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IGCSE Maths · Edexcel 4XMA1 · Unit 2 Higher · Sequences

Arithmetic Sequences and Series — Worked Solutions

Full worked solutions to all 12 questions (37 marks), set out the way an Edexcel examiner wants to see them.

Section A — Warm-up (grades 5–6)
  1. (2 marks)For the sequence 5, 9, 13, 17, …
    (a) find an expression for the nth term
    (b) find the 30th term.
    Solution
    1. a = 5, d = 4: un = 5 + 4(n − 1) = 4n + 1.
    2. u30 = 4(30) + 1 = 121.
    (a) 4n + 1 (b) 121
  2. (2 marks)An arithmetic sequence has first term 20 and common difference −3. Find the 15th term.
    Solution
    1. u15 = 20 + 14(−3) = 20 − 42.
    −22
  3. (2 marks)Is 150 a term of the sequence 4, 11, 18, 25, …? Show how you decide.
    Solution
    1. un = 7n − 3.
    2. 7n − 3 = 150 → n = 153 ÷ 7 = 21.86…, not an integer.
    No
  4. (2 marks)Find the sum of the integers from 1 to 100.
    Solution
    1. a = 1, l = 100, n = 100.
    2. S = 1002(1 + 100) = 50 × 101.
    5050
Section B — Core (grades 6–7)
  1. (3 marks)Find the sum of the first 25 terms of the series 2 + 5 + 8 + …
    Solution
    1. a = 2, d = 3, n = 25.
    2. S25 = 252[4 + 24(3)] = 252 × 76.
    950
  2. (3 marks)Find the sum 6 + 10 + 14 + … + 86.
    Solution
    1. 6 + 4(n − 1) = 86 → n = 21.
    2. S21 = 212(6 + 86) = 966.
    966
  3. (4 marks)The 3rd term of an arithmetic sequence is 11 and the 8th term is 31. Find the sum of the first 15 terms.
    Solution
    1. a + 2d = 11 and a + 7d = 31.
    2. 5d = 20 → d = 4, a = 3.
    3. S15 = 152[6 + 14(4)] = 152 × 62.
    465
  4. (4 marks)Maya saves $500 in month 1, and each month she saves $150 more than the month before.
    (a) How much does she save in month 12?
    (b) What is the total she has saved after 12 months?
    Solution
    1. (a) u12 = 500 + 11(150) = 2150.
    2. (b) S12 = 122(500 + 2150) = 6 × 2650 = 15 900.
    (a) $2150 (b) $15 900
Section C — Stretch (grades 8–9)
  1. (4 marks)How many terms of the series 3 + 7 + 11 + … are needed for the sum to be 465?
    Solution
    1. Sn = n2[6 + 4(n − 1)] = n(2n + 1).
    2. n(2n + 1) = 465 → 2n² + n − 465 = 0.
    3. (n − 15)(2n + 31) = 0, so n = 15 (reject the negative).
    15 terms
  2. (4 marks)Find the sum of all the multiples of 7 between 100 and 300.
    Solution
    1. First multiple: 105 (7 × 15). Last: 294 (7 × 42).
    2. Number of terms = 42 − 15 + 1 = 28.
    3. S = 282(105 + 294) = 14 × 399.
    5586
  3. (3 marks)Prove that the sum of the first n odd numbers is n².
    Solution
    1. The odd numbers 1, 3, 5, … have a = 1 and d = 2.
    2. Sn = n2[2 + 2(n − 1)] = n2 × 2n = n².
    Sn = n²
  4. (4 marks)The sum of the first n terms of an arithmetic series is Sn = 3n² + 2n. Find the first term and the common difference.
    Solution
    1. u1 = S1 = 3 + 2 = 5.
    2. S2 = 12 + 4 = 16, so u2 = 16 − 5 = 11.
    3. d = 11 − 5 = 6.
    a = 5, d = 6
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