IGCSE Maths 0580 Sequences Past Paper Questions and Worked Solutions

By /
IGCSE Maths (0580) · Algebra and Graphs · 2.7

Every sequences question from 20 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 2.7. 20 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/12/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Algebra and Graphs past paper questions.

0580 Sequences past paper questions: Core (Papers 1 and 3)

16 marksCore

May/June 2026 · Paper 1 · 0580/12/M/J/26 · Q21

These are the first four diagrams in a sequence made using grey squares and white squares.

0580/12 May/June 2026 Q21 – diagrams 1 to 4 of grey and white square patternsdiagram 1diagram 2diagram 3diagram 4

(a) Draw diagram 5.

(b) Complete the table for diagram 4 and diagram n: number of grey squares (1, 2, 3, …), white squares (0, 2, 6, …, n(n − 1)), total squares (1, 4, 9, …).

(c) Explain why a diagram in this sequence cannot have a total of 120 squares.

Show solution
Part (a)
  1. Each new diagram adds one more grey square to the row and one more column of white squares above and below.
  2. Draw diagram 5 with a row of 5 grey squares and the white columns stepping down on each side.
0580/12 May/June 2026 Q21(a) solution – diagram 5 drawn on the griddiagram 5
Part (b)
  1. Grey squares go 1, 2, 3, … so diagram 4 has 4 grey squares and diagram n has n.
  2. White squares are n(n − 1). Diagram 4: 4 × 3 = 12
  3. Total for diagram 4 = 4 + 12 = 16
  4. Total for diagram n = n + n(n − 1) = n + n² − n = n²
Answer 4 n 12 16 n²
Part (c)
  1. The totals are 1, 4, 9, 16, … which are square numbers (n²).
  2. 10² = 100 and 11² = 121, so 120 is not n² for any whole number n.
  3. 120 is not a square number
Answer 120 is not a square number
26 marksCore

May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q25

(a) These are the first four terms of a sequence: 3, 7, 11, 15. (i) Find the nth term. (ii) Show that 603 is a term in the sequence.

(b) The nth term of a different sequence is n² − 6. Find the first three terms of this sequence.

Show solution
Part (a)(i)
  1. The terms go up by 4, so the nth term starts with 4n.
  2. 4n gives 4, 8, 12, 16. The sequence is 3, 7, 11, 15, which is 1 less each time.
  3. nth term = 4n − 1
Answer 4n − 1
Part (a)(ii)
  1. Set the nth term equal to 603: 4n − 1 = 603.
  2. Add 1: 4n = 604. Divide by 4: n = 151.
  3. 151 is a whole number, so 603 is the 151st term
Answer 603 is the 151st term
Part (b)
  1. Substitute n = 1, 2 and 3 into n² − 6.
  2. n = 1: 1 − 6 = −5. n = 2: 4 − 6 = −2. n = 3: 9 − 6 = 3.
  3. −5, −2, 3
Answer −5, −2, 3
33 marksCore

May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q15

The nth term of a sequence is 3n − 1.

(a) Write down the first 3 terms in this sequence.

(b) 327 and 331 are not in the sequence. Write down all the numbers from this list that are in the sequence: 325, 326, 327, 328, 329, 330, 331, 332

Show solution
Part (a)
  1. Substitute n = 1, 2 and 3 into 3n − 1.
  2. n = 1: 3 − 1 = 2. n = 2: 6 − 1 = 5. n = 3: 9 − 1 = 8.
  3. 2, 5, 8
Answer 2, 5, 8
Part (b)
  1. Every term is 1 less than a multiple of 3 (3n − 1).
  2. Multiples of 3 near the list: 327, 330 and 333.
  3. One less than each: 326, 329 and 332.
  4. 326, 329, 332
Answer 326, 329, 332
45 marksCore

May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q8

(a) In this sequence the first term and the sixth term are missing: …, 17, 13, 8, 2, … Fill in the first term and the sixth term.

(b) These are the first 4 terms in a different sequence: 3⁄2, 6⁄2, 9⁄2, 12⁄2, … Write down the nth term of the sequence.

(c) The term-to-term rule for another sequence is “multiply by −2”. The first term is −1. Write down the next three terms.

