Practise 20 exam-style questions on Generating sequences for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
20 questions
1–5 marks each
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20 questions, marked as you go
Sample questions with model answers
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Question 1
1 mark
Here are the first four terms of a sequence. 3, 7, 15, 31
Write down the term-to-term rule for the sequence.
Model answer
\(3 \times 2 + 1 = 7\), \(7 \times 2 + 1 = 15\), \(15 \times 2 + 1 = 31\) The rule is: multiply by 2, then add 1.
Explanation
The differences (4, 8, 16) double each time, which suggests multiplying by 2. Doubling each term gives one less than the next term, so the rule is "multiply by 2 and add 1". Check that it works for every pair of terms.
Is 100 a term of the sequence? You must show how you get your answer.
Model answer
\(3n^2 - 2 = 100\) \(3n^2 = 102\) \(n^2 = 34\) 34 is not a square number, so \(n\) is not a whole number. No, 100 is not a term. (The 5th term is 73 and the 6th term is 106.)
Explanation
Solve the equation for \(n\). Positions must be whole numbers, and \(\sqrt{34}\) is between 5 and 6, so there is no term equal to 100.
Mark scheme
1 mark
M1 for \(3n^2 - 2 = 100\) leading to \(n^2 = 34\)
or working out the 5th and 6th terms, 73 and 106
1 mark
C1 for "No" with a reason, e.g. 34 is not a square number so \(n\) is not a whole number
oe, e.g. 100 lies between the 5th term (73) and the 6th term (106)
Question 3
3 marks
Sequence A has \(n\)th term \(5n + 2\) Sequence B has \(n\)th term \(80 - 3n\)
Is there a position \(n\) at which the two sequences have the same term? You must show your working.
Model answer
\(5n + 2 = 80 - 3n\) \(8n = 78\) \(n = 9.75\) Positions must be whole numbers, so no. (At \(n = 9\): A is 47 and B is 53; at \(n = 10\): A is 52 and B is 50.)
Explanation
If the sequences had the same term at the same position, that \(n\) would solve \(5n + 2 = 80 - 3n\). The solution is not a whole number, so there is no such position: sequence A overtakes sequence B between the 9th and 10th terms.
Mark scheme
1 mark
M1 for forming the equation \(5n + 2 = 80 - 3n\)
1 mark
A1 for \(n = 9.75\)
oe, e.g. \(\dfrac{78}{8}\)
1 mark
C1 for "No" with a reason, e.g. 9.75 is not a whole number, so it is not a position
accept a comparison of the 9th terms (47, 53) and 10th terms (52, 50)
Question 4
4 marks
Each pattern in a sequence is made from white tiles and grey tiles. Pattern number \(n\) has \(n^2 + 3n\) tiles altogether. Pattern number \(n\) has \(2n + 4\) white tiles.
Which pattern number has exactly 26 grey tiles?
Model answer
Grey tiles \(= n^2 + 3n - (2n + 4) = n^2 + n - 4\) \(n^2 + n - 4 = 26\) \(n^2 + n - 30 = 0\) \((n + 6)(n - 5) = 0\) \(n = 5\) (a pattern number cannot be negative) Pattern number 5
Explanation
Grey tiles are the total minus the white tiles; put the white-tile expression in a bracket so both terms are subtracted. Solve the quadratic and reject the negative solution. Check: pattern 5 has 40 tiles, of which 14 are white and 26 are grey.
Mark scheme
1 mark
P1 for an expression for the number of grey tiles, \(n^2 + 3n - (2n + 4)\) \((= n^2 + n - 4)\)
correct answer with no working scores full marks
1 mark
P1 for forming the equation \(n^2 + n - 4 = 26\)
1 mark
P1 for solving, \((n + 6)(n - 5) = 0\)
or a systematic trial of pattern numbers
1 mark
A1 for pattern number 5
cao
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