Growth and Decay - TCE Mathematics Applications (Tasmania)

Syllabus dot point

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

TCEMathematics ApplicationsUnit 48 min readUpdated 2 Jun 2026

Quick answer

You must model quantities that change over time as either linear (constant amount added each step) or geometric (constant percentage change each step), expressing each as a recurrence relation and as a closed-form rule, and use them to solve problems.

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).

Linear versus geometric change

Linear (arithmetic): a constant value $d$ is added each step, giving a straight-line graph. Geometric: a constant ratio $R$ multiplies each step, giving a curved (exponential) graph.

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.

Closed-form rules

Linear (arithmetic): $t_n = t_0 + nd$ .

Geometric: $t_n = t_0 R^n$ .

For a percentage change of $p\%$ per step, $R = 1 + \dfrac{p}{100}$ (use a minus for decay).

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

Geometric decay of a drug

A patient has $200$ mg of a drug. Each hour the body removes $25\%$ of what remains. How much is left after $5$ hours?

Set up and apply the rule

Removing $25\%$ leaves $75\%$ , so $R = 0.75$ , $t_0 = 200$ .
Recurrence: $t_{n+1} = 0.75\, t_n$ .
Closed form: $t_n = 200(0.75)^n$ .

Evaluate at n = 5

t_5 = 200(0.75)^5 = 200 \times 0.2373046875 = 47.46 \text{ mg}.

After 5 hours about $47.5$ mg remains. Because the ratio is constant, the amount removed each hour shrinks, but the percentage removed stays the same.

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.

Linear versus geometric: which model?

A population is recorded as $50, 55, 60, 65$ in successive years.

Test differences and ratios

Differences: $55-50 = 5$ , $60-55 = 5$ , $65-60 = 5$ . Constant, so this is linear.
Ratios: $55/50 = 1.10$ , $60/55 = 1.09$ . Not constant, so it is not geometric.

The rule is $t_n = 50 + 5n$ (with $t_0 = 50$ ). The population grows by a fixed 5 each year, not by a fixed percentage.

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).

Differences for linear, ratios for geometric

When given a table, compute first differences and first ratios. Constant differences mean linear; constant ratios mean geometric. This one check tells you which formula to use.

Using the wrong starting index

The closed form $t_n = t_0 R^n$ assumes $t_0$ is the first (zeroth) term. If a problem calls the first value $t_1$ , then after $n$ steps the term is $t_1 R^{\,n-1}$ . Read carefully whether the count starts at 0 or 1 before plugging in $n$ .

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 questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice8 marks

A 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).

Practice3 marks

a) 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.