝 Square and Cube Numbers
Square numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, ...
Formula: n² where n = 1, 2, 3, ...
Formula: n² where n = 1, 2, 3, ...
Cube numbers: 1, 8, 27, 64, 125, 216, ...
Formula: n³ where n = 1, 2, 3, ...
Formula: n³ where n = 1, 2, 3, ...
Example 5
Find the 12th square number.
Solution:
12² = 144
📝 Triangular Numbers
Definition: Numbers that can form a triangle of dots. Each triangular number is the sum of consecutive integers.
nth triangular number: n(n+1)⁄2
Example 6
Find the 10th triangular number.
Solution:
10 × 11⁄2 = 110⁄2 = 55
📝 Fibonacci-Type Sequences
Definition: Each term is the sum of the two previous terms.
Example 7
Find the next 3 terms: 2, 5, 7, 12, 19, ...
Solution:
Each term = sum of previous two
19 + 12 = 31
31 + 19 = 50
50 + 31 = 81
Next terms: 31, 50, 81
📝 Quadratic Sequences (Higher)
Definition: The nth term contains n². The second difference (difference of differences) is constant.
Example 8
Find the nth term of: 3, 6, 11, 18, 27, ...
Solution:
Sequence: 3, 6, 11, 18, 27
1st diff: 3, 5, 7, 9
2nd diff: 2, 2, 2 (constant, so quadratic)
Half of 2nd diff = 1, so n² term
Compare n²: 1, 4, 9, 16, 25
Actual - n²: 2, 2, 2, 2, 2
nth term: n² + 2
🧮 Worked Examples — Exam Style
Sequence identification: check differences first (arithmetic), then ratios (geometric), then the special sets (squares, cubes, triangular, Fibonacci).
Identify and continue: 1, 4, 9, 16, 25, …
These are SQUARE numbers (1², 2², 3²…) → next: 36, 49
B1 for identification, B1 for continuation. Difference pattern (3, 5, 7, 9…) confirms it: second difference constant = 2.
Identify and continue: 160, 80, 40, 20, …
Each term is HALVED (geometric, ratio ½) → 10, 5
B1 each. Ratio check: 80/160 = 40/80 = ½ — constant ratio means geometric.
Exam wording to expect: "Find the next … terms", "Describe the rule", "Which term is the first above …?".
❓ Practice Questions
Q1: Find the nth term of: 4, 9, 14, 19, 24, ...
Q2: Find the nth term of: 2, 6, 18, 54, ...
Q3: What is the 8th triangular number?
Q4: Find the next 3 terms: 1, 1, 2, 3, 5, 8, ...
Q5: Find the nth term of: 2, 5, 10, 17, 26, ...
Q6: What type of sequence is: 100, 90, 81, 72.9, ...?
Q7: Find the next two terms of the sequence 2, 6, 12, 20, …
Q8: A sequence starts 3, 6, 12, 24. Write the rule and the 7th term.
✅ Answers
- 5n - 1
- 2 × 3n-1
- 36
- 13, 21, 34
- n² + 1
- Geometric (ratio = 0.9)
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. Differences constant 5 → arithmetic: 5n − 1
Q2. ×3 each time → geometric: 2 × 3ⁿ⁻¹
Q3. Triangular numbers: 1, 3, 6, 10, 15, 21, 28, 36
Q4. Fibonacci (add previous two): 13, 21, 34
Q5. Differences 3, 5, 7… → second difference 2 → n² + 1
Q6. Multiplied by 0.9 each time → geometric, ratio 0.9
Q7. Differences 4, 6, 8 (+2 each time); terms are n² + n → next: 30, 42
Q8. Doubling from 3 → rule ×2; 7th term = 3 × 2⁶ = 192
📌 Key Facts to Memorise
- Constant difference → arithmetic (linear nth term)
- Constant ratio → geometric (nth term a × rⁿ⁻¹)
- Squares 1, 4, 9, 16… · Cubes 1, 8, 27… · Triangular 1, 3, 6, 10, 15…
- Fibonacci-type: each term = sum of the previous two
- Second difference constant (2) → quadratic sequence (n² + adjustment)
🗺️ In the Exam
Sequence types: 2–4 marks on Paper 1. Special sequences (triangular, Fibonacci) are the recognition marks; geometric decay (×0.9) links to depreciation.