Show solution
Part (a)
  1. Find the differences between the terms you know: 17 → 13 is −4, 13 → 8 is −5, 8 → 2 is −6.
  2. The differences go down by 1 each time. So the difference before 17 is −3, and after 2 it is −7.
  3. First term: 17 + 3 = 20
  4. Sixth term: 2 − 7 = −5
Answer 20 −5
Part (b)
  1. The denominator is always 2. The numerators are 3, 6, 9, 12, which is the 3 times table: 3n.
  2. nth term = 3n⁄2
Answer 3n⁄2
Part (c)
  1. Multiply each term by −2 to get the next one.
  2. −1 × (−2) = 2. 2 × (−2) = −4. −4 × (−2) = 8.
  3. 2, −4, 8
Answer 2, −4, 8
53 marksCore

Feb/Mar 2026 · Paper 1 · 0580/12/F/M/26 · Q5

(a) These are the first four terms of a sequence: 6000, 600, 60, 6. Write down the term-to-term rule for this sequence.

(b) These are the first four terms of a different sequence: 27, 21, 15, 9. Find the next two terms in the sequence.

Show solution
Part (a)
  1. Compare the terms: 6000 → 600 → 60 → 6. Each term is one tenth of the term before.
  2. Rule: divide by 10
Answer divide by 10
Part (b)
  1. Find the difference: 21 − 27 = −6, 15 − 21 = −6, 9 − 15 = −6. The terms go down by 6.
  2. Next terms: 9 − 6 = 3, then 3 − 6 = −3.
  3. 3, −3
Answer 3, −3
69 marksCore

Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q16

These are the first four diagrams in a sequence made using grey squares and white squares.

0580/32 February/March 2026 Q16 – diagrams 1 to 4 of grey and white squares, and a grid for diagram 5

(a) Draw diagram 5 on the grid.

(b) Complete the table for diagram 4 and diagram n: grey squares 2, 4, 6, …; white squares 0, 2, 6, …, n(n − 1); total squares 2, 6, 12, …

(c) Complete this statement: the total number of squares in diagram n is the same as the number of white squares in diagram …

(d) Mae has 1000 white squares but only 44 grey squares. She makes the largest possible diagram with her squares. Work out the number of white squares she uses.

Show solution
Part (a)
  1. Each diagram adds 2 more grey squares and a new column of white squares. Diagram 5 has 10 grey squares and 5 × 4 = 20 white squares.
0580/32 February/March 2026 Q16(a) solution – diagram 5 drawn on the grid
Part (b)
  1. Grey squares go 2, 4, 6, … (up by 2), so diagram 4 has 8 and diagram n has 2n
  2. White squares for diagram 4: n(n − 1) = 4 × 3 = 12
  3. Total for diagram 4 = 8 + 12 = 20
  4. Total for diagram n = 2n + n(n − 1) = 2n + n² − n = n² + n = n(n + 1)
Answer 8 2n 12 20 n(n + 1)
Part (c)
  1. White squares in diagram m are m(m − 1). We want m(m − 1) = n(n + 1).
  2. This works when m = n + 1, because (n + 1)(n + 1 − 1) = (n + 1)n.
  3. n + 1
Answer n + 1
Part (d)
  1. The grey squares limit the size: 2n = 44, so n = 22.
  2. White squares in diagram 22 = 22 × 21 = 462. She has 1000, which is enough.
  3. She uses 462 white squares
Answer 462
72 marksCore

Oct/Nov 2025 · Paper 1 · 0580/11/O/N/25 · Q7

These are the first four terms of a sequence: 9, 5, 1, −3. (a) Find the next term. (b) Explain how you worked out your answer.

Show solution
Part (a)
  1. Find the difference between terms: 5 − 9 = −4, 1 − 5 = −4, −3 − 1 = −4.
  2. Next term = −3 − 4 = −7
Answer −7
Part (b)
  1. Each term is 4 less than the term before, so I subtracted 4.
Answer Each term is 4 less than the term before, so I subtracted 4.
84 marksCore

Oct/Nov 2025 · Paper 1 · 0580/12/O/N/25 · Q27

(a) These are the first four terms of a sequence: 2, 8, 14, 20. Find the nth term.

(b) The nth term of a different sequence is n² + 10. Find the first three terms of this sequence.

