How to Find the nth Term of IGCSE Sequences: Linear, Quadratic and Beyond
Learn to find the nth term of linear and quadratic IGCSE sequences using differences, with clear examples, practice questions and answers.
By Math School Editorial Team
Why the nth term is useful
Imagine that a pattern continues for hundreds of steps. Writing out every number would take a long time. An nth-term formula gives the value at any position directly.
For example, the sequence 5, 9, 13, 17, ... increases by four each time. The tenth term is not found by multiplying 17 by ten. Instead, we look for a rule linking the position n to the value.
The first term is the value when n = 1, the second when n = 2, and so on. In exam questions, Tₙ is often used to mean the value of the nth term.
Step 1: Identify a linear sequence
A linear sequence has a constant first difference. Consider:
5, 9, 13, 17, 21, ...
Subtract consecutive terms:
9 − 5 = 4, 13 − 9 = 4, 17 − 13 = 4.
Because the difference is always four, the formula has the form Tₙ = 4n + c.
Substitute n = 1, when T₁ = 5:
4(1) + c = 5, so c = 1.
Therefore Tₙ = 4n + 1.
Check: T₂ = 4(2) + 1 = 9, and T₅ = 4(5) + 1 = 21. Both match the sequence. The twentieth term is T₂₀ = 4(20) + 1 = 81.
A useful shortcut is to take the common difference as the coefficient of n, then adjust the constant so the first term works. But always check the formula against more than one term.
Step 2: Recognise a quadratic sequence
Now consider:
3, 8, 15, 24, 35, ...
The first differences are 5, 7, 9, 11. They are not constant. The second differences are 2, 2, 2, which are constant.
This suggests a quadratic nth-term formula:
Tₙ = an² + bn + c.
For a quadratic sequence, the constant second difference equals 2a, provided the term numbers increase by one each time. Here 2a = 2, so a = 1.
Start with Tₙ = n² + bn + c.
Use the first term: 1 + b + c = 3, so b + c = 2.
Use the second term: 4 + 2b + c = 8, so 2b + c = 4.
Subtract the first equation from the second: b = 2. Then c = 0.
The formula is Tₙ = n² + 2n.
Check the fourth term: T₄ = 4² + 2(4) = 16 + 8 = 24. The sixth term is T₆ = 36 + 12 = 48.
A second quadratic example
Find the nth term of 4, 11, 22, 37, 56, ...
First differences: 7, 11, 15, 19.
Second differences: 4, 4, 4, so 2a = 4 and a = 2.
Assume Tₙ = 2n² + bn + c.
At n = 1: 2 + b + c = 4, hence b + c = 2.
At n = 2: 8 + 2b + c = 11, hence 2b + c = 3.
Subtract: b = 1, then c = 1.
Thus Tₙ = 2n² + n + 1.
Check n = 3: 2(9) + 3 + 1 = 22. To find the sixth term: 2(36) + 6 + 1 = 79.
Step 3: Know when a pattern is not linear or quadratic
Not every sequence has constant first or second differences.
For example, 2, 9, 28, 65, 126, ... follows Tₙ = n³ + 1. Check the first three terms: 1³ + 1 = 2, 2³ + 1 = 9, and 3³ + 1 = 28. This is a simple cubic pattern.
The sequence 3, 6, 12, 24, 48, ... multiplies by two each time. It is exponential, with formula Tₙ = 3 × 2⁽ⁿ⁻¹⁾. Its differences are not constant; the ratio of consecutive terms is.
For Cambridge IGCSE Mathematics (0580), the 2025–2027 syllabus includes linear, simple quadratic and simple cubic nth terms in Core (C2.7). Extended (E2.7) also covers broader patterns, including exponential sequences and combinations. Always follow the specific syllabus for your examination year.
Four common mistakes
Mistake 1: Treating term-to-term rules as nth-term rules. "Add four" explains how to move to the next term; 4n + 1 gives any term directly.
Mistake 2: Using the second difference as a. In a quadratic sequence, the second difference is 2a, not a.
Mistake 3: Starting at n = 0 without checking. Most questions label the first term T₁. Starting at zero changes the constant.
Mistake 4: Checking only one term. Several wrong formulas can accidentally fit one value. Test at least two or three terms.
Three original practice questions
Q1. Find the nth term of 7, 12, 17, 22, ... and calculate T₁₅.
Q2. Find the nth term of 6, 13, 24, 39, ... and calculate T₅.
Q3 (Extended pattern). Find the nth term of 5, 15, 45, 135, ... and calculate T₆.
Answers
A1. The difference is 5, so Tₙ = 5n + 2. T₁₅ = 75 + 2 = 77.
A2. First differences are 7, 11, 15; second difference is 4. Tₙ = 2n² + n + 3. T₅ = 2(25) + 5 + 3 = 58.
A3. Multiply by three each time. Tₙ = 5 × 3⁽ⁿ⁻¹⁾. T₆ = 5 × 3⁵ = 1215.
How to revise sequences effectively
Begin by writing a small table with the term number and term value. Calculate first differences, then second differences if necessary. Write a formula, substitute the known term numbers and check your result. Once the method feels secure, practise recognising patterns without a prompt that tells you whether they are linear or quadratic.
A targeted diagnostic assessment can help identify whether difficulties come from negative numbers, algebraic substitution, simultaneous equations or recognising number patterns.
Source and syllabus check: Cambridge IGCSE Mathematics (0580), 2025–2027 syllabus, C2.7 and E2.7. All worked examples and questions above are original.
