Hard GCSE nth term of linear sequences Questions

Challenging, exam-style GCSE nth term of linear sequences questions with worked solutions. Stretch yourself on the hardest nth term, integer test, is it a term, justification problems.

nth terminteger testis it a termjustificationsimultaneous equationsnon-consecutive terms
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
Sequence A has first four terms 4, 10, 16, 22, …4,\ 10,\ 16,\ 22,\ \ldots Sequence B has first four terms 32, 34, 36, 38, …32,\ 34,\ 36,\ 38,\ \ldots Work out the first position at which the term of sequence A is greater than the term of sequence B in the same position.
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Worked solution

  1. Find the common difference of sequence A

    10−4=6, 16−10=6, 22−16=610 - 4 = 6,\ 16 - 10 = 6,\ 22 - 16 = 6

    A has common difference 6.

  2. Deduce the nth term of sequence A

    A=6n−2\text{A} = 6n - 2

    6n gives 6,\ 12,\ 18,\ 24, and each term of A is 2 less, so A=6n−2A = 6n - 2.

  3. Check the rule for A

    n=1: 6×1−2=6−2=4n=4: 6×4−2=24−2=22n = 1:\ 6 \times 1 - 2 = 6 - 2 = 4 \qquad n = 4:\ 6 \times 4 - 2 = 24 - 2 = 22

    It reproduces 4 and 22, the first and fourth terms of A.

  4. Find the common difference of sequence B

    34−32=2, 36−34=2, 38−36=234 - 32 = 2,\ 36 - 34 = 2,\ 38 - 36 = 2

    B has common difference 2.

  5. Deduce the nth term of sequence B

    B=2n+30\text{B} = 2n + 30

    2n gives 2,\ 4,\ 6,\ 8, and each term of B is 30 more, so B=2n+30B = 2n + 30.

  6. Check the rule for B

    n=1: 2×1+30=2+30=32n=4: 2×4+30=8+30=38n = 1:\ 2 \times 1 + 30 = 2 + 30 = 32 \qquad n = 4:\ 2 \times 4 + 30 = 8 + 30 = 38

    It reproduces 32 and 38, the first and fourth terms of B.

  7. Set up the inequality

    6n−2>2n+306n - 2 > 2n + 30

    The question asks when a term of A first beats the term of B in the SAME position, so compare the two rules at the same n.

  8. Collect the n terms on one side

    6n−(2n)>30−(−2)6n - (2n) > 30 - (-2)

    Subtracting 2n and -2 from both sides gives 4n>324n > 32.

  9. Simplify

    4n>324n > 32

    The n terms combine into 4n.

  10. Divide to find the boundary

    n>32÷4=8n > 32 \div 4 = 8

    So n must be greater than 8.

  11. Round up to the first whole position

    n=9n = 9

    A position must be a whole number, and it must be strictly greater than 8, so the first one that works is n=9n = 9.

  12. Check position 8

    A=6×8−2=48−2=46B=2×8+30=16+30=46\text{A} = 6 \times 8 - 2 = 48 - 2 = 46 \qquad \text{B} = 2 \times 8 + 30 = 16 + 30 = 46

    At position 8, A gives 46 and B gives 46, so A is not yet ahead (46 is not greater than 46).

  13. Check position 9

    A=6×9−2=54−2=52B=2×9+30=18+30=48\text{A} = 6 \times 9 - 2 = 54 - 2 = 52 \qquad \text{B} = 2 \times 9 + 30 = 18 + 30 = 48

    At position 9, A gives 52 and B gives 48, and 52 is greater than 48. This is the first position where A overtakes B.

  14. Sense check the gap between the sequences

    gap=(6n−2)−(2n+30)=4n−32\text{gap} = (6n - 2) - (2n + 30) = 4n - 32

    The gap A - B is 4n - 32, which increases by 4 each position - so once A is ahead it stays ahead.

  15. State the answer

    n=9n = 9

    The first position at which the term of A is greater than the term of B is n=9n = 9.

Answer
99
Question 2
6 markschallenging
Here are the first four terms of a sequence. 100, 91, 82, 73, …100,\ 91,\ 82,\ 73,\ \ldots Show that 00 is not a term of this sequence, and work out the first term of the sequence that is negative.
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Worked solution

  1. Find the common difference

    91−100=−9, 82−91=−9, 73−82=−991 - 100 = -9,\ 82 - 91 = -9,\ 73 - 82 = -9

    The gap between one term and the next is the same every time, so this is a linear (arithmetic) sequence. It goes down in 9s, so the common difference is d=−9d = -9.

