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Cambridge IGCSE Mathematics revision notes

Sequences Revision Notes

Cambridge IGCSE revision notes for Sequences.

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Sequences

CambridgeIGCSEMathematicsCore Algebra and graphs

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Extending Sequences and Calculating nth Terms

This note explains how to identify and extend number sequences by recognising patterns, and how to calculate the nth term for linear, quadratic, and cubic sequences. It includes methods to find the next terms in a sequence, understand relationships between terms, and derive formulas for different types of sequences with worked examples.

Syllabus source

Section 3, Topic C2.7, page 18, objective C2.7.1

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What you should be able to do

  • ✓Identify and extend number sequences by recognising patterns.
  • ✓Understand the relationships between terms in sequences.
  • ✓Calculate the nth term for linear sequences.
  • ✓Calculate the nth term for simple quadratic sequences.
  • ✓Calculate the nth term for simple cubic sequences.
1

Recognising and Extending Number Sequences

A number sequence is an ordered list of numbers following a specific pattern or rule. To extend a sequence, examine the differences or ratios between consecutive terms to identify the pattern.

For example, consider the sequence: 1, 3, 6, 10, 15, … To find the next terms, calculate the differences between terms:

3 - 1 = 2, 6 - 3 = 3, 10 - 6 = 4, 15 - 10 = 5. These differences increase by 1 each time, so the next differences are 6 and 7. Adding these to the last term: 15 + 6 = 21, then 21 + 7 = 28. Thus, the next two terms are 21 and 28.

  • Calculate differences between consecutive terms.
  • Look for constant or regularly changing differences.
  • Use the pattern in differences to find next terms.
2

Identifying Patterns and Relationships in Sequences

Sequences may be linear, quadratic, cubic, or follow other patterns. Understanding the relationship between terms helps to find a general formula (nth term) that generates any term in the sequence.

Linear sequences have a constant difference between terms. Quadratic sequences have differences of differences constant (second differences). Cubic sequences have constant third differences.

Recognising these patterns allows you to classify the sequence and apply the correct method to find the nth term.

  • Linear sequences: constant first difference.
  • Quadratic sequences: constant second difference.
  • Cubic sequences: constant third difference.
  • Use differences to classify the sequence type.
3

Calculating the nth Term of a Linear Sequence

A linear sequence increases or decreases by a constant amount each time. The nth term formula is: an = a + (n - 1)d, where:

a is the first term, d is the common difference, n is the term number.

For example, for the sequence 4, 7, 10, 13, …, the first term a = 4 and the common difference d = 3. The nth term is: an = 4 + (n - 1) × 3 = 3n + 1. To find the 10th term: an = 3(10) + 1 = 31.

  • Identify first term and common difference.
  • Substitute into an = a + (n - 1)d.
  • Calculate specific terms by substituting n.
4

Calculating the nth Term of a Simple Quadratic Sequence

Quadratic sequences have a second difference that is constant. The nth term is generally of the form an = an² + bn + c, where a, b, and c are constants.

To find a, b, and c: calculate first and second differences, use simultaneous equations from the first few terms, then solve for a, b, and c.

Example: For the sequence 2, 5, 10, 17, …, the first differences are 3, 5, 7 and the second differences are constant at 2. Since the second difference = 2a, then 2a = 2 so a = 1. Using the first term (n=1): a(1)² + b(1) + c = 2 → 1 + b + c = 2 → b + c = 1. Using the second term (n=2): 4a + 2b + c = 5 → 4 + 2b + c = 5 → 2b + c = 1. Solving these simultaneously gives b = 0 and c = 1. Thus, nth term is an = n² + 1.

  • Calculate first and second differences.
  • Use second difference to find a.
  • Form simultaneous equations using terms.
  • Solve for b and c.
  • Write nth term as an = an² + bn + c.
5

Calculating the nth Term of a Simple Cubic Sequence

Cubic sequences have a constant third difference. The nth term is generally of the form an = an³ + bn² + cn + d, where a, b, c, and d are constants.

Finding the nth term involves calculating first, second, and third differences, then using these to set up equations to solve for the constants.

Example: For the sequence 1, 8, 27, 64, 125, … (cubes of natural numbers), the nth term is an = n³. The third difference is constant, confirming the cubic pattern.

  • Calculate first, second, and third differences.
  • Use third difference to find a.
  • Set up equations using initial terms.
  • Solve for b, c, and d.
  • Write nth term as an = an³ + bn² + cn + d.

At-a-glance summary

Comparison table

StepDetailExampleExam Use
Identify patternCalculate differences between termsSequence: 1, 3, 6, 10, 15; differences: 2, 3, 4, 5Show working clearly to gain method marks
Classify sequenceCheck if first, second, or third differences are constantFirst difference constant → linear; second → quadratic; third → cubicUse correct formula type for nth term
Find nth termUse formula an = a + (n-1)d for linear; an = an² + bn + c for quadratic; an = an³ + bn² + cn + d for cubicFor 2, 5, 10, 17: nth term = n² + 1Write formula clearly and justify steps
Calculate termsSubstitute n into nth term formulaFind 10th term of 3n + 1: 3×10 + 1 = 31Answer specific term questions accurately
Check resultsVerify terms fit original sequenceCheck 4th term of n² + 1: 16 + 1 = 17 matches sequenceAvoid careless errors

Active recall

Quick check

1. What is the next term after 15 in the sequence 1, 3, 6, 10, 15?

The differences are 2, 3, 4, 5, so next difference is 6; next term is 15 + 6 = 21.

2. How do you identify a quadratic sequence?

By finding that the second differences between terms are constant.

3. Find the nth term of the linear sequence 5, 8, 11, 14.

First term a = 5, common difference d = 3; nth term an = 5 + (n - 1)×3 = 3n + 2.

4. What is the nth term of the sequence 2, 5, 10, 17?

The nth term is an = n² + 1, found by using constant second differences and solving for coefficients.

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