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Sequences and the nth Term

Continuing sequences, finding the nth term of linear and quadratic sequences, and testing whether a number is in a sequence.

  • 10 key terms
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Learning Objectives

  1. 1Recognise and continue linear, geometric and special sequences, and describe them using term-to-term rules.
  2. 2Find the nth term of a linear sequence and use it to find terms or test whether a number is in a sequence.
  3. 3Use sequences in context, such as matchstick patterns.
  4. 4Find the nth term of a quadratic sequence (Higher tier).

Patterns that follow a rule

A sequence is a list of numbers that follows a rule. You can describe the rule in two ways: a term-to-term rule that tells you how to get from one term to the next, and a position-to-term rule, the nth term, that gives any term directly from its position. The nth term is far more powerful, because it lets you find the 100th term without writing out the first 99.

Describing sequences

Each number in a sequence is called a term, and the first term is the one in position 1.

  • Term-to-term rule

    It says how to get from one term to the next. The sequence 5, 8, 11, 14 has the rule "add 3".

  • Position-to-term rule

    The nth term is a formula using \(n\), the position. If the nth term is \(3n + 2\), the 10th term is \(3 \times 10 + 2 = 32\).

  • Arithmetic and geometric

    An arithmetic sequence goes up or down by the same amount each time, called the common difference. A geometric sequence is multiplied by the same number each time, such as 3, 6, 12, 24, which doubles.

  • Special sequences

    Square numbers 1, 4, 9, 16; cube numbers 1, 8, 27, 64; triangular numbers 1, 3, 6, 10, 15; powers of 2 such as 2, 4, 8, 16; Fibonacci-type sequences where each term is the sum of the two before, such as 1, 1, 2, 3, 5, 8.

The nth term of a linear sequence

The common difference is the number in front of \(n\).

  • Use the common difference

    If the terms go up by 3, the nth term starts \(3n\).

  • Find the adjustment

    Write out \(3n\) (3, 6, 9, 12) and compare it with the sequence (5, 8, 11, 14). The sequence is always 2 more, so the nth term is \(3n + 2\).

  • Decreasing sequences

    If the terms go down by 3, the coefficient is \(-3\). For 20, 17, 14, 11 write \(-3n\) (\(-3, -6, -9, -12\)); the sequence is 23 more, so the nth term is \(-3n + 23\).

  • Check it

    Substitute \(n = 1\): \(3 \times 1 + 2 = 5\), which is the first term.

Finding the nth term

Find an expression for the nth term of the sequence 7, 11, 15, 19, ...

Show the solutionHide the solution
  1. 1 Find the common difference \(11 - 7 = 4\), so the nth term begins \(4n\).
  2. 2 Write out \(4n\) The sequence \(4n\) is 4, 8, 12, 16.
  3. 3 Find the adjustment 7 is 3 more than 4, and 11 is 3 more than 8, so add 3.
  4. 4 Write the nth term \(4n + 3\). Check: when \(n = 4\), \(4 \times 4 + 3 = 19\).

Answer\(4n + 3\)

Is a number in the sequence?

The nth term of a sequence is \(3n + 2\). Is 100 a term of the sequence? Is 101?

Show the solutionHide the solution
  1. 1 Set the nth term equal to the number \(3n + 2 = 100\).
  2. 2 Solve \(3n = 98\), so \(n = 32.67\ldots\). A position must be a whole number, so 100 is not in the sequence.
  3. 3 Try 101 \(3n + 2 = 101\) gives \(3n = 99\), so \(n = 33\). 101 is the 33rd term.

Answer100 is not a term; 101 is the 33rd term

Sequences in context

Patterns of shapes are a favourite exam question because they connect a picture to an nth term.

  • Count and compare

    Count the matchsticks, tiles or dots in the first few patterns and record them in a table.

  • Find what is added

    In a row of squares made from matchsticks, each new square needs 3 more sticks, so the nth term starts \(3n\).

  • Check the start

    Pattern 1 has 4 matchsticks, which is \(3 \times 1 + 1\), so the nth term is \(3n + 1\).

  • Explain in context

    The \(3n\) is the three sticks added for each square, and the \(+1\) is the extra stick at the start.

