2025 Paper 2, Q2
A vehicle’s value ($) depreciates geometrically each year. The table shows the value, \(t_n\), where \(n\) is the number of years since the end of 2020.
| \(n\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|
| \(t_n\) | \(83\,600\) | \(71\,060\) | \(60\,401\) |
Determine if the value at the end of 2028 will be lower than one third of the value at the end of 2021.
Losing 15% a year
Each value is 85% of the one before. Carry the sequence on to the end of 2028 and compare it with a third of where it started.
Check the ratio, then count the years
\(\tfrac{71\,060}{83\,600}=0.85\) and \(\tfrac{60\,401}{71\,060}=0.85\). Check both, because the question says geometric and you should confirm it.
\(n\) counts years since the end of 2020, so the end of 2021 is \(n=1\) and the end of 2028 is \(n=8\).
The comparison is with one third of the 2021 value, \(t_1\). It is easy to use the last value in the table by mistake.
How fast would it need to fall?
Change the yearly ratio. At 0.85 the 2028 value only just drops below the line. A slightly slower fall, around 0.86, keeps it above.
Which term of the sequence is the end of 2028?
That is the number of years from 2021 to 2028, but 2021 is already \(n=1\).
Yes. \(n\) is years since the end of 2020, so \(2028-2020=8\).
That would be the end of 2029.
QCAA marking guide · 5 marks
\[t_1=83\,600,\qquad r=\frac{71\,060}{83\,600}=0.85\]
Marker: correctly determines the values for \(t_1\) and \(r\). This may be implied by later working.
\[t_n=t_1r^{\,n-1}=83\,600\times0.85^{\,n-1}\]
Marker: shows use of an appropriate method for a geometric sequence. Listing terms, or \(t_n=rt_{n-1}\), is accepted. Follow-through marks are allowed.
\[t_8=83\,600\times0.85^{7}=26\,800.24\]
Marker: predicts the value at the end of 2028.
\[\frac{83\,600}{3}=27\,866.67\]
Marker: determines the value for the one-third comparison. Comparing \(3\times26\,800.24=80\,400.72\) with 83 600 is also accepted.
\[26\,800.24<27\,866.67\]
The value at the end of 2028 will be lower than one third of the value at the end of 2021.
Marker: provides a decision. This mark can only be awarded if prior working supports it.
How close it is
The 2028 value is $1066.42 under the one-third mark. A year earlier, at the end of 2027, it is $31 529.69, still above.
\(0.85^7\approx0.32\), which is just under \(\tfrac13\). That is the whole comparison in one number: after seven drops of 15%, about 32% of the value is left.
Geometric sequences are familiar, and the ratio is easy to find. The complexity is in reading the question carefully. You have to match the end of 2028 to the right term, and compare with a third of the first value rather than one of the others in the table. I want your final line to state both numbers and say which is lower.
Question wording and marking-guide steps are from the 2025 QCAA General Mathematics external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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