C1 June 2017 Q4
4. A company, which is making 140 bicycles each week, plans to increase its production. The number of bicycles produced is to be increased by \(d\) each week, starting from 140 in week 1, to \(140 + d\) in week 2, to \(140 + 2d\) in week 3 and so on, until the company is producing 206 in week 12.
After week 12 the company plans to continue making 206 bicycles each week.
| Scheme | Marks |
|---|---|
| \(206 = 140 + (12 - 1)\times d \Rightarrow d = \ldots\) | M1 |
| \(\left(d =\right)6\) | A1 |
| (2) |
Notes
M1: Uses \(206 = 140 + (12 - 1)\times d\) and proceeds as far as \(d = \ldots\)
A1: Correct answer only can score both marks.
| Scheme | Marks |
|---|---|
| \(S_{12} = \dfrac{12}{2}\left(140 + 206\right)\) or \(S_{12} = \dfrac{12}{2}\left(2\times 140 + (12 - 1)\times\text{"}6\text{"}\right)\) or \(S_{11} = \dfrac{11}{2}\left(140 + 206 - \text{"}6\text{"}\right)\) or \(S_{11} = \dfrac{11}{2}\left(2\times 140 + (11 - 1)\times\text{"}6\text{"}\right)\) | M1 |
| \(S = 2076\) WAY1 or \(S = 1870\) WAY 2 | A1 |
| \((52 - 12)\times 206 = \ldots\) or \((52 - 11)\times 206 = \ldots\) | M1 |
| Total \(= \text{"}2076\text{"} + \text{"}8240\text{"} = \ldots\) (WAY 1) or Total \(= \text{"}1870\text{"} + \text{"}8446\text{"} = \ldots\) (WAY 2) | ddM1 |
| 10316 | A1 |
| (5) | |
| (7 marks) |
Notes
M1: Attempts \(S_n = \dfrac{n}{2}\left(a + l\right)\) or \(S_n = \dfrac{n}{2}\left(2a + (n - 1)d\right)\) with \(n = 12\), \(a = 140, l = 206, d = \text{‘}6\text{’}\) WAY 1
Or
Attempts \(S_n = \dfrac{n}{2}\left(a + l\right)\) or \(S_n = \dfrac{n}{2}\left(2a + (n - 1)d\right)\) with \(n = 11\), \(a = 140, l = 206 - \text{‘}6\text{’}, d = \text{‘}6\text{’}\) WAY2
If they are using \(S_n = \dfrac{n}{2}\left(2a + (n - 1)d\right)\), the \(n\) must be used consistently.
A1: Correct sum (may be implied)
M1: Attempts to find \((52 - 12)\times 206\) or \((52 - 11)\times 206\). Does not have to be consistent with their \(n\) used for the first Method mark.
ddM1: Attempts to find the total by adding the sum to 12 terms with (52 - 12) lots of 206 or attempts to find the total by adding the sum to 11 terms with (52 - 11) lots of 206. I.e. consistency is now required for this mark. Dependent on both previous method marks.
A1: cao
Listing in (b)
| Week | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Bicycles | 140 | 146 | 152 | 158 | 164 | 170 | 176 |
| Total | 140 | 286 | 438 | 596 | 760 | 930 | 1106 |
| 8 | 9 | 10 | 11 | 12 | 13 | … | 52 |
| 182 | 188 | 194 | 200 | 206 | 206 | … | 206 |
| 1288 | 1476 | 1670 | 1870 | 2076 | 2282 | … | 10316 |
M1: Attempts the sum of either 12 or 11 terms of a series with first term 140 and their \(d\) up to \(140 + 11d\) or \(140 + 10d\).
A1: S = 2076 or 1870
Then follow the scheme
Special case in (b) – Treats as single AP with \(n = 52\)
\(S_n = \dfrac{52}{2}\left(2\times 140 + (52 - 1)\times\text{"}6\text{"}\right) = 15236\) Scores 11000
M1: \(S_n = \dfrac{n}{2}\left(2a + (n - 1)d\right)\) with \(n = 52\), \(a = 140\), \(d = \text{“}6\text{”}\) A1: 15236