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Maths · Algebra

Non-linear sequences

Not every sequence goes up by the same amount. Square, cube and triangular numbers, Fibonacci sequences and geometric sequences all follow rules of their own - and at Higher, quadratic sequences have an nth term you can find from their second differences.

  • 6 key terms
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Last Lesson and Before

Answer each one, then check.

  1. 1

    Last lesson: find the nth term of 5, 8, 11, ...

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    \(3n + 2\)

  2. 2

    Last lesson: work out the 10th term of \(4n - 1\).

    Show answerHide answer

    \(39\)

  3. 3

    Write down the first four square numbers.

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    1, 4, 9, 16

  4. 4

    Work out \(2 \times 2 \times 2 \times 2\).

    Show answerHide answer

    \(16 = 2^4\)

  5. 5

    From the Number chapter: simplify \(\sqrt{2} \times \sqrt{2}\).

    Show answerHide answer

    \(2\)

Learning Objectives

  1. 1Recognise square, cube and triangular numbers and Fibonacci-type sequences.
  2. 2Continue geometric sequences and find their common ratio.
  3. 3Work out terms of a geometric sequence from its first term and ratio.
  4. 4Find the nth term of a quadratic sequence. (Higher)
  5. 5Work with geometric sequences with a surd as the ratio. (Higher)

Special Sequences

  • Square numbers

    1, 4, 9, 16, 25, ... - nth term \(n^2\).

  • Cube numbers

    1, 8, 27, 64, 125, ... - nth term \(n^3\).

  • Triangular numbers

    1, 3, 6, 10, 15, ... - add 2, then 3, then 4, ...

  • Fibonacci sequence

    1, 1, 2, 3, 5, 8, ... - each term is the sum of the two before it.

  • Powers of 2

    2, 4, 8, 16, 32, ... - nth term \(2^n\).

  • Geometric sequences

    Multiply by the same number each time: 3, 6, 12, 24, ...

Fibonacci-Type Sequences

In a Fibonacci-type sequence, each term is the sum of the two terms before it.

  • Fibonacci

    1, 1, 2, 3, 5, 8, 13, 21, ...

  • Any start

    2, 5, 7, 12, 19, 31, ... follows the same rule from a different start.

  • With letters

    Starting \(a\), \(b\), the terms are \(a\), \(b\), \(a + b\), \(a + 2b\), \(2a + 3b\), ...

  • Working back

    If two neighbouring terms are 13 and 21, the one before is \(21 - 13 = 8\).

Geometric Sequences

A geometric sequence is multiplied by the same number, the common ratio \(r\), every time.

  • Growing

    3, 6, 12, 24, ... has \(r = 2\).

  • Shrinking

    64, 32, 16, 8, ... has \(r = \frac{1}{2}\).

  • Finding \(r\)

    Divide any term by the one before it: \(24 \div 12 = 2\).

  • Any term

    The nth term is (first term) \(\times\ r^{n-1}\): the first term is multiplied by \(r\) once for each step after it.

A Geometric Sequence

A geometric sequence starts 5, 15, 45, 135, ... Find the common ratio and the 8th term.

Show the solutionHide the solution
  1. 1 Divide a term by the one before \(15 \div 5 = 3\), so \(r = 3\)
  2. 2 The 8th term is 7 steps after the first \(5 \times 3^7\)
  3. 3 Work out \(3^7\) \(3^7 = 2187\)
  4. 4 Multiply \(5 \times 2187 = 10\,935\)

Answer\(r = 3\); 8th term \(= 10\,935\)

Second Differences

2, 7, 14, 23, 34, ... The first differences change, but the second differences are all 2: the sequence is quadratic.

  • 1

    Term: 2

  • 2

    Term: 7. First difference: \(+5\)

  • 3

    Term: 14. First difference: \(+7\). Second difference: \(+2\)

  • 4

    Term: 23. First difference: \(+9\). Second difference: \(+2\)

  • 5

    Term: 34. First difference: \(+11\). Second difference: \(+2\)

The nth Term of a Quadratic Sequence

The nth term has the form \(an^2 + bn + c\).

  1. 1 Second difference

    Find it: it is always \(2a\).
  2. 2 The squared part

    Halve the second difference to get \(a\), and write \(an^2\).
  3. 3 Subtract

    Take \(an^2\) away from each term.
  4. 4 The linear part

    Find the nth term of what is left (\(bn + c\)).
  5. 5 Combine and check

    \(an^2 + bn + c\). Check with \(n = 1\) and \(n = 2\).

The nth Term of a Quadratic Sequence

Find the nth term of 2, 7, 14, 23, 34, ...

