Foundation November 2019 Paper 2 Q26
26 The \(n\)th term of a sequence is \(2n^2 - 1\)
The \(n\)th term of a different sequence is \(40 - n^2\)
Show that there is only one number that is in both of these sequences. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| Shown (supported) | M1 | for method to find at least two terms, eg \(2 \times 4^2 - 1\ (= 31)\) and \(40 - 3^2\ (= 31)\) |
| M1 | for generating at least three correct terms of each sequence | |
| A1 | for generating at least the terms 1, 7, 17, 31, 49 of the first sequence and at least the terms 39, 36, 31, 24, 15, 4 of the second sequence |
Additional guidance
1 7 17 31 49 71 97 127 161 199
39 36 31 24 15 4 \(-9\)