Show solution
Part (a)
  1. The terms go up by 6 each time, so the nth term starts with 6n.
  2. 6n gives 6, 12, 18, 24. The sequence is 2, 8, 14, 20, which is 4 less each time.
  3. nth term = 6n − 4
Answer 6n − 4
Part (b)
  1. Substitute n = 1, 2 and 3 into n² + 10.
  2. n = 1: 1 + 10 = 11. n = 2: 4 + 10 = 14. n = 3: 9 + 10 = 19.
  3. 11, 14, 19
Answer 11, 14, 19
96 marksCore

Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q21

These are the first four terms of a sequence: −6, 1, 8, 15

(a) Write down the next term. (b) Write down the term-to-term rule.

(c) (i) Find the nth term. (ii) Is 688 a term in this sequence? Explain how you decide.

Show solution
Part (a)
  1. Find the differences: 1 − (−6) = 7, 8 − 1 = 7, 15 − 8 = 7.
  2. Next term = 15 + 7 = 22
Answer 22
Part (b)
  1. Term-to-term rule: Add 7
Answer Add 7
Part (c)(i)
  1. The difference is 7, so the nth term starts with 7n.
  2. 7n gives 7, 14, 21, … but the sequence is −6, 1, 8, … which is 13 less each time.
  3. nth term = 7n − 13
Answer 7n − 13
Part (c)(ii)
  1. Set the nth term equal to 688: 7n − 13 = 688.
  2. Add 13: 7n = 701. Divide by 7: n = 100.14…
  3. n must be a whole number for 688 to be in the sequence.
  4. No, because n would not be a whole number
Answer No, because n would not be a whole number
104 marksCore

Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q18

(a) The nth term of a sequence is n² − 4. Find the first 3 terms.

(b) These are the first four terms of a different sequence: 5, 2, −1, −4. Find the nth term.

Show solution
Part (a)
  1. Substitute n = 1, 2 and 3 into n² − 4.
  2. n = 1: 1 − 4 = −3. n = 2: 4 − 4 = 0. n = 3: 9 − 4 = 5.
  3. −3, 0, 5
Answer −3, 0, 5
Part (b)
  1. The terms go down by 3, so the nth term starts with −3n.
  2. −3n gives −3, −6, −9, … The sequence is 5, 2, −1, … which is 8 more each time.
  3. nth term = 8 − 3n
Answer 8 − 3n
115 marksCore

May/June 2025 · Paper 1 · 0580/11/M/J/25 · Q9

(a) These are the first four terms of a sequence: 33, 26, 19, 12.

(i) Write down the term-to-term rule for this sequence.

(ii) Work out the next two terms in this sequence.

(b) These are the first four terms of another sequence: 19, 23, 27, 31. Find the nth term.

Show solution
Part (a)(i)
  1. Find the difference between terms: 26 − 33 = −7, 19 − 26 = −7, 12 − 19 = −7.
  2. Term-to-term rule: subtract 7
Answer subtract 7
Part (a)(ii)
  1. Next term: 12 − 7 = 5. Term after that: 5 − 7 = −2.
  2. 5, −2
Answer 5, −2
Part (b)
  1. The terms go up by 4 each time, so the nth term starts with 4n.
  2. Compare 4n with the sequence: 4n gives 4, 8, 12, 16 but the sequence is 19, 23, 27, 31.
  3. Each term is 15 more than 4n, so add 15.
  4. nth term = 4n + 15
  5. Check: n = 1 gives 4 + 15 = 19 ✓
Answer 4n + 15
122 marksCore

May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q5

Find the next term in each sequence. (a) 1, 5, 10, 16, 23, … (b) 1, 2, 4, 8, 16, …

Show solution
Part (a)
  1. Find the differences: 5 − 1 = 4, 10 − 5 = 5, 16 − 10 = 6, 23 − 16 = 7.
  2. The difference goes up by 1 each time, so the next difference is 8.
  3. Next term = 23 + 8 = 31
Answer 31
Part (b)
  1. Each term is double the one before: 1, 2, 4, 8, 16.
  2. Next term = 16 × 2 = 32
Answer 32

0580 Sequences past paper questions: Extended (Papers 2 and 4)

136 marksExtended

May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q11

(a) A, B and C are linear sequences. A has 1st term 3. B has 1st term −4 and nth term Bₙ = n − 5. C has terms −1, 2, 5, 8. For these sequences, Cₙ = Aₙ + Bₙ. Work out Aₙ, the nth term of sequence A.