  2. Write down the multiples of the common difference

    −9n:−9, −18, −27, −36-9n:\quad -9,\ -18,\ -27,\ -36

    Because the common difference is -9, the nth term must start with -9n. Working out -9n for n=1n = 1, 2, 3, 4 gives -9, -18, -27, -36.

  3. Compare the multiples with the sequence

    100−(−9)=109, 91−(−18)=109, 82−(−27)=109, 73−(−36)=109100 - (-9) = 109,\ 91 - (-18) = 109,\ 82 - (-27) = 109,\ 73 - (-36) = 109

    Every term is 109 more than the matching multiple of 9, so the constant is 109.

  4. Write down the nth term

    nth term=109−9n\text{nth term} = 109 - 9n

    Putting the -9n and the constant together gives the rule 109 - 9n.

  5. Check the rule against the given terms

    n=1: 109−9×1=109−9=100n=4: 109−9×4=109−36=73n = 1:\ 109 - 9 \times 1 = 109 - 9 = 100 \qquad n = 4:\ 109 - 9 \times 4 = 109 - 36 = 73

    Substituting n=1n = 1 gives 100 and n=4n = 4 gives 73, which are the first and last terms given. The rule is correct - always test n=1n = 1, because starting the count at the wrong place is the most common mistake.

  6. Test whether zero is a term

    109−9n=0109 - 9n = 0

    If 0 appeared in the sequence there would be a position n giving exactly 0, so set the rule equal to 0 and solve.

  7. Rearrange the equation

    9n=1099n = 109

    Adding 9n to both sides gives 9n=1099n = 109.

  8. Divide to find the position

    n=109÷9=12.1111n = 109 \div 9 = 12.1111

    n=12.1111n = 12.1111.

  9. Apply the whole-number test

    n=12.1111∉{1,2,3,…}n = 12.1111 \notin \{1, 2, 3, \ldots\}

    A position must be a positive whole number, and 12.1111 is not one, so 0 is NOT a term of the sequence. The sequence jumps from 1 straight to -8.

  10. Set up an inequality for a negative term

    109−9n<0109 - 9n < 0

    A negative term is one whose value is below zero.

  11. Rearrange the inequality

    9n>1099n > 109

    Adding 9n to both sides and moving the 0 across gives 9n>1099n > 109; keeping the n term positive avoids flipping the inequality.

  12. Divide to find the boundary

    n>109÷9=12.1111n > 109 \div 9 = 12.1111

    The terms are negative once n is past 12.1111.

  13. Round up to the first whole position

    n=13n = 13

    The smallest whole number greater than 12.1111 is 13, so the first negative term is in position 13.

  14. Work out the first negative term

    n=13: 109−9×13=109−117=−8n = 13:\ 109 - 9 \times 13 = 109 - 117 = -8

    The 13th term is -8.

  15. Check the term before it is still positive

    n=12: 109−9×12=109−108=1n = 12:\ 109 - 9 \times 12 = 109 - 108 = 1

    The 12th term is 1, which is positive, so -8 really is the FIRST negative term.

Answer
−8-8
Question 3
6 markschallenging
Here are the first four terms of a sequence. 4, 11, 18, 25, …4,\ 11,\ 18,\ 25,\ \ldots Two terms next to each other in this sequence add up to 239239. Work out the two terms.
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Worked solution

  1. Find the common difference

    11−4=7, 18−11=7, 25−18=711 - 4 = 7,\ 18 - 11 = 7,\ 25 - 18 = 7

    The gap between one term and the next is the same every time, so this is a linear (arithmetic) sequence. It goes up in 7s, so the common difference is d=7d = 7.

  2. Write down the multiples of the common difference

    7n:7, 14, 21, 287n:\quad 7,\ 14,\ 21,\ 28

    Because the common difference is 7, the nth term must start with 7n. Working out 7n for n=1n = 1, 2, 3, 4 gives 7, 14, 21, 28.

  3. Compare the multiples with the sequence

    4−7=−3, 11−14=−3, 18−21=−3, 25−28=−34 - 7 = -3,\ 11 - 14 = -3,\ 18 - 21 = -3,\ 25 - 28 = -3

    Every term is 3 less than the matching multiple of 7, so the constant is -3.

  4. Write down the nth term

    nth term=7n−3\text{nth term} = 7n - 3

    Putting the 7n and the constant together gives the rule 7n - 3.

  5. Check the rule against the given terms

    n=1: 7×1−3=7−3=4n=4: 7×4−3=28−3=25n = 1:\ 7 \times 1 - 3 = 7 - 3 = 4 \qquad n = 4:\ 7 \times 4 - 3 = 28 - 3 = 25

    Substituting n=1n = 1 gives 4 and n=4n = 4 gives 25, which are the first and last terms given. The rule is correct - always test n=1n = 1, because starting the count at the wrong place is the most common mistake.

