C1 January 2008 Q11
11. The first term of an arithmetic sequence is 30 and the common difference is −1.5
The \(r\)th term of the sequence is 0.
The sum of the first \(n\) terms of the sequence is \(S_n\).
| Scheme | Marks |
|---|---|
| \(u_{25} = a + 24d = 30 + 24\times(-1.5)\) | M1 |
| \(= -6\) | A1 |
| (2) |
Notes
M: Substitution of \(a = 30\) and \(d = \pm 1.5\) into \((a + 24d)\).
Use of \(a + 25d\) (or any other variations on 24) scores M0.
M: Listing terms (found by a correct method), and picking the 25th term. (There may be numerical slips).
| Scheme | Marks |
|---|---|
| \(a + (n - 1)d = 30 - 1.5(r - 1) = 0\) | M1 |
| \(r = 21\) | A1 |
| (2) |
Notes
M: Attempting to use the term formula, equated to 0, to form an equation in \(r\) (with no other unknowns). Allow this to be called \(n\) instead of \(r\).
Here, being ‘one off’ (e.g. equivalent to \(a + nd\)), scores M1.
M: Listing terms (found by a correct method), until the zero term is seen. (There may be numerical slips).
‘Trial and error’ approaches (or where working is unclear or non-existent) score M1 A1 for 21, M1 A0 for 20 or 22, and M0 A0 otherwise.
| Scheme | Marks |
|---|---|
| \(S_{20} = \dfrac{20}{2}\left\{60 + 19(-1.5)\right\}\) or \(S_{21} = \dfrac{21}{2}\left\{60 + 20(-1.5)\right\}\) or \(S_{21} = \dfrac{21}{2}\left\{30 + 0\right\}\) | M1 A1ft |
| \(= 315\) | A1 |
| (3) | |
| (7 marks) |
Notes
M: Attempting to use the correct sum formula to obtain \(S_{20}\), \(S_{21}\), or, with their \(r\) from part (b), \(S_{r-1}\) or \(S_r\).
1st A(ft): A correct numerical expression for \(S_{20}\), \(S_{21}\), or, with their \(r\) from part (b), \(S_{r-1}\) or \(S_r\) …. but the ft is dependent on an integer value of \(r\).
Methods such as calculus to find a maximum only begin to score marks after establishing a value of \(r\) at which the maximum sum occurs.
This value of \(r\) can be used for the M1 A1ft, but must be a positive integer to score A marks, so evaluation with, say, \(n = 20.5\) would score M1 A0 A0.
M: Listing sums, or listing and adding terms (found by a correct method), at least as far as the 20th term. (There may be numerical slips).
A2 (scored as A1 A1) for 315 (clearly selected as the answer).
‘Trial and error’ approaches essentially follow the main scheme, beginning to score marks when trying \(S_{20}\), \(S_{21}\), or, with their \(r\) from part (b), \(S_{r-1}\) or \(S_r\).
If no working (or no legitimate working) is seen, but the answer 315 is given, allow one mark (scored as M1 A0 A0).
Sums: 30, 58.5, 85.5, 111, 135, 157.5, 178.5, 198, 216, 232.5, 247.5, 261, 273, 283.5, 292.5, 300, 306, 310.5, 313.5, 315, ……..