Specimen 2020 Paper 2 Q11
11 For sequence A, \(\quad n\)th term \(= \dfrac{n}{14n + 30}\)
For sequence B, \(\quad n\)th term \(= \dfrac{2}{n}\)
The \(k\)th term of sequence A equals the \(k\)th term of sequence B.
Work out the value of \(k\).
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(k^2 = 2(14k + 30)\) | M1 | oe correct equation with fractions eliminated |
| \(k^2 - 28k - 60\ (= 0)\) | M1dep | oe equation |
| \((k + 2)(k - 30)\ (= 0)\) or \(\dfrac{--28 \pm \sqrt{(-28)^2 - 4 \times 1 \times -60}}{2 \times 1}\) or \(14 \pm \sqrt{256}\) | M1 | oe correct attempt to solve their 3-term quadratic equation |
| 30 | A1 | 30 and \(-2\) is A0 |