Edexcel International GCSE Maths Sequences Past Paper Questions and Solutions (4MA1 Higher)

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Edexcel International GCSE Maths A (4MA1) · Higher · Sequences, Functions and Graphs

Sequences and Arithmetic Series: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every sequences and arithmetic series question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 17 questions worth 74 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.

13 marksHigherJune 2022 · Paper 1H (4MA1/1H) · Q1

Here are the first five terms of an arithmetic sequence: 1, 5, 9, 13, 17

(a) Find an expression, in terms of n, for the nth term of this sequence. (2)

The nth term of another arithmetic sequence is 3n + 5

(b) Find an expression, in terms of m, for the (2m)th term of this sequence. (1)

Show solution

Part (a)

  1. The terms go up by 4 each time, so the nth term starts with 4n.
  2. When n = 1, 4n = 4, but the first term is 1, so subtract 3.

Answer: 4n − 3

Part (b)

  1. The (2m)th term means replace n with 2m: 3(2m) + 5

Answer: 6m + 5

23 marksHigherJune 2024 · Paper 1H (4MA1/1H) · Q1

Here are the first four terms of an arithmetic sequence: 1, 4, 7, 10

(a) Find an expression, in terms of n, for the nth term of this sequence. (2)

The nth term of a different arithmetic sequence is 5n + 17

(b) Find the 12th term of this sequence. (1)

Show solution

Part (a)

  1. Find the common difference: 4 − 1 = 3, 7 − 4 = 3, so the sequence goes up by 3 each time.
  2. A sequence that goes up by 3 is linked to the 3 times table, so the nth term starts with 3n.
  3. When n = 1, 3n = 3, but the first term is 1. So we must subtract 2 from 3n.
  4. Check with n = 4: 3 × 4 − 2 = 10 ✓

Answer: 3n − 2

Part (b)

  1. The 12th term means n = 12, so substitute 12 for n.
  2. 5 × 12 + 17 = 60 + 17

Answer: 77

32 marksHigherJune 2021 · Paper 1H (4MA1/1H) · Q2

Here are the first 4 terms of an arithmetic sequence: 85, 79, 73, 67. Find an expression, in terms of n, for the nth term of the sequence.

Show solution
  1. The terms go down by 6 each time, so the nth term starts with −6n.
  2. When n = 1, −6n = −6, but the first term is 85. 85 − (−6) = 91, so add 91.

Answer: 91 − 6n

44 marksHigherJune 2022 · Paper 1HR (4MA1/1HR) · Q2

In a warehouse there are two types of shelves, type R and type S. They are arranged into shelving units that form a sequence of patterns: R S R, then R S R S R, then R S R S R S R, and so on. The width of each type R shelf is 2.4 m and the width of each type S shelf is 3.5 m.

(a) Work out the total width of a shelving unit that has 6 type R shelves. (2)

A shelving unit has n type R shelves. The total width of this shelving unit is W metres.

(b) Find an expression for W in terms of n. Give your answer in its simplest form. (2)

Show solution

Part (a)

  1. Each pattern starts and ends with R, so there is always one fewer S than R. 6 R shelves means 5 S shelves.
  2. Width = 6 × 2.4 + 5 × 3.5 = 14.4 + 17.5

Answer: 31.9 m

Part (b)

  1. n R shelves and (n − 1) S shelves: W = 2.4n + 3.5(n − 1)
  2. Expand and simplify: 2.4n + 3.5n − 3.5

Answer: W = 5.9n − 3.5

54 marksHigherJanuary 2022 · Paper 1HR (4MA1/1HR) · Q16

An arithmetic series has first term 1 and common difference 4. Find the sum of all terms of the series from the 41st term to the 100th term inclusive.

Show solution
  1. Sum from term 41 to term 100 = S100 − S40
  2. S100 = 100⁄2[2 × 1 + 99 × 4] = 50 × 398 = 19 900
  3. S40 = 40⁄2[2 × 1 + 39 × 4] = 20 × 158 = 3160
  4. 19 900 − 3160

Answer: 16 740

64 marksHigherJune 2020 · Paper 1HR (4MA1/1HR) · Q18

In an arithmetic series, the 6th term is 39. In the same arithmetic series, the 19th term is 7.8. Work out the sum of the first 25 terms of the arithmetic series.

Show solution
  1. 6th term: a + 5d = 39. 19th term: a + 18d = 7.8
  2. Subtract: 13d = −31.2, so d = −2.4. Then a = 39 − 5(−2.4) = 51
  3. S25 = 25⁄2[2 × 51 + 24 × (−2.4)] = 12.5 × (102 − 57.6) = 12.5 × 44.4

Answer: 555

76 marksHigherJune 2023 · Paper 1H (4MA1/1H) · Q20

The sum of the first 80 terms of an arithmetic series, S, is 470. The 75th term of S is 14.5. The sum of the first X terms of S is 171. Work out the value of X. Show your working clearly.

