C1 June 2012 Q5
5. A sequence of numbers \(a_1, a_2, a_3 \ldots\) is defined by\[\begin{aligned}a_1 &= 3\\ a_{n+1} &= 2a_n - c \qquad (n \geqslant 1)\end{aligned}\]where \(c\) is a constant.
Given that \(\displaystyle\sum_{i=1}^{4} a_i \geqslant 23\)
| Scheme | Marks |
|---|---|
| \(a_1 = 3,\ a_{n+1} = 2a_n - c,\ n \geqslant 1,\ c\) is a constant | |
| \(\{a_2 =\}\ 2\times 3 - c\) or \(2(3) - c\) or \(6 - c\) | B1 |
| (1) |
Notes
The answer to part (a) cannot be recovered from candidate’s working in part (b) or part (c).
Once the candidate has achieved the correct result you can ignore subsequent working in this part.
| Scheme | Marks |
|---|---|
| \(\{a_3 =\}\ 2\times(\text{"}6 - c\text{"}) - c\) | M1 |
| \(= 12 - 3c \quad (*)\) | A1 cso |
| (2) |
Notes
M1: For a correct substitution of their \(a_2\) which must include term(s) in \(c\) into \(2a_2 - c\) giving a result for \(a_3\) in terms of only \(c\). Candidates must use correct bracketing for this mark.
A1: for correct solution only. No incorrect working/statements seen. (Note: the answer is given!)
| Scheme | Marks |
|---|---|
| \(a_4 = 2\times(\text{"}12 - 3c\text{"}) - c \qquad \{= 24 - 7c\}\) | M1 |
| \(\left\{\displaystyle\sum_{i=1}^{4} a_i =\right\}\ 3 + (6 - c) + (12 - 3c) + (24 - 7c)\) | M1 |
| \(\text{"}45 - 11c\text{"} \geqslant 23\) or \(\text{"}45 - 11c\text{"} = 23\) | M1 |
| \(c \leqslant 2\) or \(2 \geqslant c\) | A1 cso |
| (4) | |
| (7 marks) |
Notes
1st M1: For a correct substitution of \(a_3\) which must include term(s) in \(c\) into \(2a_3 - c\) giving a result for \(a_4\) in terms of only \(c\). Candidates must use correct bracketing (can be implied) for this mark.
2nd M1: for an attempt to sum their \(a_1\), \(a_2\), \(a_3\) and \(a_4\) only.
3rd M1: for their sum (of 3 or 4 or 5 consecutive terms) \(=\) or \(\geqslant\) or \(> 23\) to form a linear inequality or equation in \(c\).
A1: for \(c \leqslant 2\) or \(2 \geqslant c\) from a correct solution only.
Beware: \(-11c \geqslant -22 \Rightarrow c \geqslant 2\) is A0.Note: \(45 - 11c \geqslant 23 \Rightarrow -11c \leqslant -22 \Rightarrow c \leqslant 2\) would be A0 cso.Note: Applying either \(S_n = \dfrac{n}{2}(2a + (n - 1)d)\) or \(S_n = \dfrac{n}{2}(a + l)\) is 2nd M0, 3rd M0.
Note: If a candidate gives a numerical answer in part (a); they will then get M0A0 in part (b); but if they use the printed result of \(a_3 = 12 - 3c\) they could potentially get M0M1M1A0 in part (c)
Note: If a candidate only adds numerical values (not in terms of \(c\)) in part (c) then they could potentially get only M0M0M1A0.
Note: For the 3rd M1 candidates will usually sum \(a_1, a_2, a_3\) and \(a_4\) or \(a_2, a_3\) and \(a_4\) or \(a_2, a_3, a_4\) and \(a_5\) or \(a_1, a_2, a_3, a_4\) and \(a_5\)