June 2018 Paper 2 Q1
1 The \(n\)th term of a sequence is \(\quad \dfrac{1420 - 5n}{1420 + 5n}\)
(a) Work out the position of the term that has the value zero. [2 marks]
(b) Write down the limiting value of the sequence as \(\ n \to \infty\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(1420 - 5n = 0\) or \(5n = 1420\) or \(\dfrac{1420}{5}\) | M1 | oe eg \(5(284 - n) = 0\) |
| 284 | A1 |
Additional guidance
| \(\dfrac{1420 - 5n}{1420 + 5n} = 0\) | Zero |
| \(1420 - 5n = 0(1420 + 5n)\) | Zero |
| \(n = 284\) | M1A1 |
| \(1420 - 5n = 0\) and \(1420 + 5n = 0\) with correct equation not selected | Zero |
| \(\pm 284\) is A0 | |
| Embedded answer | M1A0 |
| Answer | Mark | Comments |
|---|---|---|
| \(-1\) | B1 |
Additional guidance
| \(-\dfrac{5}{5}\) | B0 |
| \(-1 \quad n \to \infty\) | B1 |
| \(-1 \to \infty\) | B0 |
| \(x \to -1\) (any letter other than \(n\)) | B1 |