Show solution
  1. Sum of 80 terms: 80⁄2[2a + 79d] = 470, so 2a + 79d = 11.75
  2. 75th term: a + 74d = 14.5, so 2a + 148d = 29
  3. Subtract: 69d = 17.25, so d = 0.25. Then a = 14.5 − 74 × 0.25 = 14.5 − 18.5 = −4
  4. Sum of X terms: X⁄2[2(−4) + (X − 1)(0.25)] = 171
  5. Multiply by 2: X(−8 + 0.25X − 0.25) = 342, so 0.25X2 − 8.25X − 342 = 0
  6. Multiply by 4: X2 − 33X − 1368 = 0, which factorises as (X − 57)(X + 24) = 0
  7. X must be a positive whole number, so X = 57.

Answer: X = 57

86 marksHigherJanuary 2021 · Paper 1HR (4MA1/1HR) · Q21

The nth term of an arithmetic series is un where un > 0 for all n. The sum to n terms of the series is Sn. Given that u4 = 6 and that S11 = (u6)2 + 18, find the value of u20.

Show solution
  1. u4 = a + 3d = 6
  2. S11 = 11⁄2(2a + 10d) = 11(a + 5d) = 11u6
  3. So 11u6 = (u6)2 + 18. Let w = u6: w2 − 11w + 18 = 0, so (w − 2)(w − 9) = 0
  4. If u6 = 2: d = (2 − 6) ÷ 2 = −2 and a = 12. Then the terms decrease and eventually become negative, which is not allowed.
  5. So u6 = 9: d = (9 − 6) ÷ 2 = 1.5 and a = 6 − 3 × 1.5 = 1.5
  6. u20 = a + 19d = 1.5 + 28.5

Answer: 30

94 marksHigherJanuary 2023 · Paper 1H (4MA1/1H) · Q22

The first term of an arithmetic series is (2t + 1) where t > 0. The nth term of this arithmetic series is (14t − 5). The common difference of the series is 3. The sum of the first n terms of the series can be written as p(qt − 1)r where p, q and r are integers. Find the value of p, the value of q and the value of r. Show clear algebraic working.

Show solution
  1. nth term: (2t + 1) + 3(n − 1) = 14t − 5, so 3(n − 1) = 12t − 6 and n = 4t − 1
  2. Sum = n⁄2(first term + last term) = 4t − 1⁄2 × (2t + 1 + 14t − 5) = 4t − 1⁄2 × (16t − 4)
  3. 16t − 4 = 4(4t − 1), so Sum = 4t − 1⁄2 × 4(4t − 1) = 2(4t − 1)2

Answer: p = 2, q = 4, r = 2

105 marksHigherJune 2021 · Paper 1H (4MA1/1H) · Q23

The sum of the first N terms of an arithmetic series, S, is 292. The 2nd term of S is 8.5. The 5th term of S is 13. Find the value of N. Show clear algebraic working.

Show solution
  1. 2nd term: a + d = 8.5. 5th term: a + 4d = 13
  2. Subtract: 3d = 4.5, so d = 1.5 and a = 7
  3. N⁄2[2 × 7 + (N − 1) × 1.5] = 292, so N(14 + 1.5N − 1.5) = 584
  4. 1.5N2 + 12.5N − 584 = 0. Multiply by 2: 3N2 + 25N − 1168 = 0
  5. Factorise: (3N + 73)(N − 16) = 0. N must be a positive whole number.

Answer: N = 16

114 marksHigherJune 2024 · Paper 1HR (4MA1/1HR) · Q23

Here are the first three terms of an arithmetic sequence: (4x − 14), (x + 2), (7x − 9). Find, as an integer, the sum of the first 40 terms of the sequence. Show clear algebraic working.

Show solution
  1. In an arithmetic sequence the difference between consecutive terms is constant: (x + 2) − (4x − 14) = (7x − 9) − (x + 2)
  2. −3x + 16 = 6x − 11, so 27 = 9x and x = 3
  3. The terms are 4(3) − 14 = −2, 3 + 2 = 5 and 7(3) − 9 = 12. So a = −2 and d = 7.
  4. Sum of the first n terms: Sn = n⁄2[2a + (n − 1)d]
  5. S40 = 40⁄2[2(−2) + 39 × 7] = 20 × (−4 + 273) = 20 × 269

Answer: 5380

124 marksHigherJune 2020 · Paper 1H (4MA1/1H) · Q24

Here are the first five terms of an arithmetic sequence: 8, 15, 22, 29, 36. Work out the sum of all the terms from the 50th term to the 100th term inclusive.

