## Question 8:

### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Arithmetic sequence with $a = 9300$, $a_{10} = a + 9d = 9300 + 9d = 3900$ | M1 | Using formula for term of AP with values substituted |
| $d = -600$ | A1 | soi; e.g. $d = 600$ stated but later subtracted |
| $a_{20} = a + 19d = 9300 - 19 \times 600 = -2100$, so 20th term is negative | A1 [3] | Allow for $-2100$ without comment; also allow for earlier negative term found and comment that $20^{th}$ is less |

### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S$ is increasing as long as extra terms are positive; first negative term when $a_n < 0$ | M1 | Attempt to find first negative term with $9300$ and their $d$ |
| $9300 - 600(n-1) < 0 \Rightarrow \frac{9300}{600} < n - 1 \Rightarrow n > 16.5$, so maximum sum after 16 terms | A1 | Allow $n > 16$ or $n \geq 17$ oe; also allow M1A1 for establishing $a_{16} = 300$ and $a_{17} = -300$ |
| $S_{16} = \frac{16}{2}(2 \times 9300 - 600(16-1)) = 76800$ | M1, A1 [4] | Using sum with their $n$ and their $d$; cao |
| **Alternative:** $S = \frac{n}{2}(2 \times 9300 - 600(n-1)) \left[= 9600n - 300n^2\right]$ | M1 | Using sum formula with $9300$ and their $d$ substituted |
| $\frac{dS}{dn} = 9600 - 600n = 0$, max sum when $n = 16$ | M1, A1 | Setting derivative to zero and solve; allow for $n = 16$; also allow M1 for other methods, e.g. $S_n = k - 300(n-16)^2$ |
| $S_{16} = \frac{16}{2}(2 \times 9300 - 600(16-1)) = 76800$ | A1 | cao |
| **Second alternative:** $S_{15} = 76500$, $S_{16} = 76800$, $S_{17} = 76500$, max total is $76800$ | M1, M1, A1, A1 | Using sum formula; evaluating $S_{15}$, $S_{16}$, $S_{17}$; award for $S_{16}$ as maximum; cao. FT their $d$: three consecutive totals around their maximum |

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