  6. Write an expression for a term and the next term

    term n=7n−3,term (n+1)=7(n+1)+(−3)\text{term } n = 7n - 3, \qquad \text{term } (n+1) = 7(n + 1) + (-3)

    The two terms are consecutive, so if the first is at position n the second is at position n + 1. Using one letter for both is what makes this solvable.

  7. Expand the second expression

    7(n+1)+(−3)=7n+47(n + 1) + (-3) = 7n + 4

    Multiplying out gives 7n + 4.

  8. Add the two expressions

    (7n−3)+(7n+4)=14n+1(7n - 3) + (7n + 4) = 14n + 1

    Collecting like terms: 7n+7n=14n7n + 7n = 14n, and the constants give 1.

  9. Form an equation using the given total

    14n+1=23914n + 1 = 239

    The two terms add up to 239, so set the expression equal to 239.

  10. Solve for n

    14n=238⇒n=238÷14=1714n = 238 \quad \Rightarrow \quad n = 238 \div 14 = 17

    n=17n = 17, a positive whole number, so such a pair really does exist.

  11. Work out the first term of the pair

    n=17: 7×17−3=119−3=116n = 17:\ 7 \times 17 - 3 = 119 - 3 = 116

    The 17th term is 116.

  12. Work out the second term of the pair

    n=18: 7×18−3=126−3=123n = 18:\ 7 \times 18 - 3 = 126 - 3 = 123

    The 18th term is 123.

  13. Check the two terms add to the given total

    116+123=239116 + 123 = 239

    They add to 239, exactly as required.

  14. Check the two terms are consecutive

    123−116=7123 - 116 = 7

    They differ by 7, the common difference, so they really are next to each other in the sequence.

  15. State the two terms

    116 and 123116 \text{ and } 123

    The two consecutive terms are 116 and 123.

Answer
116 and 123116 \text{ and } 123
Question 4
6 markschallenging
Sequence A has first four terms 2, 9, 16, 23, …2,\ 9,\ 16,\ 23,\ \ldots Sequence B has first four terms 5, 9, 13, 17, …5,\ 9,\ 13,\ 17,\ \ldots Work out the smallest number greater than 3030 that is a term of both sequences.
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Worked solution

  1. Find the common difference of sequence A

    9−2=7, 16−9=7, 23−16=79 - 2 = 7,\ 16 - 9 = 7,\ 23 - 16 = 7

    A has common difference 7.

  2. Deduce the nth term of sequence A

    A=7n−5\text{A} = 7n - 5

    7n gives 7,\ 14,\ 21,\ 28, and each term of A is 5 less than that, so A=7n−5A = 7n - 5.

  3. Check the rule for A

    n=1: 7×1−5=7−5=2n=4: 7×4−5=28−5=23n = 1:\ 7 \times 1 - 5 = 7 - 5 = 2 \qquad n = 4:\ 7 \times 4 - 5 = 28 - 5 = 23

    It reproduces the first and fourth terms of A, 2 and 23.

  4. Find the common difference of sequence B

    9−5=4, 13−9=4, 17−13=49 - 5 = 4,\ 13 - 9 = 4,\ 17 - 13 = 4

    B has common difference 4.

  5. Deduce the nth term of sequence B

    B=4n+1\text{B} = 4n + 1

    4n gives 4,\ 8,\ 12,\ 16, and each term of B is 1 more than that, so B=4n+1B = 4n + 1.

  6. Check the rule for B

    n=1: 4×1+1=4+1=5n=4: 4×4+1=16+1=17n = 1:\ 4 \times 1 + 1 = 4 + 1 = 5 \qquad n = 4:\ 4 \times 4 + 1 = 16 + 1 = 17

    It reproduces the first and fourth terms of B, 5 and 17.

  7. Understand what a common term means

    7n+(−5)=4m+(1)7n + (-5) = 4m + (1)

    A number in BOTH sequences need not be in the same position in each, so use two different letters: position n in A and position m in B.

  8. List terms of A

    A:2, 9, 16, 23, 30, 37, 44, 51, 58, 65, …\text{A}: 2,\ 9,\ 16,\ 23,\ 30,\ 37,\ 44,\ 51,\ 58,\ 65,\ \ldots

    Writing out the terms of A makes the search concrete.

  9. List terms of B

    B:5, 9, 13, 17, 21, 25, 29, 33, 37, 41, …\text{B}: 5,\ 9,\ 13,\ 17,\ 21,\ 25,\ 29,\ 33,\ 37,\ 41,\ \ldots

    Now the two lists can be compared.

