Sequences – Year 10 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Difference 4 (1), \(\mathbf{4n+5}\) (1) The common difference is 4, and the zero term is \(9-4=5\). Check: \(4\times2+5=13\).
   2. \(4n+5=150\), \(4n=145\), \(n=36.25\) (1), not a whole number so 150 is not a term (1) Set the \(n\)th term equal to 150 and solve. A term number must be a whole number, so 150 is not a term of the sequence.

   The 36th term is 149 and the 37th is 153.

2. Answer to Question 2 (4 marks)

   1. Difference 2 (1), \(2n+2\) (1) Each extra table adds 2 seats, and the zero term is \(4-2=2\). Check: \(2\times3+2=8\).
   2. \(2n+2\ge50\), \(n\ge24\) (1), so 24 tables (1) Solve the inequality: \(2n\ge48\), so \(n\ge24\). Exactly 24 tables seat 50 people.

   In (a), check one more pattern: four tables give \(2\times4+2=10\) seats, since each end seats one person and each long side seats one per table. In (b), the answer must be a whole number of tables, and 23 tables would seat only 48 people.

3. Answer to Question 3 (4 marks)

   1. Ratio 3 (1); next terms 135, 405 (1) Divide any term by the one before: \(15\div5=3\). Then \(45\times3=135\) and \(135\times3=405\).
   2. \(5\times3^7\) (1) \(=5\times2187=\) 10 935 (1) The \(n\)th term is \(5\times3^{n-1}\), so the 8th term is \(5\times3^7\). The 8th term is 10 935.

   A common error in (b) is to work out \(5\times3^8\), using the term number as the power; the first term already uses \(3^0\). Another error is \(15^7\). Check: listing gives 5, 15, 45, 135, 405, 1215, 3645, 10 935.

4. Answer to Question 4 (4 marks)

   1. Difference 6 (1), £\((6n+14)\) (1) The common difference is 6, and the zero term is \(20-6=14\). Check: \(6\times3+14=32\).
   2. Week 10: \(74\). Sum \(=\dfrac{10}{2}(20+74)\) (1) \(=\) £470 (1) In week 10 she saves \(6\times10+14=74\) pounds. Use the sum formula with first term 20 and last term 74. She has saved £470.

   Check (b) by adding: \(20+26+32+\dots+74\) has ten terms with an average of \(47\), and \(47\times10=470\).

5. Answer to Question 5 (4 marks)

   1. First differences 7, 11, 15; second differences 4, 4 (constant) (1)
   2. \(2n^2\) gives 2, 8, 18, 32; remainders 2, 3, 4, 5 so \(n+1\) (1). \(2n^2+n+1\) (1) Half of the second difference 4 gives \(2n^2\).
   3. \(2\times225+15+1=\) 466 (1) Substitute \(n=15\): \(15^2=225\). The 15th term is 466.

   In (a), a constant second difference proves the sequence is quadratic. Check (b) with \(n=4\): \(2\times16+4+1=37\), which matches the fourth term given in the question.

Total : 20 marks