Hard GCSE Generating sequences Questions

Challenging, exam-style GCSE Generating sequences questions with worked solutions. Stretch yourself on the hardest position-to-term rule, is it a term, reasoning, solving an equation problems.

position-to-term ruleis it a termreasoningsolving an equationterm-to-term ruleworking backwards
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
The nnth term of a sequence is 2n+52n + 5. Sam says, "Every term of this sequence is odd." Is Sam right? Explain your answer, and work out the 5050th term.
Show worked solution

Worked solution

  1. Generate the first term

    2×1+5=72 \times 1 + 5 = 7

    The first term is 77, which is odd.

  2. Generate the second term

    2×2+5=92 \times 2 + 5 = 9

    The second term is 99, also odd.

  3. Generate the third term

    2×3+5=112 \times 3 + 5 = 11

    The third term is 1111, odd again.

  4. Notice that examples are not a proof

    7, 9, 11, …7,\ 9,\ 11,\ \ldots

    Three odd terms do not prove that ALL of them are odd — you need a reason.

  5. Look at the first part of the rule

    2n is always even2n \text{ is always even}

    22 multiplied by any whole number gives an even number.

  6. Add the 55

    even+5=odd\text{even} + 5 = \text{odd}

    An even number plus an odd number is always odd.

  7. Answer the first part

    Sam is right\text{Sam is right}

    Every term of the sequence is odd.

  8. Start the 50th50th term

    2×50=1002 \times 50 = 100

    Substitute n=50n = 50 into the rule.

  9. Finish the 50th50th term

    100+5=105100 + 5 = 105

    The 50th50th term is 105105.

  10. Check it against the rule

    105 is odd105 \text{ is odd}

    Consistent with the argument above.

  11. Name the common error

    n=50 is even, so the term must be evenn = 50 \text{ is even, so the term must be even}

    This is the trap: students often assume the term copies the parity of the position.

  12. Explain why that is wrong

    the +5 changes the parity\text{the } +5 \text{ changes the parity}

    2n is even whatever n is, so the +5 always tips the term to odd.

  13. Test an odd position too

    2×51+5=1072 \times 51 + 5 = 107

    The 51st term is odd as well — the parity of n makes no difference.

  14. Describe the sequence

    7, 9, 11, 13, …7,\ 9,\ 11,\ 13,\ \ldots

    The sequence is simply the odd numbers from 77 upwards.

  15. State the full answer

    Sam is right; the 50th term is 105\text{Sam is right; the 50th term is } 105

    Every term is odd, and the 50th50th term is 105105.

Answer
Sam is right; the 50th term is 105\text{Sam is right; the 50th term is } 105
Question 2
5 markschallenging
The nnth term of sequence A is 3n+13n + 1. The nnth term of sequence B is 2n+92n + 9. Work out the position of the first term of A that is greater than the term of B in the same position.
Show worked solution

Worked solution

  1. Write down both rules

    3n+1and2n+93n + 1 \quad \text{and} \quad 2n + 9

    The same position n is used in each sequence.

  2. Compare the first terms

    4vs114 \quad \text{vs} \quad 11

    At n=1n = 1, B is well ahead.

  3. Compare the 5th5th terms

    16vs1916 \quad \text{vs} \quad 19

    At n=5n = 5, B is still ahead — but by less.

  4. Explain why A catches up

    A rises by 3,B rises by 2A \text{ rises by } 3, \quad B \text{ rises by } 2

    A gains 1 on B every term, so A must overtake eventually.

  5. Write down the condition

    3n+1>2n+93n + 1 > 2n + 9

    You want the term of A to be BIGGER.

  6. Subtract 2n2n from both sides

    n+1>9n + 1 > 9

    Collect the n terms.

  7. Subtract 11 from both sides

    n>8n > 8

    Collect the numbers.

  8. Choose the right whole number

    n=9n = 9

    n must be a whole number strictly greater than 8, so the first one is 9 — not 8.

  9. Check n=8n = 8: sequence A

    3×8+1=253 \times 8 + 1 = 25

    Test the position just below.

