**(a)** Use $n^{\text{th}}$ term $= a + (n-1)d$ with $d = 10$; $a = 150$ and $n = 8$, or $a = 160$ and $n = 7$, or $a = 170$ and $n = 6$: $= 150+7 \times 10$ or $160 + 6 \times 10$ or $170 + 5 \times 10$ | M1 |

$= 220^*$ (Or gives clear list – see note) | A1* |

| | **(2 marks)** |

| **Or:** If answer $220$ is assumed and $130 - (n-1)10 = 220$ or variation is solved for $n=$ then $n = 8$, so $2007$ is the year (must conclude the year). | M1 |

| | A1* | **(2 marks)** |

**(b)** Use $S_n = \frac{n}{2}\{2a + (n-1)10\}$ Or $S_n = \frac{n}{2}\{a + l\}$ and $l = a + (n-1)10$ | M1 |

$= 7(300 + 13 \times 10)$ or $7(150 + 280)$ | A1 |

$= 7 \times 430$ | A1 |

$= 3010$ | |

| | **(3 marks)** |

**(c)** Cost in year $n = 900+(n-1) \times 20$ | M1 |

Sales in year $n = 150+(n-1) \times 10$ | |

Cost $= 3 \times$ Sales $\Rightarrow 900+(n-1) \times 20 = 3 \times (150+(n-1) \times 10)$ | M1 |

$900 - 20n + 20 = 450 + 30n - 30$ | |

$500 = 50n$ | M1 |

$n = 10$ | M1 |

Year is $2009$ | A1 |

As $n$ is not defined they may work correctly from another base year to get the answer $2009$ and their $n$ may not equal $10$. If doubtful – send to review. |

| | **(4 marks)** |

| **Notes:** | |
|---|---|
| **(a)** | M1: Attempt to use $n^{\text{th}}$ term $= a + (n-1)d$ with $d = 10$, and correct combination of $a$ and $n$ i.e. $a = 150$ and $n = 8$ or $a = 160$ and $n = 7$, or $a = 170$ and $n = 6$. A1*: Shows that $220$ computers are sold in $2007$ with no errors. Note that this is a given solution, so needed $150+7 \times 10$ or $160 + 6 \times 10$ or $170 + 5 \times 10$. Accept a correct list showing all values and years for both marks. Just $150, 160, 170, 180, 190, 200, 210, 220$ is M1A0. Need some reference to years as well as the list of numbers of computers for A1. |
| **(b)** | M1: Attempts to use $S_n = \frac{n}{2}\{2a + (n-1)d\}$ with $d = 10$, and correct combination of $a$ and $n$ i.e. $a = 150$ and $n = 14$, or $a = 160$ and $n = 13$, or $a = 170$ and $n = 12$. A1: Uses $S_n = \frac{n}{2}\{2a + (n-1)d\}$ with $a = 150, d = 10$ and $n = 14$ [N.B. $S_n = \frac{n}{2}\{a + l\}$ needs $l = a + (n - 1)d$ as well]. NB A0 for $a = 160$ and $n = 13$ or $a = 170$ and $n = 12$ unless they then add the first, or first two terms respectively. A1: Cao $3010$. This answer (with no working) implies correct method M1A1A1. |
| **(c)** | M1: Writes down an expression for the cost $= 900+(n-1) \times 20$ or writes $900 + (n-1) \times d$ and allow use of $20$ instead of $-20$ and allow $n$ (consistently) instead of $n - 1$ for this mark. Ignore £ signs in equation. M1: **Attempts to write down an equation in $n$ for statement "cost $= 3 \times$ sales"** or writes $900 + (n-1) \times 20 = 3 \times(150 + (n-1) \times 10)$. Accept the $3$ on the wrong side and allow use of $20$ instead of $-20$ and allow $n$ (consistently) instead of $n - 1$ for this mark. Ignore £ signs in equation. M1: Solves the correct linear equation in $n$ to achieve $n = 10$ (for those using $n - 1$) or $n = 9$ (for those using $n$). Ignore £ signs. A1: Cso Year $2009$ (A0 for the answer Year 10 if $2009$ is not given). Special case: **Just answer or trial and improvement with no equation leading to answer scores SC 0,0,1,1.** Equations satisfying the method mark descriptors followed by trial and improvement could get all four marks. |

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