Higher June 2021 Paper 1 Q23
23 The sum of the first \(N\) terms of an arithmetic series, \(S\), is 292
The 2nd term of \(S\) is 8.5
The 5th term of \(S\) is 13
Find the value of \(N\).
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
| \(a + d = 8.5\), \(a + 4d = 13\) oe | M1 |
| \(a = 7\), \(d = 1.5\) | A1 |
\(\dfrac{N}{2}(2 \times 7 + (N - 1)1.5) = 292\) (eg \(3N^2 + 25N - 1168\ [= 0]\) or \(1.5N^2 + 12.5N - 584\ [= 0]\)) | M1 |
eg \((3N + 73)(N - 16)\ [= 0]\) \([N =]\ \dfrac{-25 \pm \sqrt{25^2 - 4 \times 3 \times -1168}}{2 \times 3}\) | M1 |
| Working required Answer: 16 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for at least 1 correct equation or for \(d = 1.5\)
A1: both values correct
M1: A correct equation for the total of the first \(N\) terms of the series with \(a\) and \(d\) substituted in.
The mark can be gained by using their values of \(a\) and \(d\) even if no previous marks awarded.
M1: A correct method dep on the previous M1 for solving their 3 term quadratic equation using any correct method (allow one sign error and some simplification – allow as far as \(\dfrac{-25 \pm \sqrt{625 + 14016}}{6}\)) oe (may be ± or just +) or if factorising, allow brackets which expanded give 2 out of 3 terms correct, or if completing the square allow as far as the stage \(3\left(\left(N + \dfrac{25}{6}\right)^2 - \dfrac{25^2}{6^2}\right) - 1168\ (= 0)\)
A1: dep on M2