Quadratic sequences (Higher tier)

When the first differences are not constant, look at the differences of the differences.

  • Second difference

    If the second difference is constant, the sequence is quadratic and the nth term begins \(an^2\), where \(a\) is half the second difference.

  • Subtract

    Subtract \(an^2\) from each term. What is left is a linear sequence.

  • Find its nth term

    Use the usual method for a linear sequence, and add it to \(an^2\).

  • Check

    Substitute \(n = 1\), \(2\) and \(3\) into your answer.

Finding a quadratic nth term

Find the nth term of the sequence 4, 10, 18, 28, 40, ...

Show the solutionHide the solution
  1. 1 First differences \(6, 8, 10, 12\).
  2. 2 Second differences \(2, 2, 2\). This is constant, so the sequence is quadratic, with \(a = 2 \div 2 = 1\).
  3. 3 Subtract \(n^2\) \(n^2\) is 1, 4, 9, 16, 25, and the sequence minus these is 3, 6, 9, 12, 15.
  4. 4 Find the linear nth term 3, 6, 9, 12, 15 is \(3n\).
  5. 5 Combine The nth term is \(n^2 + 3n\). Check: when \(n = 3\), \(9 + 9 = 18\).

Answer\(n^2 + 3n\)

Two ways to describe a sequence

Term-to-term rule

  • Tells you how to get from one term to the next
  • Needs the previous term, so it is slow for the 100th term
  • Example: add 3 each time

Position-to-term rule (nth term)

  • Gives any term directly from its position
  • Lets you find the 100th term in one step
  • Example: \(3n + 2\)

Sequences from patterns and tables

A table is the quickest way to turn a pattern of shapes into an nth term.

  • Record the pattern

    Make a table with the pattern number \(n\) in one row and the number of matchsticks, tiles or dots in the next.

  • Find the difference

    If each new pattern adds the same amount, that amount is the coefficient of \(n\).

  • Fix the start

    Compare \(3n\) with the actual numbers to find the extra, such as \(+1\).

  • Explain in context

    The part with \(n\) is what is added for each new shape, and the number on its own is what is there to begin with.

Finding the first term above a value

The nth term of a sequence is \(3n + 2\). Find the first term of the sequence that is greater than 100.

Show the solutionHide the solution
  1. 1 Write an inequality \(3n + 2 > 100\).
  2. 2 Solve it \(3n > 98\), so \(n > 32.67\ldots\).
  3. 3 Choose the smallest whole number \(n = 33\), because the position must be a whole number.
  4. 4 Work out the term \(3 \times 33 + 2 = 101\).

Answer101

Finding the nth term from two terms

The 3rd term of an arithmetic sequence is 11 and the 7th term is 27. Find the nth term.

Show the solutionHide the solution
  1. 1 Find the common difference From the 3rd to the 7th term there are 4 steps, and \(27 - 11 = 16\), so the difference is \(16 \div 4 = 4\).
  2. 2 Start the nth term The nth term begins \(4n\), and \(4 \times 3 = 12\).
  3. 3 Adjust The 3rd term is 11, which is 1 less than 12, so the nth term is \(4n - 1\).
  4. 4 Check When \(n = 7\), \(4 \times 7 - 1 = 27\).

Answer\(4n - 1\)

Geometric sequences and the nth term (Higher tier)

A geometric sequence is multiplied by the same number each time.

  • Common ratio

    The number you multiply by. In 3, 6, 12, 24 the common ratio is 2.

  • The nth term

    For a first term \(a\) and common ratio \(r\), the nth term is \(ar^{n-1}\). For the sequence above it is \(3 \times 2^{n-1}\).

  • Example

    The 5th term is \(3 \times 2^4 = 3 \times 16 = 48\).

  • Decreasing sequences

    A common ratio between 0 and 1 gives a decreasing sequence. In 80, 40, 20 the ratio is \(\tfrac{1}{2}\).

A quadratic sequence with a fractional coefficient (Higher tier)

Find the nth term of the triangular numbers 1, 3, 6, 10, 15.

