June 2017 Paper 1 Q8
8 The first four terms of a quadratic sequence are
\[10 \qquad 33 \qquad 64 \qquad 103 \qquad \ldots\]Work out an expression for the \(n\)th term. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| 2nd difference = 8 or \(a = 4\) | M1 | sight of \(4n^2\) implies this mark |
| subtract their \(4n^2\) or sight of three of 6 17 28 39 | M1 | subtracting 4 16 36 64 the coefficient of their \(4n^2\) will come from half the value of their 2nd difference |
| subtract their \(11n\) or \(b = 11\) or tests \(4n^2 + 11n\) and compares to original sequence or sight of three of 15 38 69 108 | M1dep | dep on 2nd M mark |
| \(4n^2 + 11n - 5\) | A1 | |
| Alternative method 2 | ||
| Any three of these \(a + b + c = 10\) \(4a + 2b + c = 33\) \(9a + 3b + c = 64\) \(16a + 4b + c = 103\) | M1 | |
| Any two of these \(3a + b = 23\) \(5a + b = 31 \quad 7a + b = 39\) | M1dep | |
| \(a = 4\) and \(b = 11\) | A1 | |
| \(4n^2 + 11n - 5\) | A1 | |
| Alternative method 3 | ||
| \(a = 4\) | M1 | |
| \(3a + b = 33 - 10\) and substitutes their \(a\) in this equation | M1 | oe |
| \(b = 11\) | A1 | |
| \(4n^2 + 11n - 5\) | A1 | |
Additional guidance
SC3 for \(4n^2 - 11n + 5\)
Condone \(4x^2 + 11x - 5\) or eg \(4x^2 + 11n - 5\) (mixed letters)