(b) These are the first five terms of another sequence: 1⁄2, 1, 2, 4, 8. (i) Find the next term. (ii) Find the nth term.

Show solution
Part (a)
  1. Find the nth term of C: −1, 2, 5, 8 go up by 3, so Cₙ = 3n + something. 3 × 1 = 3, and the first term is −1, so Cₙ = 3n − 4.
  2. Cₙ = Aₙ + Bₙ, so Aₙ = Cₙ − Bₙ.
  3. Aₙ = (3n − 4) − (n − 5) = 3n − 4 − n + 5 = 2n + 1
  4. Check: n = 1 gives 2 + 1 = 3, which is the first term of A ✓
Answer 2n + 1
Part (b)(i)
  1. Each term is double the one before: 8 × 2 = 16
Answer 16
Part (b)(ii)
  1. Write the terms as powers of 2: 1⁄2 = 2⁻¹, 1 = 2⁰, 2 = 2¹, 4 = 2², 8 = 2³.
  2. The power is always 2 less than the term number: n = 1 gives 2⁻¹.
  3. nth term = 2ⁿ⁻²
Answer 2ⁿ⁻²
144 marksExtended

May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q14

(a) The nth term of a sequence is n² − 3. Write down the first three terms.

(b) These are the first four terms of another sequence: 4, 12, 36, 108. Find the nth term.

Show solution
Part (a)
  1. Substitute n = 1, 2 and 3 into n² − 3.
  2. n = 1: 1 − 3 = −2. n = 2: 4 − 3 = 1. n = 3: 9 − 3 = 6.
  3. −2, 1, 6
Answer −2, 1, 6
Part (b)
  1. Divide each term by the one before: 12 ÷ 4 = 3, 36 ÷ 12 = 3, 108 ÷ 36 = 3. The sequence multiplies by 3 each time.
  2. The first term is 4, so the nth term is 4 multiplied by 3 a total of (n − 1) times.
  3. nth term = 4 × 3ⁿ⁻¹
  4. Check: n = 4 gives 4 × 27 = 108 ✓
Answer 4 × 3ⁿ⁻¹
152 marksExtended

Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q22

These are the first five terms of a sequence: 1⁄6, 1, 6, 36, 216. Find an expression for the nth term of this sequence.

Show solution
  1. Each term is 6 times the one before, so write them as powers of 6.
  2. 1⁄6 = 6⁻¹, 1 = 6⁰, 6 = 6¹, 36 = 6², 216 = 6³.
  3. For n = 1 the power is −1, for n = 2 it is 0: the power is always n − 2.
  4. 6ⁿ⁻²
Answer 6ⁿ⁻²
165 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q16

Find the nth term of each sequence. Sequence A: 8, 13, 18, 23, 28. Sequence B: 3⁄2, 4⁄3, 5⁄4, 6⁄5, 7⁄6. Sequence C: 1⁄4, 1⁄2, 1, 2, 4.

Show solution
  1. Sequence A: the terms go up by 5, so start with 5n. 5n gives 5, 10, 15, … and the sequence is 3 more each time.
  2. nth term of A = 5n + 3
  3. Sequence B: the numerators are 3, 4, 5, … (n + 2) and the denominators are 2, 3, 4, … (n + 1).
  4. nth term of B = n + 2⁄n + 1
  5. Sequence C: each term doubles. Write them as powers of 2: 1⁄4 = 2⁻², 1⁄2 = 2⁻¹, 1 = 2⁰, …
  6. The power is 3 less than the term number (n = 1 gives −2).
  7. nth term of C = 2ⁿ⁻³
Answer 5n + 3 n + 2⁄n + 1 2ⁿ⁻³
173 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q20

These are the first five terms of a sequence: 48, 24, 12, 6, 3. (a) Find the next term. (b) Find the nth term.

