How to Solve: Sequences (nth Term, Quadratic, Special Sequences) – GCSE & A-Level Maths Guide

How to Solve: Sequences (nth Term, Quadratic, Special Sequences) – GCSE & A-Level Maths Guide

Introduction

Mastering sequences unlocks 8–12 marks on your GCSE/A-Level exam—enough to boost your grade by a full level. Whether it’s predicting the next term in a pattern, finding the 50th term of a quadratic sequence, or spotting a hidden Fibonacci, sequences appear in every exam paper. Get this right, and you’ll save time for harder questions.

What You Need To Know First

Before diving in, make sure you’re solid on: 1. Linear equations – Rearranging and solving for n. 2. Quadratic equations – Expanding brackets and solving an² + bn + c. 3. Basic algebra – Substituting values into formulas.

If any of these feel shaky, pause and review them first.

Key Vocabulary

Formulas To Know

1. Linear (Arithmetic) Sequences

Formula: nth term = a + (n – 1)d - a = first term - d = common difference (difference between terms) - n = term number

MEMORISE THIS – It’s not given on the exam sheet.

2. Quadratic Sequences

Formula: nth term = an² + bn + c - a, b, c = constants to find - n = term number

Given on exam sheet (but you must know how to find a, b, c).

3. Special Sequences

MEMORISE square, cube, and triangular numbers—examiners love testing these!

Step-by-Step Method

How to Find the nth Term of a Linear Sequence

Step 1: Write out the sequence and label the term numbers (n). Step 2: Find the common difference (d) by subtracting consecutive terms. Step 3: Write the formula: nth term = a + (n – 1)d Step 4: Substitute a (first term) and d into the formula. Step 5: Simplify (if needed) to get nth term = dn + (a – d).

Example: Find the nth term of 5, 8, 11, 14… 1. n: 1, 2, 3, 4 2. d = 8 – 5 = 3 3. nth term = a + (n – 1)d → 5 + (n – 1)3 4. Simplify: 5 + 3n – 3 → 3n + 2

Answer: nth term = 3n + 2

How to Find the nth Term of a Quadratic Sequence

Step 1: Write out the sequence and label n. Step 2: Find the first differences (differences between terms). Step 3: Find the second differences (differences of the first differences). Step 4: If the second difference is constant, it’s a quadratic sequence. Step 5: The coefficient a = (second difference) ÷ 2. Step 6: Write an² and subtract it from the original sequence to get a new sequence. Step 7: Find the nth term of this new sequence (it will be linear). Step 8: Combine an² with the linear nth term to get the final formula.

Example: Find the nth term of 2, 5, 10, 17, 26… 1. n: 1, 2, 3, 4, 5 2. First differences: 3, 5, 7, 9 3. Second differences: 2, 2, 2 → Constant! 4. a = 2 ÷ 2 = 1 5. an² = n² 6. Subtract n² from original sequence: 2–1=1, 5–4=1, 10–9=1, 17–16=1, 26–25=1 → New sequence: 1, 1, 1, 1, 1 7. nth term of new sequence = 1 (constant) 8. Final nth term = n² + 1

Answer: nth term = n² + 1

How to Spot Special Sequences

Step 1: Check if the sequence matches a known pattern (squares, cubes, Fibonacci). Step 2: If not, calculate differences to see if it’s linear or quadratic. Step 3: If differences don’t settle, check for geometric sequences (multiplying by a fixed number).

Example: Is 1, 4, 9, 16, 25… a special sequence? - These are square numbers → nth term = n²

Worked Examples

Example 1 – Basic Linear Sequence

Question: Find the nth term of 7, 10, 13, 16… Working: 1. n: 1, 2, 3, 4 2. d = 10 – 7 = 3 3. nth term = a + (n – 1)d → 7 + (n – 1)3 4. Simplify: 7 + 3n – 3 → 3n + 4

Answer: nth term = 3n + 4 What we did and why: We found the common difference (d) and used the linear formula to express any term in terms of n.

Example 2 – Medium Quadratic Sequence

Question: Find the nth term of 3, 8, 15, 24, 35… Working: 1. n: 1, 2, 3, 4, 5 2. First differences: 5, 7, 9, 11 3. Second differences: 2, 2, 2 → Constant! 4. a = 2 ÷ 2 = 1 5. an² = n² 6. Subtract n² from original: 3–1=2, 8–4=4, 15–9=6, 24–16=8, 35–25=10 → New sequence: 2, 4, 6, 8, 10 7. nth term of new sequence = 2n (linear) 8. Final nth term = n² + 2n

Answer: nth term = n² + 2n What we did and why: We confirmed it was quadratic (constant second difference), found a, then combined it with the linear part.

Example 3 – Exam-Style (Disguised)

Question: A sequence has the nth term 4n – 1. Find the 20th term. Working: 1. Substitute n = 20 into 4n – 1. 2. 4(20) – 1 = 80 – 1 = 79

Answer: 79 What we did and why: We directly substituted n into the given formula—no need to overcomplicate!

Common Mistakes

Exam Traps

1-Minute Recap (Night Before the Exam)

"Right, listen up—this is your 60-second sequence survival guide.

1. Linear sequences? Find the difference (d), plug into a + (n – 1)d, simplify.
2. Quadratic sequences? Check second differences—if constant, a = (second diff) ÷ 2. Then subtract an² and find the linear part.
3. Special sequences? Memorise squares (n²), cubes (n³), triangular (n(n+1)/2), and Fibonacci (add the two before).
4. Exam traps? Watch for negative differences, hidden patterns, and non-integer d.
5. Always label n from 1—never start at 0!

Now go smash those 12 marks!