Hint
The weekly amounts form an AP, but the threshold in part (c) concerns their running total.
Worked solution
(a)
\(u_{12}=40+11(5)=95\) tokens.
Answer: 95 tokens.
(b)
\(\displaystyle S_{12}= \frac{12(40+ 95)}{2}= 810\) tokens.
Answer: 810 tokens.
(c)
\(\displaystyle S_{n}= n\{80+ \frac{5(n- 1)\}}{2}= (\frac{5}{2})n(n+ 15)\). Then \(\displaystyle S_{13}=910<1000\), while \(\displaystyle S_{14}=1015\ge 1000\). Since every weekly amount is positive, the totals strictly increase. The first qualifying week is 14.
Answer: Week 14.
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 2 marks
- 1 mark: Use eleven increases after week 1.
- 1 mark: Give 95 tokens.
Alternative methods: Listing correctly indexed weekly amounts is accepted.
Common error: Counting twelve increases after the initial amount.
(b) · 2 marks
- 1 mark: Use the AP sum over twelve weekly amounts.
- 1 mark: Give 810 tokens.
Alternative methods: Adding the weekly amounts directly is accepted.
Common error: Reporting only the week-12 collection.
(c) · 3 marks
- 1 mark: Form the correct cumulative-sum expression or equivalent increasing totals.
- 1 mark: Identify week 14.
- 1 mark: Check both adjacent totals and justify that no earlier week qualifies.
Alternative methods: Solving the quadratic inequality and selecting the correct integer with a neighbour check is accepted.
Common error: Using the single-week amount for the cumulative threshold.
Total: 7 marks.