QCE General Mathematics - Unit 3 - Growth and decay in sequences
Geometric Sequences | QCE General Mathematics
Learn recursion, common ratios, nth-term rules and practical growth or decay models for QCE General Mathematics.
Updated 2026-05-18 - 5 min read
QCAA official coverage - General Mathematics 2025 v1.3
Exact syllabus points covered
- Use recursion to generate a geometric sequence.
- Display the terms of a geometric sequence in both tabular and graphical form and demonstrate that geometric sequences can be used to model exponential growth and decay in discrete situations.
- Use the rule for the $n$th term of a geometric sequence: $t_n=t_1r^{n-1}$, where $t_n$ is the $n$th term, $t_1$ is the first term, $n$ is the term number and $r$ is the common ratio.
- Use geometric sequences to model and analyse practical situations involving geometric growth and decay (use of logarithms not required), e.g. modelling the growth of a bacterial population that doubles in size each hour, calculating the value of an item using the diminishing-value method of depreciation.
A geometric sequence changes by multiplying by the same factor each time. That factor is called the common ratio. Geometric sequences model discrete exponential growth and decay, such as a population doubling each hour or a car losing a fixed percentage of value each year.
Original Sylligence diagram for general arithmetic geometric sequences.
Recursive form
Recursive form is:
$ t_{n+1}=rt_n $
where $r$ is the common ratio. If $t_1=80$ and $r=1.25$, the terms are:
$ 80,\ 100,\ 125,\ 156.25,\ldots $
Each term is $125\%$ of the previous term, so the sequence grows by $25\%$ per step.
The nth-term rule
The direct rule is:
$ t_n=t_1r^{n-1} $
The exponent is $n-1$ because the first term has had zero multiplications by $r$.
Worked example
Common traps
Check whether the question names the initial value as $t_0$ or $t_1$. The syllabus formula uses $t_1$, but technology tables may start at $n=0$.
Checking for geometric behaviour
To test whether a sequence is geometric, divide consecutive terms:
$ \frac{t_2}{t_1},\quad \frac{t_3}{t_2},\quad \frac{t_4}{t_3},\ldots $
If the ratio is constant, the sequence is geometric. The ratio can be greater than $1$, between $0$ and $1$, or negative.
| Common ratio | Behaviour | |---:|---| | $r>1$ | growth | | $0<r<1$ | decay toward zero | | $r=1$ | constant sequence | | $r<0$ | signs alternate |
Depreciation as geometric decay
Diminishing-value depreciation is geometric because each period removes a percentage of the current value.
If an item depreciates by $25\%$ each year, its multiplier is:
$ r=1-0.25=0.75 $
The value after $n-1$ yearly steps is:
$ t_n=t_1(0.75)^{n-1} $
Depth: identifying geometric structure
A geometric sequence has a constant ratio:
$ r=\frac{t_{n+1}}{t_n} $
Each term is found by multiplying the previous term by the same number.
| Ratio | Behaviour | |---:|---| | $r>1$ | exponential growth | | $0<r<1$ | exponential decay | | $r=1$ | constant sequence | | $r<0$ | terms alternate sign |
In QCE General Mathematics finance and depreciation contexts, the ratio is usually positive.
Rule forms
The recursive rule is:
$ t_{n+1}=rt_n $
The explicit rule is:
$ t_n=t_1r^{n-1} $
Use the explicit rule when finding a term far into the sequence.
Percentage change and multipliers
Convert percentage change to a multiplier before writing the sequence rule.
| Description | Multiplier | |---|---:| | increase by $6\%$ | $1.06$ | | increase by $2.5\%$ | $1.025$ | | decrease by $12\%$ | $0.88$ | | decrease by $3.4\%$ | $0.966$ |
This multiplier is the common ratio.
Arithmetic versus geometric
If a value increases by the same amount each period, it is arithmetic. If it increases by the same percentage each period, it is geometric.
Depth: geometric sums
When repeated percentage changes are accumulated, a question may ask for the total of several terms. The sum of the first $n$ terms is:
$ S_n=t_1\left(\frac{r^n-1}{r-1}\right) $
for $r\ne1$. An equivalent form is:
$ S_n=t_1\left(\frac{1-r^n}{1-r}\right) $
which is often convenient when $0<r<1$.
When a context starts at time zero, label the starting value as $t_0$ or $V_0$ if that is clearer. Then after $n$ completed periods the model is usually $V_n=V_0r^n$. This avoids the common one-period shift that happens when students force every problem into the $t_n=t_1r^{n-1}$ form.