  10. Check n=8n = 8: sequence B

    2×8+9=252 \times 8 + 9 = 25

    At n=8n = 8 the two terms are EQUAL, not bigger.

  11. Check n=9n = 9: sequence A

    3×9+1=283 \times 9 + 1 = 28

    Now test position 99.

  12. Check n=9n = 9: sequence B

    2×9+9=272 \times 9 + 9 = 27

    B's 9th term is 27.

  13. Compare them

    28>2728 > 27

    A's 9th term is bigger than B's 9th term by 1.

  14. Confirm it is the first

    at n=8 they are equal\text{at } n = 8 \text{ they are equal}

    Since A only overtakes at n=9n = 9, position 99 is the first one that works.

  15. State the answer

    n=9n = 9

    The 9th term of A is the first to beat B.

Answer
n=9n = 9
Question 3
6 markschallenging
A sequence has term-to-term rule 'halve the previous term, then add 44'. The 44th term is 99. Work out the first term.
Show worked solution

Worked solution

  1. Write down the rule

    an+1=an2+4a_{n+1} = \frac{a_n}{2} + 4

    Forwards: halve, then add 44.

  2. Write down the inverse rule

    an=(an+1−4)×2a_n = (a_{n+1} - 4) \times 2

    Backwards: undo the LAST operation first, so subtract 44, then double.

  3. Start undoing from the 4th4th term

    (9−4)×2(9 - 4) \times 2

    Work back to the 3rd3rd term.

  4. Do the subtraction

    9−4=59 - 4 = 5

    Undo the +44.

  5. Do the doubling

    5×2=105 \times 2 = 10

    The 3rd3rd term is 1010.

  6. Undo again

    (10−4)×2(10 - 4) \times 2

    Work back to the 2nd2nd term.

  7. Do the subtraction

    10−4=610 - 4 = 6

    Undo the +44.

  8. Do the doubling

    6×2=126 \times 2 = 12

    The 2nd2nd term is 1212.

  9. Undo once more

    (12−4)×2(12 - 4) \times 2

    Work back to the 1st1st term.

  10. Do the subtraction

    12−4=812 - 4 = 8

    Undo the +44.

  11. Do the doubling

    8×2=168 \times 2 = 16

    The 1st1st term is 1616.

  12. Check forwards to the 2nd2nd term

    16÷2+4=1216 \div 2 + 4 = 12

    1616 halves to 88, plus 44 is 1212.

  13. Check forwards to the 3rd3rd term

    12÷2+4=1012 \div 2 + 4 = 10

    1212 halves to 66, plus 44 is 1010.

  14. Check forwards to the 4th4th term

    10÷2+4=910 \div 2 + 4 = 9

    1010 halves to 55, plus 44 is 99 — this matches the question.

  15. State the answer

    1616

    The sequence is 1616, 1212, 1010, 99, so the first term is 1616.

Answer
1616
Question 4
6 markschallenging
The nnth term of sequence A is 4n+14n + 1. Sequence B has first term 33 and term-to-term rule 'add 44'. Explain why sequences A and B have no terms in common.
Show worked solution

Worked solution

  1. Generate the first five terms of A

    5, 9, 13, 17, 215,\ 9,\ 13,\ 17,\ 21

    Substituting n=1n = 1 to 55 into 4n+14n + 1 gives 55, 99, 1313, 1717, 2121.

  2. Generate the first five terms of B

    3, 7, 11, 15, 193,\ 7,\ 11,\ 15,\ 19

    Starting at 33 and adding 44 gives 33, 77, 1111, 1515, 1919.

  3. Notice both go up in 4s4s

    common difference 4\text{common difference } 4

    The two sequences rise at exactly the same rate, but they start in different places.

  4. Describe A's terms

    5=4+1,9=8+1,13=12+15 = 4 + 1, \quad 9 = 8 + 1, \quad 13 = 12 + 1

    Every term of A is 11 more than a multiple of 44.

  5. Describe B's terms

    3=0+3,7=4+3,11=8+33 = 0 + 3, \quad 7 = 4 + 3, \quad 11 = 8 + 3

    Every term of B is 3 more than a multiple of 4.

