Higher June 2018 Paper 3 Q16
16 The \(n\)th term of a sequence is given by \(an^2 + bn\) where \(a\) and \(b\) are integers.
The 2nd term of the sequence is \(-2\)
The 4th term of the sequence is 12
(a) Find the 6th term of the sequence. (4)
Here are the first five terms of a different quadratic sequence.
0 2 6 12 20
(b) Find an expression, in terms of \(n\), for the \(n\)th term of this sequence. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 42 | P1 | for process to find an equation in \(a\) and \(b\), eg \(a \times 2^2 + b \times 2 = -2\) (\(4a + 2b = -2\)) or \(a \times 4^2 + b \times 4 = 12\) (\(16a + 4b = 12\)) |
| P1 | for process to find a pair of simultaneous equations and eliminate one unknown, eg \(16a + 8b = -8\) and \(16a + 4b = 12\) and subtraction or \(16a + 4b = 12\) and \(8a + 4b = -4\) and subtraction | |
| A1 | for \(a = 2\) and \(b = -5\) | |
| A1 | cao |
Additional guidance
Allow one arithmetic error in elimination,
eg \(16a + 8b = -8\) and \(16a + 4b = 12\) leading to \(4b = 20\) but no subtraction sign seen
| Answer | Mark | Mark scheme |
|---|---|---|
| \(n^2 - n\) | M1 | for correct method, eg \(n^2\) seen as a term |
| A1 | for \(n^2 - n\) oe |