C1 January 2013 Q7
7. Lewis played a game of space invaders. He scored points for each spaceship that he captured.
Lewis scored 140 points for capturing his first spaceship.
He scored 160 points for capturing his second spaceship, 180 points for capturing his third spaceship, and so on.
The number of points scored for capturing each successive spaceship formed an arithmetic sequence.
Sian played an adventure game. She scored points for each dragon that she captured. The number of points that Sian scored for capturing each successive dragon formed an arithmetic sequence.
Sian captured \(n\) dragons and the total number of points that she scored for capturing all \(n\) dragons was 8500.
Given that Sian scored 300 points for capturing her first dragon and then 700 points for capturing her \(n\)th dragon,
| Scheme | Marks |
|---|---|
| Lewis; arithmetic series, \(a = 140,\ d = 20\). | |
| \(T_{20} = 140 + (20 - 1)(20);\ = 520\) Or lists 20 terms to get to 520 OR \(120 + (20)(20)\) | M1; A1 |
| (2) |
Notes
M1: Attempt to use formula for 20th term of Arithmetic series with first term 140 and \(d\) = 20. Normal formula rules apply – see General principles at the start of the mark scheme re “Method Marks”
Or: uses \(120 + 20n\) with \(n = 20\)
Or: Listing method : Lists 140, 160, 180, 200, 220, 240, 260, 280, … 520. M1A1 if correct M0A0 if wrong. (So 2 marks or zero)
A1: For 520
Method 1
| Scheme | Marks |
|---|---|
| Either: Uses \(\tfrac{1}{2}n(2a + (n - 1)d)\) | M1 |
| \(\dfrac{20}{2}\left(2\times 140 + (20 - 1)(20)\right)\) | A1 |
| 6600 | A1 |
| (3) |
Notes
M1: An attempt to apply \(\tfrac{1}{2}n(2a + (n - 1)d)\) or \(\tfrac{1}{2}n(a + l)\) with their values for \(a\), \(n\), \(d\) and \(l\)
A1: Uses \(a\) = 140, \(d\) = 20, \(n\) = 20 in their formula (two alternatives given above) but ft on their value of l from (a) if they use Method 2.
A1: 6600 cao
Or: Listing method : Lists 140, 160, 180, 200, 220, 240, 260, 280, … 520 and adds
6600 gets M1A1A1- any other answer gets M1 A0A0 provided there are 20 numbers, the first is 140 and the last is 520.
Method 2
| Scheme | Marks |
|---|---|
| Or: Uses \(\tfrac{1}{2}n(a + l)\) | M1 |
| \(\dfrac{20}{2}\left(140 + \text{"}520\text{"}\right)\) ft 520 | A1 |
| 6600 | A1 |
| Scheme | Marks |
|---|---|
| Sian; arithmetic series, \(a = 300,\ l = 700,\ S_n = 8500\) | |
| Either: Attempt to use \(8500 = \dfrac{n}{2}(a + l)\) | M1 |
| \(8500 = \dfrac{n}{2}(300 + 700)\) | A1 |
| \(\Rightarrow n = 17\) | A1 |
| (3) | |
| (8 marks) |
Notes
First method
M1: Attempt to use \(S_n = \dfrac{n}{2}(a + l)\) with their values for \(a\), and \(l\) and \(S\) =8500
A1: Uses formula with correct values
A1: Finds exact value 17
Alternative method
| Scheme | Marks |
|---|---|
| Or: May use both \(8500 = \tfrac{1}{2}n(2a + (n - 1)d)\) and \(l = a + (n - 1)d\) and eliminate \(d\) | M1 |
| \(8500 = \dfrac{n}{2}(600 + 400)\) | A1 |
| \(\Rightarrow n = 17\) | A1 |
M1: If both formulae \(8500 = \tfrac{1}{2}n(2a + (n - 1)d)\) and \(l = a + (n - 1)d\) are used, then \(d\) must be eliminated before this mark is awarded by valid work. Should not be using \(d\) = 400. This would be M0.
A1: Correct equation in \(n\) only
then A1 for 17 exactly
Trial and error methods: Finds \(d\) = 25 and \(n\) = 17 and list from 300 to 700 with total checked – 3/3