Show the solutionHide the solution
  1. 1 First differences \(2, 3, 4, 5\).
  2. 2 Second differences \(1, 1, 1\), so the coefficient of \(n^2\) is half of 1, which is \(\tfrac{1}{2}\).
  3. 3 Subtract \(\tfrac{1}{2}n^2\) \(\tfrac{1}{2}n^2\) is 0.5, 2, 4.5, 8, 12.5. The sequence minus this is 0.5, 1, 1.5, 2, 2.5, which is \(\tfrac{1}{2}n\).
  4. 4 Combine The nth term is \(\tfrac{1}{2}n^2 + \tfrac{1}{2}n = \dfrac{n(n + 1)}{2}\). Check: when \(n = 5\), \(\dfrac{5 \times 6}{2} = 15\).

Answer\(\tfrac{1}{2}n^2 + \tfrac{1}{2}n\)

Common sequence mistakes

What students write

  • The nth term of 5, 8, 11 is \(n + 3\)
  • The nth term of 5, 8, 11 is \(3n\)
  • 100 is in the sequence \(3n + 2\) because \(n\) is about 33

What is correct

  • The nth term uses the common difference as the coefficient of \(n\)
  • Compare \(3n\) with the terms to find the adjustment, which gives \(3n + 2\)
  • \(n\) must be a whole number, and \(n = 32.67\ldots\) is not, so 100 is not in the sequence

Test yourself

  1. 1

    What is the next term of 2, 6, 18, 54?

    Show answerHide answer

    162, because each term is multiplied by 3.

  2. 2

    What is the nth term of 3, 7, 11, 15?

    Show answerHide answer

    \(4n - 1\).

  3. 3

    What is the 10th term of the sequence with nth term \(2n + 5\)?

    Show answerHide answer

    25.

  4. 4

    What is the common difference of 5, 11, 17, 23?

    Show answerHide answer

    6.

  5. 5

    Is 24 a term of the sequence with nth term \(3n\)?

    Show answerHide answer

    Yes, because \(3n = 24\) gives \(n = 8\), which is a whole number.

Exam technique: sequences

Sequence questions are short and follow the same pattern from paper to paper.

  • Show the differences

    Write the differences above or below the terms, as that is where the method mark is.

  • Test your nth term

    Substitute \(n = 1\) and \(n = 2\) before you move on.

  • Use a whole-number test

    To decide whether a number is in a sequence, solve for \(n\) and check it is a positive whole number.

  • Finish with a sentence

    Say "No, because \(n\) is not a whole number" rather than leaving the working unexplained.

Summary and exam focus

  • A term-to-term rule uses the previous term; the nth term uses the position.
  • The coefficient of \(n\) in a linear sequence is the common difference.
  • To find the nth term, write out the multiples of the difference, find the adjustment and test with \(n = 1\).
  • To test whether a number is in a sequence, set the nth term equal to it and solve; \(n\) must be a positive whole number.
  • For a quadratic sequence, half the second difference is the coefficient of \(n^2\); subtract and find the linear remainder.

Exam focus

Here are the first four terms of an arithmetic sequence: 9, 15, 21, 27. (a) Find an expression for the nth term. (b) Is 150 a term in this sequence? You must show how you decide. (4 marks) (4 marks)

For part (b), set your nth term equal to 150 and solve for \(n\); if \(n\) is not a whole number, the answer is no, and you must say so. Answering "no" with no working scores nothing, because the working is what is being tested.

Key terms

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

Common ratio
The number that each term of a geometric sequence is multiplied by to get the next term.
Sequence
A list of numbers that follow a rule.
Term
One of the numbers in a sequence.
Term-to-term rule
A rule that tells you how to get from one term to the next.
nth term
A formula that gives the term in position \(n\) of a sequence.
Arithmetic sequence
A sequence that increases or decreases by the same amount each time.
Common difference
The constant amount added to get from one term of an arithmetic sequence to the next.
Geometric sequence
A sequence in which each term is multiplied by the same number to get the next.
Fibonacci-type sequence
A sequence in which each term is the sum of the two terms before it.
Quadratic sequence
A sequence whose nth term includes an \(n^2\) term, so its second difference is constant.

