C1 June 2009 Q5
5. A 40-year building programme for new houses began in Oldtown in the year 1951 (Year 1) and finished in 1990 (Year 40).
The numbers of houses built each year form an arithmetic sequence with first term \(a\) and common difference \(d\).
Given that 2400 new houses were built in 1960 and 600 new houses were built in 1990, find
| Scheme | Marks |
|---|---|
| \(a + 9d = 2400 \qquad a + 39d = 600\) | M1 |
| \(d = \dfrac{-1800}{30} \qquad d = -60 \qquad\) (accept \(\pm 60\) for A1) | M1 A1 |
| (3) |
Notes
Note:
If the sequence is considered ‘backwards’, an equivalent solution may be given using \(d = 60\) with \(a = 600\) and \(l = 2940\) for part (b). This can still score full marks. Ignore labelling of (a) and (b)
1st M1: for an attempt to use 2400 and 600 in \(a + (n - 1)d\) formula. Must use both values
i.e. need \(a + pd = 2400\) and \(a + qd = 600\) where \(p = 8\) or 9 and \(q = 38\) or 39 (any combination)
2nd M1: for an attempt to solve their 2 linear equations in \(a\) and \(d\) as far as \(d = \ldots\)
A1: for \(d = \pm 60\). Condone correct equations leading to \(d = 60\) or \(a + 8d = 2400\) and \(a + 38d = 600\) leading to \(d = -60\). They should get penalised in (b) and (c).
NB This is a “one off” ruling for A1. Usually an A mark must follow from their work.
ALT 1st M1 for \((30d) = \pm\,(2400 - 600)\)
2nd M1 for \((d =) \pm\dfrac{(2400 - 600)}{30}\)
A1 for \(d = \pm 60\)
\(a + 9d = 600,\ a + 39d = 2400\) only scores M0 BUT if they solve to find \(d = \pm 60\) then use ALT scheme above.
| Scheme | Marks |
|---|---|
| \(a - 540 = 2400 \qquad a = 2940\) | M1 A1 |
| (2) |
Notes
Note:
If the sequence is considered ‘backwards’, an equivalent solution may be given using \(d = 60\) with \(a = 600\) and \(l = 2940\) for part (b). This can still score full marks. Ignore labelling of (a) and (b)
M1: for use of their \(d\) in a correct linear equation to find \(a\) leading to \(a = \ldots\)
A1: their \(a\) must be compatible with their \(d\) so \(d = 60\) must have \(a = 600\) and \(d = -60\), \(a = 2940\)
So for example they can have \(2400 = a + 9(60)\) leading to \(a = \ldots\) for M1 but it scores A0
Any approach using a list scores M1A1 for a correct \(a\) but M0A0 otherwise
| Scheme | Marks |
|---|---|
| Total \(= \dfrac{1}{2}n\left\{2a + (n - 1)d\right\} = \dfrac{1}{2}\times 40\times\left(5880 + 39\times -60\right)\) (ft values of \(a\) and \(d\)) | M1 A1ft |
| \(= \underline{70\,800}\) | A1cao |
| (3) | |
| (8 marks) |
Notes
M1: for use of a correct \(\mathrm{S}_n\) formula with \(n = 40\) and at least one of \(a\), \(d\) or \(l\) correct or correct ft.
1st A1ft: for use of a correct \(\mathrm{S}_{40}\) formula and both \(a, d\) or \(a, l\) correct or correct follow through
ALT Total \(= \dfrac{1}{2}n\left\{a + l\right\} = \dfrac{1}{2}\times 40\times\left(2940 + 600\right)\) (ft value of \(a\)) M1 A1ft
2nd A1: for 70800 only