Show the solutionHide the solution
  1. 1 Second difference 2, so \(a = 1\) \(n^2\)
  2. 2 \(n^2\) gives 1, 4, 9, 16, 25
  3. 3 Sequence minus \(n^2\) 1, 3, 5, 7, 9
  4. 4 The nth term of 1, 3, 5, 7, 9 \(2n - 1\)
  5. 5 Combine \(n^2 + 2n - 1\)

Answer\(n^2 + 2n - 1\) (check \(n = 3\): \(9 + 6 - 1 = 14\))

A Harder Quadratic Sequence

Find the nth term of 4, 13, 26, 43, 64, ...

Show the solutionHide the solution
  1. 1 First differences 9, 13, 17, 21
  2. 2 Second difference 4, so \(a = 2\) \(2n^2\)
  3. 3 \(2n^2\) gives 2, 8, 18, 32, 50
  4. 4 Sequence minus \(2n^2\) 2, 5, 8, 11, 14, with nth term \(3n - 1\)
  5. 5 Combine \(2n^2 + 3n - 1\)

Answer\(2n^2 + 3n - 1\) (check \(n = 2\): \(8 + 6 - 1 = 13\))

Geometric Sequences with Surds

The common ratio can be a surd.

  • Example

    \(\sqrt{2}\), 2, \(2\sqrt{2}\), 4, \(4\sqrt{2}\), ... has \(r = \sqrt{2}\), because \(\sqrt{2} \times \sqrt{2} = 2\).

  • Every other term

    Two steps multiply by \((\sqrt{2})^2 = 2\), so every second term doubles.

  • Finding a term

    The 7th term of \(\sqrt{3}\), 3, \(3\sqrt{3}\), ... is \(\sqrt{3} \times (\sqrt{3})^6 = \sqrt{3} \times 27 = 27\sqrt{3}\).

Case study

Fibonacci and His Rabbits

In 1202 the Italian mathematician Leonardo of Pisa, later known as Fibonacci, published Liber Abaci, the book that helped bring Hindu-Arabic numerals to Europe. In it he set a puzzle about a pair of rabbits that breeds a new pair every month. The number of pairs month by month gives 1, 1, 2, 3, 5, 8, 13, ... - the sequence now named after him. Divide each term by the one before (\(8 \div 5 = 1.6\), \(13 \div 8 = 1.625\), \(21 \div 13 = 1.615\ldots\)) and the answers close in on \(1.618\ldots\), the golden ratio.

1202 Liber Abaci is published
1.618... The golden ratio, which the ratio of neighbouring terms approaches

What Kind of Sequence?

For each sequence, say whether it is linear, geometric, Fibonacci-type or quadratic, and write down the next two terms. (a) 2, 6, 18, 54, ... (b) 3, 4, 7, 11, 18, ... (c) 4, 9, 14, 19, ... (d) 1, 4, 9, 16, ... (e) 80, 40, 20, 10, ... (f) 3, 6, 11, 18, 27, ... Higher: find the nth term of (f).

1. Check the differences.

2. Check the ratios.

3. Check whether each term is the sum of the two before.

A good answer shows: (a) Geometric, \(r = 3\): 162, 486. (b) Fibonacci-type: 29, 47. (c) Linear: 24, 29. (d) Quadratic (square numbers): 25, 36. (e) Geometric, \(r = \frac{1}{2}\): 5, 2.5. (f) Quadratic: 38, 51; nth term \(n^2 + 2\).

Can I...?

  1. 1Recognise square, cube and triangular numbers.
  2. 2Continue a Fibonacci-type sequence.
  3. 3Find the common ratio of a geometric sequence.
  4. 4Work out a term of a geometric sequence.
  5. 5Find second differences. (Higher)
  6. 6Find the nth term of a quadratic sequence. (Higher)
  7. 7Use a surd as a common ratio. (Higher)
  8. 8Work backwards in a Fibonacci-type sequence.

Summary & Exam Focus

  • Special sequences: square, cube, triangular, Fibonacci, powers of 2.
  • Geometric: multiply by the common ratio \(r\) each time.
  • Fibonacci-type: add the two previous terms.
  • (Higher) Quadratic: constant second difference; \(a\) = second difference \(\div\ 2\), then find the linear part of what is left.

Exam focus

Find an expression for the nth term of the quadratic sequence 3, 9, 19, 33, 51, ... (3 marks) (3 marks)

Write the first and second differences under the sequence before you do anything else. They tell you what kind of sequence it is and give you the first part of a quadratic nth term.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Triangular numbers
1, 3, 6, 10, 15, ... - the numbers of dots in triangular patterns.
Fibonacci-type sequence
Each term is the sum of the two terms before it.
Geometric sequence
Each term is multiplied by the same number to get the next.
Common ratio
The number a geometric sequence is multiplied by each time.
Quadratic sequence (Higher)
A sequence whose second differences are all the same; its nth term includes \(n^2\).
Second difference (Higher)
The difference between neighbouring first differences.

