§-Syllabus dot point

TASMathematics ApplicationsSyllabus dot point

How do we model quantities that grow or decay by a constant factor over time?

Model and analyse linear and geometric growth and decay using recurrence relations and rules.

Linear (arithmetic) versus geometric (exponential) growth and decay, recurrence relations, closed-form rules, and applications in TCE Mathematics Applications.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Growth and decay extend the financial recurrence ideas to any repeating process: populations, drug concentrations, radioactive material, or savings. The key question is whether each step adds a fixed amount (linear) or multiplies by a fixed factor (geometric).

Recurrence relations

A recurrence relation defines each term from the previous one and a starting value:

Linear: $t_{n+1} = t_n + d$ , with $t_0$ given. Here $d > 0$ is growth, $d < 0$ is decay.

Geometric: $t_{n+1} = R\, t_n$ , with $t_0$ given. Here $R > 1$ is growth, $0 < R < 1$ is decay.

The closed form lets you jump straight to any term without listing all the ones before it.

Recognising the model

If the differences between consecutive terms are constant, the data is linear. If the ratios between consecutive terms are constant, the data is geometric. Always test which one fits before choosing a rule.

Comparing the two models

Over the long run, geometric growth always overtakes linear growth, because multiplying repeatedly outpaces adding repeatedly. Geometric decay approaches zero but never quite reaches it, whereas linear decay hits zero (and would go negative if the rule kept applying).

Finding how long a change takes

A frequent task asks not for a value but for the time to reach a target: how long for an investment to double, or for a drug to fall below a threshold. Set the closed form equal to the target and solve for $n$ . Because $n$ is in the exponent, this needs logarithms or a guess-and-check on the calculator. For example, to find when $t_0 R^n$ first exceeds a target $T$ , solve $R^n = \frac{T}{t_0}$ , then $n = \frac{\log(T / t_0)}{\log R}$ , and round up to the next whole period because the target is reached part way through.

Time for an investment to double

An investment of $\$5000$ grows at $6\%$ per year ( $R = 1.06$ ). After how many whole years does it first exceed $\$10\,000$ ?

Step 1: Write the geometric closed-form equation and set it equal to the target

Because the investment grows by a constant percentage each year, it follows the geometric rule $t_n = t_0 R^n$ . We substitute $t_0 = 5000$ , $R = 1.06$ , and the target $t_n = 10000$ to obtain the equation $5000(1.06)^n = 10000$ . This gives us a single equation in the unknown $n$ .

Step 2: Isolate the exponential and apply logarithms to solve for $n$

Dividing both sides by $5000$ isolates the exponential: $(1.06)^n = 2$ . Because $n$ sits in the exponent, we take $\log$ of both sides and use the power rule to bring $n$ down as a coefficient, giving $n = \frac{\log 2}{\log 1.06} = \frac{0.3010}{0.0253} = 11.9$ . The result is not a whole number, so we cannot stop here.

Step 3: Round up to the next whole year and verify with the closed form

Because the investment compounds annually, only complete years produce growth, so we must round up to the nearest whole number. After $12$ years the investment first exceeds $\$10\,000$ . We verify by substituting back: $5000(1.06)^{12} = \$10\,061.41$ , which is just above the target, confirming that year 12 is correct and year 11 would fall just short.

Sum of a finite geometric series

Many growth-and-decay problems ask for a running total, such as the total output over several years. The sum of the first $n$ terms of a geometric sequence is $S_n = \dfrac{a(r^n - 1)}{r - 1}$ (equivalently $\dfrac{a(1 - r^n)}{1 - r}$ for $r < 1$ ). When $|r| < 1$ the total over infinitely many periods converges to $S_\infty = \dfrac{a}{1 - r}$ , which is what a "adding forever" question is testing.

A complete answer states the type of model, gives both the recurrence and the closed form, solves for time using logarithms where asked, and rounds sensibly to the context (whole people for populations, two decimals for milligrams).

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20248 marksA table shows the declining births per year in a large country: 2021

= 15\,644\,946

(

t_1

), 2022

= 15\,269\,467

(

t_2

), 2023

= 14\,903\,000

(

t_3

). a) Is the sequence arithmetic or geometric? Show working. b) Find the explicit rule. c) Predict the total births over the 10 years from 2021 (inclusive), to the nearest whole number. d) Adding the births forever, what is the total?

Show worked answer →

a) (2 marks) Test ratios: $\frac{t_2}{t_1} = \frac{15269467}{15644946} = 0.976$ and $\frac{t_3}{t_2} = \frac{14903000}{15269467} = 0.976$ . The ratio is constant (the differences are not), so the sequence is geometric with $r = 0.976$ .

b) (2 marks) Geometric rule $t_n = a r^{\,n-1}$ with $a = 15644946$ and $r = 0.976$ : $t_n = 15644946 \times (0.976)^{n-1}$ .

c) (2 marks) Sum of the first 10 terms $S_n = \frac{a(1 - r^n)}{1 - r} = \frac{15644946(1 - 0.976^{10})}{1 - 0.976} \approx 140\,590\,150$ births (nearest whole number).

d) (2 marks) Since $|r| < 1$ , the infinite sum converges: $S_\infty = \frac{a}{1 - r} = \frac{15644946}{0.024} = 651\,872\,236$ . Adding forever does not give infinity; it approaches this fixed value (though in reality births cannot continue indefinitely).

TCE 20233 marksa) i. Growth of organisms in a petri dish follows

100, 150, 225, \dots

Find an equation to model the sequence. ii. A herd of 1500 wildebeest grows by

15\%

each year, but loses 40 to lion attacks each year. Find an equation to model the situation.

Show worked answer →

a) i. (1 mark) Ratios: $\frac{150}{100} = 1.5$ and $\frac{225}{150} = 1.5$ , so this is geometric with $a = 100$ and $r = 1.5$ . Explicit rule: $t_n = 100 \times (1.5)^{n-1}$ .

a) ii. (2 marks) Each year the herd grows by $15\%$ (multiply by $1.15$ ) then loses 40. This both multiplies and subtracts a fixed amount, so it is a first-order linear recurrence: $t_{n+1} = 1.15\,t_n - 40$ , with $t_0 = 1500$ . Markers want the multiplier $1.15$ and the subtraction of 40, not a pure geometric rule.