Show solution
  1. a = 8 and d = 7
  2. Sum from term 50 to term 100 = S100 − S49 (take away the first 49 terms, not 50, so the 50th term is kept)
  3. S100 = 100⁄2[2 × 8 + 99 × 7] = 50 × 709 = 35 450
  4. S49 = 49⁄2[2 × 8 + 48 × 7] = 24.5 × 352 = 8624
  5. 35 450 − 8624

Answer: 26 826

134 marksHigherJanuary 2021 · Paper 1H (4MA1/1H) · Q24

An arithmetic series has first term a and common difference d. The sum of the first 2n terms of the series is four times the sum of the first n terms of the series. Find an expression for a in terms of d. Show your working clearly.

Show solution
  1. S2n = 2n⁄2[2a + (2n − 1)d] = n[2a + (2n − 1)d]
  2. Sn = n⁄2[2a + (n − 1)d], so 4Sn = 2n[2a + (n − 1)d]
  3. Set equal and divide by n: 2a + 2nd − d = 4a + 2nd − 2d
  4. The 2nd terms cancel: −d = 2a − 2d, so 2a = d

Answer: a = d⁄2

145 marksHigherJanuary 2023 · Paper 1HR (4MA1/1HR) · Q24

An arithmetic sequence has first term 8 and common difference 11. The sequence has k terms, where k > 21. The sum of the last 20 terms of the sequence is 10 170. Find the value of k. Show clear algebraic working.

Show solution
  1. nth term = 8 + 11(n − 1) = 11n − 3
  2. The last 20 terms run from term (k − 19) to term k.
  3. First of these: 11(k − 19) − 3 = 11k − 212. Last: 11k − 3
  4. Sum = 20⁄2 × (11k − 212 + 11k − 3) = 10(22k − 215) = 10 170
  5. 22k − 215 = 1017, so 22k = 1232

Answer: k = 56

155 marksHigherJune 2025 · Paper 1H (4MA1/1H) · Q24

An arithmetic series has 30 terms. The first term is a. The common difference is d. The 20th term is 123. The sum of the 30 terms is 2880. Work out the value of a and the value of d. Show clear algebraic working.

Show solution
  1. The nth term is a + (n − 1)d, so the 20th term gives a + 19d = 123
  2. The sum formula Sn = n⁄2[2a + (n − 1)d] gives 15(2a + 29d) = 2880, so 2a + 29d = 192
  3. Double the first equation: 2a + 38d = 246
  4. Subtract: (2a + 38d) − (2a + 29d) = 246 − 192, so 9d = 54 and d = 6
  5. Substitute back: a = 123 − 19 × 6 = 123 − 114 = 9

Answer: a = 9, d = 6

166 marksHigherJune 2025 · Paper 1HR (4MA1/1HR) · Q24

The first 3 terms of an arithmetic series are (2x + 5), (3y − 4), (4x − 2) where x and y are constants. The sum of the first 9 terms of the series is 216. Find the value of x and the value of y. Show clear algebraic working.

Show solution
  1. In an arithmetic series the middle term is the mean of its neighbours, so 2(3y − 4) = (2x + 5) + (4x − 2), which gives 6y − 8 = 6x + 3, so 6y = 6x + 11
  2. First term a = 2x + 5. The common difference is half the gap from term 1 to term 3: d = (4x − 2) − (2x + 5)⁄2 = 2x − 7⁄2
  3. Sum of 9 terms: S9 = 9⁄2[2a + 8d] = 9(a + 4d) = 216, so a + 4d = 24
  4. Substitute: (2x + 5) + 2(2x − 7) = 24, so 6x − 9 = 24 and x = 5.5
  5. Then 6y = 6(5.5) + 11 = 44, so y = 44⁄6 = 22⁄3
  6. Check: the terms are 16, 18, 20, … and S9 = 9⁄2(2 × 16 + 8 × 2) = 216 ✓

Answer: x = 5.5, y = 22⁄3

175 marksHigherJune 2022 · Paper 1HR (4MA1/1HR) · Q25

The sum of the first 10 terms of an arithmetic series is 4 times the sum of the first 5 terms of the same series. The 8th term of this series is 45. Find the first term of this series. Show clear algebraic working.

Show solution
  1. S10 = 10⁄2[2a + 9d] = 5(2a + 9d) and S5 = 5⁄2[2a + 4d]
  2. S10 = 4S5: 5(2a + 9d) = 10(2a + 4d), so 10a + 45d = 20a + 40d
  3. Simplify: 5d = 10a, so d = 2a
  4. 8th term: a + 7d = 45. Substitute d = 2a: a + 14a = 45, so 15a = 45

Answer: a = 3

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