  6. Write B's mth term

    bm=3+4(m−1)=4m−1b_m = 3 + 4(m - 1) = 4m - 1

    The mth term of B is 3 plus (m - 1) jumps of 4.

  7. Check B's formula

    m=1: 3,m=3: 11m = 1: \ 3, \quad m = 3: \ 11

    It reproduces 33 and 1111 correctly.

  8. Suppose the two sequences share a term

    4n+1=4m−14n + 1 = 4m - 1

    Assume a term of A equals a term of B and see what happens.

  9. Collect the multiples of 44

    4m−4n=24m - 4n = 2

    Rearranging gives 4m−4n=24m - 4n = 2.

  10. Factorise the left-hand side

    4(m−n)=24(m - n) = 2

    The left side is a multiple of 44.

  11. Divide by 22

    2(m−n)=12(m - n) = 1

    The left side is now even.

  12. Spot the contradiction

    even≠odd\text{even} \neq \text{odd}

    The left side is even and the right side is 11, which is odd — impossible for whole numbers.

  13. State the conclusion of the algebra

    no whole numbers m,n work\text{no whole numbers } m, n \text{ work}

    There is no position in A and no position in B giving the same term.

  14. Give the remainder version of the argument

    A leaves remainder 1, B leaves remainder 3\text{A leaves remainder } 1, \text{ B leaves remainder } 3

    When divided by 4, every term of A leaves remainder 1 and every term of B leaves remainder 3.

  15. Conclude

    A and B share no terms\text{A and B share no terms}

    A number cannot leave two different remainders when divided by 44, so the sequences never meet.

Answer
No common terms: A leaves remainder 1 and B leaves remainder 3 on division by 4\text{No common terms: A leaves remainder } 1 \text{ and B leaves remainder } 3 \text{ on division by } 4
Question 5
5 markschallenging
A sequence is defined by xn+1=2xn−3x_{n+1} = 2x_n - 3 with x1=4x_1 = 4. Write down the first five terms of the sequence.
Show worked solution

Worked solution

  1. Read the rule

    xn+1=2xn−3x_{n+1} = 2x_n - 3

    Double the previous term, then subtract 33.

  2. Write down the first term

    x1=4x_1 = 4

    The first term is GIVEN — do not apply the rule to get it.

  3. Double the first term

    2×4=82 \times 4 = 8

    Start working out x2.

  4. Subtract 33

    8−3=58 - 3 = 5

    So x2=5x2 = 5.

  5. Double the second term

    2×5=102 \times 5 = 10

    Now work out x3.

  6. Subtract 33

    10−3=710 - 3 = 7

    So x3=7x3 = 7.

  7. Double the third term

    2×7=142 \times 7 = 14

    Now work out x4.

  8. Subtract 33

    14−3=1114 - 3 = 11

    So x4=11x4 = 11.

  9. Double the fourth term

    2×11=222 \times 11 = 22

    Now work out x5.

  10. Subtract 33

    22−3=1922 - 3 = 19

    So x5=19x5 = 19.

  11. Write the five terms in order

    4, 5, 7, 11, 194,\ 5,\ 7,\ 11,\ 19

    These are the first five terms.

  12. Check the differences

    1, 2, 4, 81,\ 2,\ 4,\ 8

    The gaps double each time, which is what a "multiply by 22" rule does — a good sign the terms are right.

  13. Watch out for the position trap

    2n−3⇒−1, 1, 3, 5, 72n - 3 \Rightarrow -1,\ 1,\ 3,\ 5,\ 7

    Using the POSITION n instead of the previous term would give -1, 1, 3, 5, 7 — a completely different sequence.

  14. Watch out for the bracket trap

    2(xn−3)⇒4, 2, −2, −102(x_n - 3) \Rightarrow 4,\ 2,\ -2,\ -10

    Subtracting 33 BEFORE doubling would give 44, 22, −2-2, −10-10 — the order of operations matters.

  15. State the answer

    4, 5, 7, 11, 194,\ 5,\ 7,\ 11,\ 19

    The first five terms are 44, 55, 77, 1111 and 1919.

Answer
4, 5, 7, 11, 194,\ 5,\ 7,\ 11,\ 19

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