Depreciation models in sequences (QCE General Mathematics Unit 3)

What this dot point is asking

When a business buys a car, a machine or a computer, its value falls over time: it depreciates. The QCAA General Mathematics 2025 syllabus uses depreciation as a key application of Topic 4, Growth and decay in sequences:

• straight-line depreciation (the value falls by the same amount each period) is modelled by an arithmetic sequence, and
• diminishing-value depreciation (the value falls by the same percentage each period) is modelled by a geometric sequence.

You must generate the values by recursion, display them in a table and a graph, use the $n$th-term rules, and solve practical problems: the value after a given time, when the value falls to a given level, the rate of depreciation, and which method gives the lower value.

The answer

Straight-line depreciation: an arithmetic sequence

The asset loses the same dollar amount each year, so the values form an arithmetic sequence with a negative common difference.

A coffee machine bought for $12 600 that depreciates by $1400 per year has values

Recursion: $t_{n+1} = t_n - 1400$, with $t_1 = 12\,600$.

Rule: $t_n = t_1 + (n - 1)d = 12\,600 + (n - 1)(-1400)$.

The graph of value against time is a set of points on a straight line falling towards zero, which is where the name comes from. The machine would reach zero when $12\,600 = 1400 \times 9$, after 9 years.

Diminishing-value depreciation: a geometric sequence

The asset loses the same percentage of its current value each year, so it keeps a fixed fraction: the values form a geometric sequence with common ratio

A car bought for $38 000 that depreciates by 15% per year keeps $85\%$ of its value each year, so $r = 0.85$:

Recursion: $t_{n+1} = 0.85t_n$, with $t_1 = 38\,000$.

Rule: $t_n = t_1r^{n-1} = 38\,000 \times 0.85^{n-1}$.

The dollar loss shrinks every year ($5700, then $4845, then $4118.25) because 15% of a smaller value is smaller. The graph is a curve that falls steeply at first and then flattens, getting closer to zero without ever reaching it.

The term-number trap

The syllabus rules use $t_1$ for the first term. If $t_1$ is the value when the asset is bought (time 0), then:

So the value after 5 years of diminishing value is $t_6 = t_1r^5$, and the value after 9 years of straight-line depreciation is $t_{10} = t_1 + 9d$. The tractor question above tests exactly this: from the start of 2012 to the start of 2021 is 9 years, which is the 10th term.

A useful check: the power of $r$ (or the multiple of $d$) always equals the number of years of depreciation.

Recognising the method from values

Given a list of values, test both patterns:

• Constant difference (subtract consecutive values): arithmetic, straight-line depreciation. The difference is the yearly loss.
• Constant ratio (divide consecutive values): geometric, diminishing-value depreciation. The rate is $1 - r$.

For $24 000, $21 600, $19 440: differences $-2400$ and $-2160$ are not equal, but the ratios are both 0.9, so it is diminishing value at 10% per year.

Solving practical problems

• Value after $k$ years. Use the rule with term number $k + 1$ (if $t_1$ is the purchase price).
• When is the value first below a threshold? For straight-line, solve the linear equation and round up. For diminishing value, the syllabus does not require logarithms: use trial and improvement (or a table of values) and give the first whole year that meets the condition.
• Finding the rate. Straight-line: loss per year $= \frac{\text{total loss}}{\text{years}}$. Diminishing value: $r^{\text{years}} = \frac{\text{later value}}{\text{earlier value}}$, then take the root; for two years, $r = \sqrt{\frac{32\,000}{50\,000}} = 0.8$, a 20% rate.
• Comparing methods. Tabulate both and look for when one becomes lower than the other.

A third method, unit cost depreciation (a fixed amount per kilometre or per hour of use), is also arithmetic when usage is steady, but it is not named in the QCAA syllabus.