  10. Find the smallest number in both lists

    9 is in A and in B9 \ \text{is in A and in B}

    The smallest number appearing in both sequences is 9.

  11. Find how often common terms repeat

    7×4=287 \times 4 = 28

    After a shared value, the next shared value comes 28 later: it must be a further whole number of 7s (to stay in A) and of 4s (to stay in B), and the smallest number that is both is 7×4=287 \times 4 = 28.

  12. Build the list of common terms

    9, 37, 65, 93, …9,\ 37,\ 65,\ 93,\ \ldots

    The common terms go up in 28s from 9.

  13. Pick the first common term above the limit

    37>3037 > 30

    Working along that list, 37 is the first common term greater than 30.

  14. Check it is a term of A

    7n−5=37 ⇒ n=6(7×6−5=42−5=37)7n - 5 = 37 \ \Rightarrow \ n = 6 \quad (7 \times 6 - 5 = 42 - 5 = 37)

    n=6n = 6 is a positive whole number, so 37 is the 6th term of A.

  15. Check it is a term of B

    4n+1=37 ⇒ m=9(4×9+1=36+1=37)4n + 1 = 37 \ \Rightarrow \ m = 9 \quad (4 \times 9 + 1 = 36 + 1 = 37)

    m=9m = 9 is a positive whole number, so 37 is also the 9th term of B. Both tests pass, so 37 is the answer.

Answer
3737
Question 5
6 markschallenging
A sequence is linear. Its 33rd term is 2020 and its 1010th term is −1-1. Is −40-40 a term of this sequence? If it is, write down its position. You must show all your working.
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Worked solution

  1. Set up the general form of the nth term

    nth term=dn+c\text{nth term} = dn + c

    The sequence is linear (arithmetic), so its nth term is dn + c, where d is the common difference and c is a constant. Two facts are given, so two equations can be formed.

  2. Use the 3rd term to form the first equation

    3d+c=203d + c = 20

    Substituting n=3n = 3 into dn + c must give the 3rd term, 20.

  3. Use the 10th term to form the second equation

    10d+c=−110d + c = -1

    Substituting n=10n = 10 into dn + c must give the 10th term, -1.

  4. Subtract the first equation from the second

    (10d+c)−(3d+c)=−1−20(10d + c) - (3d + c) = -1 - 20

    The constant c cancels, leaving 7d=−217d = -21. This works because the terms are 7 places apart, so they differ by 7 common differences.

  5. Work out the common difference

    7d=−21⇒d=−21÷7=−37d = -21 \quad \Rightarrow \quad d = -21 \div 7 = -3

    The common difference is d=−3d = -3.

  6. Substitute back to find the constant

    3×−3+c=20⇒−9+c=20⇒c=293 \times -3 + c = 20 \quad \Rightarrow \quad -9 + c = 20 \quad \Rightarrow \quad c = 29

    Putting d=−3d = -3 into the first equation gives c=29c = 29.

  7. Write down the nth term

    nth term=29−3n\text{nth term} = 29 - 3n

    The rule for the sequence is 29 - 3n.

  8. Check the 3rd term

    n=3: 29−3×3=29−9=20n = 3:\ 29 - 3 \times 3 = 29 - 9 = 20

    The rule gives 20, which is the 3rd term stated in the question.

  9. Check the 10th term

    n=10: 29−3×10=29−30=−1n = 10:\ 29 - 3 \times 10 = 29 - 30 = -1

    The rule gives -1, which is the 10th term stated in the question. Both given facts are reproduced, so 29 - 3n is right.

  10. Set the nth term equal to the number

    29−3n=−4029 - 3n = -40

    If -40 is a term of the sequence, there must be a position n for which the rule gives -40. Solving this equation finds that position - if it exists.

  11. Rearrange the equation

    3n=29−(−40)=693n = 29 - (-40) = 69

    Collect the n term on one side: 3n=693n = 69.

  12. Divide to find the position

    n=69÷3=23n = 69 \div 3 = 23

    Dividing gives n=23n = 23.

  13. Check that the position is a positive whole number

    n=23∈{1,2,3,…}n = 23 \in \{1, 2, 3, \ldots\}

    n=23n = 23 is a positive whole number, so there really is a 23rd term - a position must be a counting number, and 23 is one.

  14. State the conclusion

    Yes: 29−3×23=29−69=−40\text{Yes: } 29 - 3 \times 23 = 29 - 69 = -40

    Substituting n=23n = 23 back into the rule gives -40, confirming that -40 is the 23rd term of the sequence.

  15. Answer the question in words

    −40=term 23-40 = \text{term } 23

    Yes - -40 is a term of the sequence, and it is the 23rd term.

Answer
2323

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