**(a)** $500 + (500 + 200) = 1200$ or $S_2 = \frac{1}{2} \times 2[1000 + 200] = 1200$ | B1 |

**(b)** Using $a = 500, d = 200$ with $n = 7, 8$ or $9$ and "$a + (n-1)d$" or "listing" | M1 |

$500 + (7 \times 200) = (\pounds)1900$ | A1 |

**(c)** Using $\frac{1}{2}n\{2a + (n-1)d\}$ or $\frac{1}{2}n[a + l]$, or listing and "summing" terms | M1 |

$S_8 = \frac{1}{8}\{2 \times 500 + 7 \times 200\}$ or $S_8 = \frac{1}{8}\{500 + 1900\}$, or all terms in list correct | A1 |

$= (\pounds)9600$ | A1 |

**(d)** $\frac{1}{2}n\{2 \times 500 + (n-1) \times 200\} = 32000$ | M1 A1 | M1: General $S_n$, equated to 32000; M1: Simplify to 3 term quadratic

$n^2 + 4n - 320 = 0$ (or equiv.) |  |

$(n+20)(n-16) = 0$ | M1 | M1: Attempt to solve 3 t.q.

$n = \ldots$ |  |

$n = 16,$ Age is 26 | A1 A1 |

**Total: 13 marks**

| Guidance: (b) Correct answer with no working: Allow both marks. (c) Some working must be seen to score marks: Minimum working: $500 + 700 + 900 + \ldots (+ 1900) = \ldots$ scores M1 (A1). (d) Allow $\geq$ or $>$ throughout, apart from "Age 26". A common misread here is 3200. This gives $n = 4$ and age 14, and can score M1 A0 M1 A0 M1 A1 A1 with the usual misread rule. Alternative: (Listing sums) $(500, 1200, 2100, 3200, 4500, 6000, 7700, 9600,) 11700, 14000, 16500, 19200, 22100, 25200, 28500, 32000$. List at least up to 32000 [M3]; All values correct [A2]; $n = 16$ (perhaps implied by age) [A1 cso]; Age 26 [A1 cso]. If there is a mistake in the list, e.g. 16th sum = 32100, possible marks are: M3 A0 A0 A0. Alternative: (Trial and improvement) Use of $S_n$ formula with $n = 16$ (and perhaps other values) [M3]; Accurately achieving 32000 for $n = 16$ [A3]; Age 26 [A1].

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