Questions and answers

17 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Find 2 marks Core

Find an expression for the nth term of the sequence \(5, 12, 19, 26, \ldots\)

Mark scheme — 2 marks available

  • \(7n\) seen — M1
  • \(7n - 2\) — A1

Model answer

The difference is 7, so the nth term begins \(7n\). \(7n\) is \(7, 14, 21, 28\), and each term is 2 less, so the nth term is \(7n - 2\).

2. Exam question Work out 5 marks Core

Here are the first three patterns in a sequence. The patterns are made from matchsticks. (a) Work out the number of matchsticks in Pattern 10. [1 mark] (b) Find an expression, in terms of \(n\), for the number of matchsticks in Pattern \(n\). [2 marks] (c) Which pattern number uses exactly 70 matchsticks? [2 marks]

The first three patterns in a sequence: one square, two squares and three squares in a row, made from matchsticks.

Mark scheme — 5 marks available

  • (a) 31, or \(3 \times 10 + 1\) — B1
  • (b) \(3n\) seen — M1
  • (b) \(3n + 1\) — A1
  • (c) \(3n + 1 = 70\) or \(3n = 69\) — M1
  • (c) Pattern 23 — A1

Model answer

(b) Each new square needs 3 more matchsticks and Pattern 1 has 4, so the nth term is \(3n + 1\). (a) \(3 \times 10 + 1 = 31\). (c) \(3n + 1 = 70\), so \(3n = 69\) and \(n = 23\). It is Pattern 23.

3. Exam question Work out 4 marks Core

The nth term of a sequence is \(2n^2 + 1\). (a) Work out the first three terms of the sequence. [2 marks] (b) Is 51 a term of this sequence? You must show how you decide. [2 marks]

Mark scheme — 4 marks available

  • (a) Two terms correct — M1
  • (a) 3, 9, 19 — A1
  • (b) \(2n^2 = 50\) or \(n^2 = 25\) — M1
  • (b) Yes, it is the 5th term — A1

Model answer

(a) \(2 \times 1^2 + 1 = 3\), \(2 \times 2^2 + 1 = 9\) and \(2 \times 3^2 + 1 = 19\). (b) \(2n^2 + 1 = 51\) gives \(2n^2 = 50\), so \(n^2 = 25\) and \(n = 5\). Yes, 51 is the 5th term.

4. Exam question Work out 3 marks Stretch

Here are the first four terms of a geometric sequence. \(2\) \(6\) \(18\) \(54\) (a) Write down the next term. [1 mark] (b) Work out the 7th term. [2 marks]

Mark scheme — 3 marks available

  • (a) 162 — B1
  • (b) \(2 \times 3^6\) or continues the sequence to 486 and 1458 — M1
  • (b) 1458 — A1

Model answer

(a) Each term is multiplied by 3, so the next term is \(54 \times 3 = 162\). (b) The 7th term is \(2 \times 3^6 = 2 \times 729 = 1458\).

5. Exam question Work out 2 marks Core

The nth term of a sequence is \(3n + 5\). Sally says that 80 is a term in this sequence. Is Sally correct? You must show how you decide.

Mark scheme — 2 marks available

  • \(3n + 5 = 80\) or \(3n = 75\) — M1
  • Yes, with \(n = 25\) — A1

Model answer

Solve \(3n + 5 = 80\): \(3n = 75\), so \(n = 25\). Since 25 is a whole number, 80 is the 25th term, so Sally is correct.

6. Exam question Find 3 marks Stretch

Here are the first four terms of a quadratic sequence. \(3\) \(10\) \(21\) \(36\) Find an expression for the nth term.

Mark scheme — 3 marks available

  • Second difference 4 found, or \(2n^2\) seen — M1
  • Subtracts \(2n^2\) to leave 1, 2, 3, 4 — M1
  • \(2n^2 + n\) — A1

Model answer

The first differences are 7, 11, 15 and the second difference is 4, so the nth term starts \(2n^2\), which is 2, 8, 18, 32. Subtracting leaves 1, 2, 3, 4, which is \(n\). The nth term is \(2n^2 + n\).

