Question 1 Report
A school displays trophies from sports day in a growing block arrangement; the first three patterns below hold 9, 13 and 17 trophies.
An arithmetic sequence with a formula \(a_1 + (n-1)d\) uses the first term \(a_1\) and the common difference \(d\); a maximum capacity limits how large the pattern number can be, since beyond a certain point the pattern needs more trophies than the cabinet can hold.
The trophies rise from 9 by 4 each pattern, so the common difference is \[d = 4\] Using the first term 9 and this common difference, the \(n\)th term is \(9+(n-1)(4)\), which expands to \[4n + 5\] [2 marks]
Substituting \(n=8\) gives \[4(8) + 5 = 32 + 5 = 37 \text{ trophies}\] [1 mark]
Setting \(4n+5 \le 100\) and rearranging gives \(4n \le 95\), so \[n \le 23.75\] The greatest whole number satisfying this is \(n=23\), so Pattern 23 is the largest pattern that fits in the cabinet. [2 marks]
Pattern 25 needs \(4(25)+5=105\) trophies, which is more than the cabinet's limit of 100 trophies, so Pattern 25 could not be displayed. [1 mark]
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