Practice questions

Have a go at each one before you open its answer.

  1. Question 1 Non-calculator 1 mark

    Here are the first six terms of a Fibonacci sequence: 1, 1, 2, 3, 5, 8. Write down the next term.

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    Model answer

    13

    Mark scheme

    • 13 — B1
  2. Question 2 Non-calculator 2 marks

    Here are the first four terms of a geometric sequence: 2, 6, 18, 54. (a) Write down the next term. (b) Work out the 6th term.

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    Model answer

    (a) \(54 \times 3 = 162\) (b) \(162 \times 3 = 486\)

    Mark scheme

    • (a) 162 — B1
    • (b) 486 — B1
  3. Question 3 Non-calculator · Higher 3 marks

    The first two terms of a Fibonacci-type sequence are \(a\) and \(b\). The 3rd term is 7 and the 5th term is 18. Find the values of \(a\) and \(b\).

    Show answerHide answer

    Model answer

    The terms are \(a\), \(b\), \(a + b\), \(a + 2b\), \(2a + 3b\). So \(a + b = 7\) and \(2a + 3b = 18\). Doubling the first: \(2a + 2b = 14\), so \(b = 4\) and \(a = 3\).

    Mark scheme

    • Terms written in terms of \(a\) and \(b\), at least to \(a + 2b\) — M1
    • \(a + b = 7\) and \(2a + 3b = 18\) — M1
    • \(a = 3\) and \(b = 4\) — A1
  4. Question 4 Non-calculator · Higher 3 marks

    Find an expression for the nth term of the quadratic sequence 3, 9, 19, 33, 51, ...

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    Model answer

    First differences 6, 10, 14, 18; second difference 4, so \(2n^2\). \(2n^2\) gives 2, 8, 18, 32, 50; the sequence minus \(2n^2\) is 1, 1, 1, 1, 1. nth term \(2n^2 + 1\).

    Mark scheme

    • Second difference of 4, or \(2n^2\) seen — M1
    • Sequence minus \(2n^2\) found, e.g. 1, 1, 1, ... — M1
    • \(2n^2 + 1\) — A1
  5. Question 5 Non-calculator · Higher 3 marks

    Find an expression for the nth term of the quadratic sequence 0, 5, 12, 21, 32, ...

    Show answerHide answer

    Model answer

    First differences 5, 7, 9, 11; second difference 2, so \(n^2\). \(n^2\) gives 1, 4, 9, 16, 25; the sequence minus \(n^2\) is \(-1\), 1, 3, 5, 7, with nth term \(2n - 3\). nth term \(n^2 + 2n - 3\).

    Mark scheme

    • Second difference of 2, or \(n^2\) seen — M1
    • \(-1\), 1, 3, 5, 7 or \(2n - 3\) found — M1
    • \(n^2 + 2n - 3\) — A1
  6. Question 6 Non-calculator · Higher 2 marks

    The first term of a geometric sequence is \(\sqrt{3}\) and the common ratio is \(\sqrt{3}\). Show that the 5th term is \(9\sqrt{3}\).

    Show answerHide answer

    Model answer

    5th term \(= \sqrt{3} \times (\sqrt{3})^4 = \sqrt{3} \times 9 = 9\sqrt{3}\)

    Mark scheme

    • \(\sqrt{3} \times (\sqrt{3})^4\), or the terms \(\sqrt{3}\), 3, \(3\sqrt{3}\), 9, \(9\sqrt{3}\) listed — M1
    • Correct working showing \(9\sqrt{3}\) — A1

Quick check

  1. What are the next two terms of the triangular numbers 1, 3, 6, 10, ...?

    1. A14, 18
    2. B15, 21
    3. C16, 22
    4. D13, 16
    Show answerHide answer

    B: 15, 21

    The differences go \(+2\), \(+3\), \(+4\), then \(+5\) and \(+6\): 15 and 21.

  2. What is the common ratio of the geometric sequence 81, 27, 9, 3, ...?

    1. A\(3\)
    2. B\(-54\)
    3. C\(-3\)
    4. D\(\frac{1}{3}\)
    Show answerHide answer

    D: \(\frac{1}{3}\)

    \(27 \div 81 = \frac{1}{3}\). Each term is a third of the one before.

  3. (Higher) What is the nth term of 2, 5, 10, 17, 26, ...?

    1. A\(n^2 + 1\)
    2. B\(3n - 1\)
    3. C\(2n^2\)
    4. D\(n^2 + n\)
    Show answerHide answer

    A: \(n^2 + 1\)

    Second difference 2 gives \(n^2\): 1, 4, 9, 16, 25. The sequence is 1 more each time: \(n^2 + 1\).

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