7. Exam question Work out 4 marks Stretch

Here are the first three terms of a sequence. \(17\) \(13\) \(9\) (a) Find an expression for the nth term. [2 marks] (b) Work out the first term in the sequence that is negative. [2 marks]

Mark scheme — 4 marks available

  • (a) \(-4n\) seen — M1
  • (a) \(-4n + 21\) — A1
  • (b) \(-4n + 21 < 0\), or finds the terms 5 and 1 then \(-3\) — M1
  • (b) \(-3\) — A1

Model answer

(a) The sequence goes down by 4, so it begins \(-4n\). \(-4n\) is \(-4, -8, -12\), and each term is 21 more, so the nth term is \(-4n + 21\). (b) \(-4n + 21 < 0\) gives \(n > 5.25\), so the first negative term is when \(n = 6\): \(-4 \times 6 + 21 = -3\).

8. Multiple choice 1 mark Stretch

The nth term of a sequence is \(3n + 2\). What is the first term that is greater than 100?

  1. A 104
  2. B 98
  3. C 101 Correct
  4. D 100

Why: \(3n + 2 > 100\) gives \(n > 32.67\ldots\), so \(n = 33\) and the term is \(3 \times 33 + 2 = 101\).

9. Multiple choice 1 mark Core

What is the common ratio of the geometric sequence 5, 10, 20, 40, ...?

  1. A 10
  2. B 2 Correct
  3. C 5
  4. D 15

Why: Each term is the previous term multiplied by 2.

10. Multiple choice 1 mark Easier

What is the next term in the sequence 2, 5, 8, 11, ...?

  1. A 15
  2. B 17
  3. C 13
  4. D 14 Correct

Why: The sequence goes up by 3 each time, so the next term is \(11 + 3 = 14\).

11. Multiple choice 1 mark Core

What is the nth term of the sequence 4, 7, 10, 13, ...?

  1. A \(4n + 3\)
  2. B \(3n + 4\)
  3. C \(3n + 1\) Correct
  4. D \(n + 3\)

Why: The difference is 3, so it begins \(3n\). The sequence is 1 more than \(3n\), so the nth term is \(3n + 1\).

12. Multiple choice 1 mark Core

Which sequence has nth term \(2n - 3\)?

  1. A \(-1, 1, 3, 5\) Correct
  2. B \(2, 4, 6, 8\)
  3. C \(5, 7, 9, 11\)
  4. D \(-3, -1, 1, 3\)

Why: Substituting \(n = 1, 2, 3, 4\) gives \(-1, 1, 3, 5\).

13. Multiple choice 1 mark Stretch

What is the nth term of the sequence 20, 17, 14, 11, ...?

  1. A \(-3n + 23\) Correct
  2. B \(-3n + 20\)
  3. C \(23n - 3\)
  4. D \(3n + 17\)

Why: The sequence goes down by 3, so it begins \(-3n\). Adding 23 gives 20 when \(n = 1\).

14. Multiple choice 1 mark Core

Which of these numbers is a term in the sequence with nth term \(4n + 1\)?

  1. A 63
  2. B 61 Correct
  3. C 62
  4. D 64

Why: \(4n + 1 = 61\) gives \(n = 15\). The other numbers do not give a whole number for \(n\).

15. Multiple choice 1 mark Core

Which of these is a geometric sequence?

  1. A \(3, 6, 9, 12\)
  2. B \(3, 6, 12, 24\) Correct
  3. C \(1, 4, 9, 16\)
  4. D \(1, 1, 2, 3\)

Why: Each term is multiplied by 2: \(3 \times 2 = 6\), \(6 \times 2 = 12\), \(12 \times 2 = 24\).

16. Multiple choice 1 mark Core

The first four terms of a Fibonacci-type sequence are 1, 3, 4, 7. What is the 5th term?

  1. A 11 Correct
  2. B 12
  3. C 14
  4. D 10

Why: Each term is the sum of the two before it, so the 5th term is \(4 + 7 = 11\).

17. Multiple choice 1 mark Stretch

A sequence has a constant second difference of 2. Its nth term begins with which term?

  1. A \(4n^2\)
  2. B \(n^2\) Correct
  3. C \(2n\)
  4. D \(2n^2\)

Why: The coefficient of \(n^2\) is half the second difference, so it is \(1\), giving \(n^2\).