June 2022 Paper 1 Q9
9
(a) A quadratic sequence starts \(\qquad -2 \qquad -1 \qquad 4 \qquad 13\)
Work out an expression for the \(n\)th term. [3 marks]
(b) A different quadratic sequence has \(n\)th term \(\quad n^2 + 10n\)
Use an algebraic method to work out how many terms in the sequence are less than 2000
Do not use trial and improvement.
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 (grid) | ||
| \(\begin{array}{ccccc} 1 & & 5 & & 9 \\ & 4 & & 4 & \end{array}\) and \(2n^2\) | M1 | |
| \(-4 \quad -9 \quad (-14 \quad -19)\) and \(-5n\;(+\,c)\) | M1dep | subtract \(2n^2\) |
| \(2n^2 - 5n + 1\) | A1 | |
| Alternative method 2 (simultaneous equations) | ||
| Any 3 of: \(a + b + c = -2\) \(4a + 2b + c = -1\) \(9a + 3b + c = 4\) \(16a + 4b + c = 13\) | M1 | using \(n^{\text{th}}\) term \(= an^2 + bn + c\) |
| \(3a + b = 1\) or \(5a + b = 5\) | M1dep | or any other equation with an unknown eliminated |
| \(b = -5\), \(c = 1\) so \(2n^2 - 5n + 1\) | A1 | |
| Alternative method 3 (using terms) | ||
| \(\begin{array}{ccccc} 1 & & 5 & & 9 \\ & 4 & & 4 & \end{array}\) so \(a = 2\) | M1 | using \(n^{\text{th}}\) term \(= an^2 + bn + c\) |
| \(3a + b = 1\) and \(a = 2\) substituted in this equation | M1dep | oe |
| \(b = -5\), \(c = 1\) so \(2n^2 - 5n + 1\) | A1 | |
Additional guidance
| Condone other letters used eg \(2x^2 - 5x + 1\) or even \(2n^2 - 5x + 1\) | |
| After finding \(a = 2\) they may find the 0th term to get \(c = 1\) | M2 |
| \(2n^2 + 5n - 1\) from Alt 1 but subtracting the wrong way round | SC2 |
Notes
(corrected from the printed mark scheme: the fourth equation in Alternative method 2 is printed as \(16a + 9b + c = 13\); for \(n = 4\) it is \(16a + 4b + c = 13\))
| Answer | Mark | Comments |
|---|---|---|
| \(n^2 + 10n - 2000 \lt 0\) | M1 | the correct inequality needed for this mark and must be written in this form |
| \((n - 40)(n + 50)\) or \((n + 5)^2 - 25 - 2000\) or \(\dfrac{-10 \pm \sqrt{10^2 - 4 \times 1 \times -2000}}{2}\) | M1 | oe inequality not needed for this mark condone + instead of \(\pm\) as the negative solution has no meaning here |
| 39 | A1 |
Additional guidance
| Do not accept T&I | |
| Incorrect use of inequalities can be recovered by a correct use of inequalities later in the method such as \(n \lt 40\) near the end | M2 |
| Incorrect use of inequalities can be recovered for full marks. An answer of 39 after a method that uses an incorrect inequality or = shows inequality has been recovered | M2A1 |
| An incorrect solution with incorrect use of inequalities can only be awarded the second M mark | M0M1A0 |
| Correct answer not coming from correct working will not gain any marks | M0A0 |
| For students who try to complete the square accept \((n + 5)^2 \lt 2025\) as an oe giving M2 but \((n + 5)^2 = 2025\) would only gain M0M1 unless recovered in the answer |