Links to: nth term (A25), generating sequences (A23), growth and decay (R16).
⚡ Targeted Drills
Cover the answers, work each one, then mark yourself. If any takes over a minute, revisit the matching section above.
- 2, 5, 8, 11 … → arithmetic (+3) · 3, 9, 27 … → geometric (×3)
- Next two: 1, 2, 4, 7, 11 … → 16, 22 (differences 1,2,3,4,5…)
- 16th triangular number → 136 (16 × 17 ÷ 2)
- Rule for 5, 15, 45 … → ×3 → next 135
- 1, 1, 2, 3, 5, 8 … → Fibonacci → 13
🧭 Board Focus — OCR
Spec J560. Three papers of 100 marks each: Paper 1 non-calculator, Papers 2 and 3 calculator, at Foundation/Higher tiers. OCR marks problem-solving explicitly, and its papers tend to front-load multi-step worded contexts — practise extracting the mathematics from prose before calculating.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Arithmetic: constant difference → an + b form
- Geometric: constant ratio → a × rn-1 form
- Fibonacci: each term = sum of previous two
- For quadratic: check 2nd difference is constant
- Triangular: n(n+1)/2
🧠 Problem-Solving Strategies
Problem-Solving
Identify the sequence type: constant difference = arithmetic (nth term: dn + c), constant ratio = geometric (nth term: a × r^(n-1)), sum of previous two = Fibonacci. For quadratic: second difference is constant, half of it = coefficient of n².
Multi-Step Problem
A bacteria population starts at 500 and triples every hour. (a) Write the population as a sequence. (b) Find the nth term. (c) After how many hours does it exceed 100,000?
Solution:
(a) 500, 1500, 4500, 13500, 40500, ...
(b) Geometric: a = 500, r = 3. nth term = 500 × 3^(n-1)
(c) 500 × 3^(n-1) > 100000 → 3^(n-1) > 200. n-1 > log₃(200) ≈ 4.82. So n > 5.82, meaning after 6 hours (7th term).
⚠️ Common Errors
Watch Out!
1. Wrong: Treating a geometric sequence as arithmetic Correct: Check whether you ADD or MULTIPLY to get the next term — arithmetic adds, geometric multiplies
2. Wrong: The nth term of a geometric sequence is a × r^n Correct: It is a × r^(n-1) — the first term has n = 1, so r^(1-1) = r⁰ = 1, giving just a
3. Wrong: The 10th triangular number is 10 × 11 = 110 Correct: The 10th triangular number is (10 × 11)/2 = 55 (must divide by 2)
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A ball is dropped from 80 cm. Each bounce reaches 3/4 of the previous height. (a) Write the first 5 heights as a sequence. (b) What type of sequence is this? Find the nth term. (c) After which bounce is the height below 10 cm?
(a) 80, 60, 45, 33.75, 25.3125
(b) Geometric sequence with a = 80, r = 3/4. nth term = 80 × (3/4)^(n-1)
(c) 80 × (3/4)^(n-1) < 10 → (3/4)^(n-1) < 1/8 → (n-1) > log(1/8)/log(3/4) ≈ 8.00. So n > 9. After the 9th bounce the height is below 10 cm.
Mark scheme: (a) 2 marks. (b) 2 marks for type and formula. (c) 2 marks for solving inequality.
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A salary starts at £24,000 and increases by £1,500 each year.
(a) Write the nth term for the salary in year n.
(b) After how many years does the salary exceed £40,000?
(c) A different job starts at £24,000 and increases by 5% each year. Which job pays more in year 10?
Answers: (a) Arithmetic: S = 24000 + 1500(n-1) = 1500n + 22500. (b) 1500n + 22500 > 40000 → n > 11.67, so after 12 years. (c) Job 1 year 10: £37,500. Job 2 year 10: 24000 × 1.05^9 ≈ £37,218. Job 1 pays slightly more in year 10.
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