nth term of linear sequences Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE nth term of linear sequences questions. See exactly how to solve problems on common difference, linear sequence, decreasing sequence, nth term.

common differencelinear sequencedecreasing sequencenth termmultiplescommon difference of 1
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
Here are the first four terms of a sequence. 3, 5, 7, 9, 3,\ 5,\ 7,\ 9,\ \ldots Write down the common difference of the sequence.

Worked solution

  1. Find the gaps between consecutive terms

    53=2, 75=2, 97=25 - 3 = 2,\ 7 - 5 = 2,\ 9 - 7 = 2

    Subtract each term from the one after it.

  2. Check the gaps are all the same

    d=2d = 2

    Every gap is 22, so the sequence is linear and the common difference is 22.

  3. State the common difference

    22

    The terms increase by the same amount each time.

Answer
22
Question 2
1 markeasy
Here are the first four terms of a sequence. 20, 17, 14, 11, 20,\ 17,\ 14,\ 11,\ \ldots Write down the common difference of the sequence.

Worked solution

  1. Find the gaps between consecutive terms

    1720=3, 1417=3, 1114=317 - 20 = -3,\ 14 - 17 = -3,\ 11 - 14 = -3

    Subtract each term from the one after it.

  2. Check the gaps are all the same

    d=3d = -3

    Every gap is 3-3, so the sequence is linear and the common difference is 3-3.

  3. State the common difference

    3-3

    The terms are decreasing, so the common difference is negative.

Answer
3-3
Question 3
1 markeasy
Here are the first four terms of a sequence. 2, 4, 6, 8, 2,\ 4,\ 6,\ 8,\ \ldots Work out the nnth term of the sequence.

Worked solution

  1. Find the common difference

    42=2, 64=2, 86=24 - 2 = 2,\ 6 - 4 = 2,\ 8 - 6 = 2

    The gap between one term and the next is the same every time, so this is a linear (arithmetic) sequence. It goes up in 2s2s, so the common difference is d=2d = 2.

  2. Write down the multiples of the common difference

    2n:2, 4, 6, 82n:\quad 2,\ 4,\ 6,\ 8

    Because the common difference is 22, the nthn\mathrm{th} term must start with 2n2n. Working out 2n2n for n=1n = 1, 22, 33, 44 gives 22, 44, 66, 88.

  3. Write down the nthn\mathrm{th} term

    nth term=2nn\mathrm{th}\text{ term} = 2n

    Putting the 2n2n and the constant together gives the rule 2n2n.

Answer
2n2n
Question 4
1 markeasy
Here are the first four terms of a sequence. 5, 10, 15, 20, 5,\ 10,\ 15,\ 20,\ \ldots Work out the nnth term of the sequence.

Worked solution

  1. Find the common difference

    105=5, 1510=5, 2015=510 - 5 = 5,\ 15 - 10 = 5,\ 20 - 15 = 5

    The gap between one term and the next is the same every time, so this is a linear (arithmetic) sequence. It goes up in 5s5s, so the common difference is d=5d = 5.

  2. Write down the multiples of the common difference

    5n:5, 10, 15, 205n:\quad 5,\ 10,\ 15,\ 20

    Because the common difference is 55, the nthn\mathrm{th} term must start with 5n5n. Working out 5n5n for n=1n = 1, 22, 33, 44 gives 55, 1010, 1515, 2020.

  3. Write down the nthn\mathrm{th} term

    nth term=5nn\mathrm{th}\text{ term} = 5n

    Putting the 5n5n and the constant together gives the rule 5n5n.

Answer
5n5n
Question 5
2 markseasy
Here are the first four terms of a sequence. 4, 5, 6, 7, 4,\ 5,\ 6,\ 7,\ \ldots Work out the nnth term of the sequence.

Worked solution

  1. Find the common difference

    54=1, 65=1, 76=15 - 4 = 1,\ 6 - 5 = 1,\ 7 - 6 = 1

    The gap between one term and the next is the same every time, so this is a linear (arithmetic) sequence. It goes up in 1s1s, so the common difference is d=1d = 1.

  2. Write down the multiples of the common difference

    n:1, 2, 3, 4n:\quad 1,\ 2,\ 3,\ 4

    Because the common difference is 1, the nthn\mathrm{th} term must start with n. Working out n for n=1n = 1, 22, 33, 44 gives 11, 22, 33, 44.

  3. Adjust by the constant and write the nthn\mathrm{th} term

    n gives 1, 2, 3, 4; add 3n+3n \text{ gives } 1,\ 2,\ 3,\ 4; \text{ add } 3 \text{: } n + 3

    Each term of the sequence is 3 more than the matching multiple of 1, so the nthn\mathrm{th} term is n + 3. Check n=1n = 1: 1+3=1+3=41 + 3 = 1 + 3 = 4, the first term.

Answer
n+3n + 3

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