C1 January 2009 Q9
9. The first term of an arithmetic series is \(a\) and the common difference is \(d\).
The 18th term of the series is 25 and the 21st term of the series is \(32\frac{1}{2}\).
The sum of the first \(n\) terms of the series is 2750.
| Scheme | Marks |
|---|---|
| \(a + 17d = 25\) or equiv. (for 1st B1), \(a + 20d = 32.5\) or equiv. (for 2nd B1), | B1, B1 |
| (2) |
Notes
Mark parts (a) and (b) as ‘one part’, ignoring labelling.
Alternative:
1st B1: \(d = 2.5\) or equiv. or \(d = \dfrac{32.5 - 25}{3}\). No method required, but \(a = -17.5\) must not be assumed.
2nd B1: Either \(a + 17d = 25\) or \(a + 20d = 32.5\) seen, or used with a value of \(d\)…
or for ‘listing terms’ or similar methods, ‘counting back’ 17 (or 20) terms.
| Scheme | Marks |
|---|---|
| Solving (Subtract) \(3d = 7.5\) so \(\underline{\boldsymbol{d = 2.5}}\) | M1 |
| \(a = 32.5 - 20\times 2.5\) so \(\underline{\boldsymbol{a = -17.5}}\) (*) | A1cso |
| (2) |
Notes
Mark parts (a) and (b) as ‘one part’, ignoring labelling.
M1: In main scheme: for a full method (allow numerical or sign slips) leading to solution for \(d\) or \(a\) without assuming \(a = -17.5\)
In alternative scheme: for using a \(d\) value to find a value for \(a\).
A1: Finding correct values for both \(a\) and \(d\) (allowing equiv. fractions such as \(d = \dfrac{15}{6}\)), with no incorrect working seen.
In the main scheme, if the given \(a\) is used to find \(d\) from one of the equations, then allow M1A1 if both values are checked in the 2nd equation.
| Scheme | Marks |
|---|---|
| \(2750 = \dfrac{n}{2}\left[-35 + \tfrac{5}{2}(n - 1)\right]\) | M1A1ft |
| \(\left\{\ 4\times 2750 = n(5n - 75)\ \right\}\) | |
| \(4\times 550 = n(n - 15)\) | M1 |
| \(\underline{n^2 - 15n = 55\times 40}\) (*) | A1cso |
| (4) |
Notes
1st M1: for attempt to form equation with correct \(S_n\) formula and 2750, with values of \(a\) and \(d\).
1st A1ft: for a correct equation following through their \(d\).
2nd M1: for expanding and simplifying to a 3 term quadratic.
2nd A1: for correct working leading to printed result (no incorrect working seen).
| Scheme | Marks |
|---|---|
| \(n^2 - 15n - 55\times 40 = 0\) or \(n^2 - 15n - 2200 = 0\) | M1 |
| \((n - 55)(n + 40) = 0 \qquad n = \ldots\) | M1 |
| \(\underline{\boldsymbol{n = 55}}\) (ignore \(-40\)) | A1 |
| (3) | |
| (11 marks) |
Notes
1st M1: forming the correct 3TQ \(= 0\). Can condone missing “\(= 0\)” but all terms must be on one side.
First M1 can be implied (perhaps seen in (c), but there must be an attempt at (d) for it to be scored).
2nd M1: for attempt to solve 3TQ, by factorisation, formula or completing the square (see general marking principles at end of scheme). If this mark is earned for the ‘completing the square’ method or if the factors are written down directly, the 1st M1 is given by implication.
A1: for \(n = 55\) dependent on both Ms. Ignore \(-40\) if seen.
No working or ‘trial and improvement’ methods in (d) score all 3 marks for the answer 55, otherwise no marks.