Show solution
Part (a)
  1. Each term is half of the term before: 48 → 24 → 12 → 6 → 3.
  2. Next term = 3 ÷ 2 = 1.5
Answer 1.5
Part (b)
  1. This is a geometric sequence with common ratio 1⁄2, so the nth term has the form a × (1⁄2)ⁿ.
  2. When n = 1 the term is 48, so a × 1⁄2 = 48 and a = 96.
  3. nth term = 96 × (1⁄2)ⁿ
  4. Check: n = 3 gives 96 × 1⁄8 = 12 ✓
Answer 96 × (1⁄2)ⁿ
184 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q10

Find the nth term of each sequence. (a) 17, 9, 1, −7, −15, … (b) 3, 12, 27, 48, 75, …

Show solution
Part (a)
  1. The terms go down by 8 each time, so the nth term starts with −8n.
  2. When n = 1, −8n = −8. The first term is 17, so add 17 − (−8) = 25.
  3. nth term = 25 − 8n
  4. Check: n = 2 gives 25 − 16 = 9 ✓
Answer 25 − 8n
Part (b)
  1. Each term is a multiple of 3. Divide each by 3: 1, 4, 9, 16, 25. These are the square numbers n².
  2. nth term = 3n²
Answer 3n²
192 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q1

The nth term of a sequence is 5 − 2n. (a) Find the 6th term. (b) Find the greatest number in this sequence.

Show solution
Part (a)
  1. Substitute n = 6 into 5 − 2n: 5 − 2 × 6 = 5 − 12 = −7
Answer −7
Part (b)
  1. Write out the first few terms: n = 1 gives 3, n = 2 gives 1, n = 3 gives −1, …
  2. The terms go down by 2 each time, so the first term is the greatest: 3
Answer 3
203 marksExtended

May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q7

These are the first 4 terms of a sequence: 11, 8, 5, 2.

(a) Find the next term of this sequence.

(b) Find the nth term of this sequence.

Show solution
Part (a)
  1. The terms go down by 3 each time: 11, 8, 5, 2.
  2. Next term = 2 − 3 = −1
Answer −1
Part (b)
  1. The difference is −3, so the nth term starts with −3n.
  2. −3n gives −3, −6, −9, … The sequence is 11, 8, 5, … which is 14 more each time.
  3. nth term = 14 − 3n
  4. Check: n = 4 gives 14 − 12 = 2 ✓
Answer 14 − 3n

Question papers used on this page

All sequences questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/12/M/J/260580_s26_qp_12May/June 2026 · Paper 1 (Core)Q21
0580/21/M/J/260580_s26_qp_21May/June 2026 · Paper 2 (Extended)Q11
0580/22/M/J/260580_s26_qp_22May/June 2026 · Paper 2 (Extended)Q14
0580/31/M/J/260580_s26_qp_31May/June 2026 · Paper 3 (Core)Q25
0580/32/M/J/260580_s26_qp_32May/June 2026 · Paper 3 (Core)Q15
0580/33/M/J/260580_s26_qp_33May/June 2026 · Paper 3 (Core)Q8
0580/12/F/M/260580_m26_qp_12Feb/Mar 2026 · Paper 1 (Core)Q5
0580/32/F/M/260580_m26_qp_32Feb/Mar 2026 · Paper 3 (Core)Q16
0580/42/F/M/260580_m26_qp_42Feb/Mar 2026 · Paper 4 (Extended)Q22
0580/11/O/N/250580_w25_qp_11Oct/Nov 2025 · Paper 1 (Core)Q7
0580/12/O/N/250580_w25_qp_12Oct/Nov 2025 · Paper 1 (Core)Q27
0580/13/O/N/250580_w25_qp_13Oct/Nov 2025 · Paper 1 (Core)Q21
0580/23/O/N/250580_w25_qp_23Oct/Nov 2025 · Paper 2 (Extended)Q16
0580/31/O/N/250580_w25_qp_31Oct/Nov 2025 · Paper 3 (Core)Q18
0580/41/O/N/250580_w25_qp_41Oct/Nov 2025 · Paper 4 (Extended)Q20
0580/42/O/N/250580_w25_qp_42Oct/Nov 2025 · Paper 4 (Extended)Q10
0580/43/O/N/250580_w25_qp_43Oct/Nov 2025 · Paper 4 (Extended)Q1
0580/11/M/J/250580_s25_qp_11May/June 2025 · Paper 1 (Core)Q9
0580/13/M/J/250580_s25_qp_13May/June 2025 · Paper 1 (Core)Q5
0580/22/M/J/250580_s25_qp_22May/June 2025 · Paper 2 (Extended)Q7

Related 0580 topics to practise next

More IGCSE Maths 0580 Algebra and Graphs questions

Browse IGCSE Maths questions by topic

See every topic and subtopic on one page →

Book a free trial class

Tell us a little about your requirement.