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Arithmetic series questions

Add the same amount at every step and you have an arithmetic sequence. Add the sequence up and a two-line trick collapses the whole thing. The sum formula is printed in the booklet, but the specification also wants you to be able to prove it, and the proof is the best part of the topic.

7 original questions · 23 marks · the arithmetic series notes · Sequences and series

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

• 2 marks · warm-up

  An arithmetic sequence has first term 7 and common difference 5. Find the 15th term.

  Worked answer

  u₁₅ = 7 + 14 × 5 = 77. M1 for a + 14d, A1 for 77. Fourteen steps reach the 15th term, not fifteen. The bracket in a + (n − 1)d is where this topic catches people, and 7 + 15 × 5 = 82 is the commonest wrong answer it produces.

• 3 marks · find · warm-up

  Find the nth term of the sequence 4, 11, 18, …, and determine whether 200 is a term.

  Worked answer

  a = 4 and d = 7, so u_n = 4 + 7(n − 1) = 7n − 3. Setting 7n − 3 = 200 gives n = 29, a positive whole number, so 200 is the 29th term. B1 for u_n = 7n − 3, M1 for setting it equal to 200, A1 for the 29th term. A fractional n would have settled it the other way, and the working that produces it is the proof. Simplifying to 7n − 3 rather than leaving 4 + 7(n − 1) is expected.

• 4 marks

  The 4th term of an arithmetic sequence is 18 and the 9th term is 38. Find the first term and common difference, and hence the 25th term.

  Worked answer

  From the 4th term to the 9th is five steps, so 5d = 38 − 18 = 20 and d = 4. Then a + 3d = 18 gives a = 6, and u₂₅ = 6 + 24 × 4 = 102. M1 A1 for d = 4, A1 for a = 6, A1 for the 25th term. Counting the steps between two given terms beats writing out simultaneous equations, and it removes the commonest error, which is dividing the difference by six rather than five.

• 3 marks · find

  Find the sum of the first 30 terms of the series 8 + 11 + 14 + …

  Worked answer

  a = 8 and d = 3, so S₃₀ = 15 × (16 + 29 × 3) = 15 × 103 = 1545. The last-term form agrees, since l = 8 + 29 × 3 = 95 and 15 × (8 + 95) = 1545. Two dressings of one formula give a check that costs a single line. M1 for the sum formula with a, d and n in place, A1 for the correct substitution, A1 for 1545. Marks go to the substitution as much as the total, so write the formula out with a, d and n in place before reaching for the calculator.

• 4 marks · prove

  Prove that the sum of the first n natural numbers is n(n + 1)/2, and evaluate the sum of the first 200.

  Worked answer

  Write S = 1 + 2 + … + n forwards and backwards and add the two lines: each column pairs to n + 1, and there are n columns, so 2S = n(n + 1) and S = n(n + 1)/2. For n = 200: 200 × 201/2 = 20 100. M1 for reversing and adding, A1 for 2S = n(n + 1), A1 for the printed result, B1 for the value of the sum. The pairing argument is the required proof, and it is two lines long.

• 3 marks

  A saver deposits £20 in the first month, then increases the deposit by £4 each month. Find the total saved after 24 months.

  Worked answer

  The deposits are arithmetic with a = 20 and d = 4, so S₂₄ = 12 × (40 + 23 × 4) = 12 × 132 = £1584. M1 for the sum formula with a = 20 and d = 4, A1 for the substitution, A1 for the total. This is a sum rather than a term, because every month's deposit stays in the pot. Answering £112, the 24th deposit alone, is the misreading the phrase total saved is there to prevent.

• 4 marks · find · stretch

  Find the smallest number of terms of the series 5 + 9 + 13 + … whose sum exceeds 1000.

  Worked answer

  S_n = n(2 × 5 + 4(n − 1))/2 = n(2n + 3) must exceed 1000. Trying the boundary: S₂₁ = 945, too small, and S₂₂ = 1034, past the mark. So 22 terms are needed. M1 for S_n in terms of n, M1 for comparing it with 1000, A1 for locating the boundary, A1 for 22 terms. Solving the quadratic 2n² + 3n − 1000 > 0 gives n > 21.6, the same verdict; either route must end with whole terms, rounded up.

The same practice on paper: the printable workbook for this topic, questions and a worked answer book.

Practise arithmetic series one question at a time

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