Geometric sequences, series and limiting behaviour
Master geometric sequences, series and limiting behaviour in Year 11 VCE Specialist Mathematics. A geometric sequence multiplies by the same common ratio \(r\) at every step, so its \(n\)th term is \(t_n=a\,r^{\,n-1}\), and when \(|r|<1\) the terms shrink toward \(0\). It sits in the Discrete mathematics area of study of the VCE Mathematics Study Design (VCAA), within the Sequences and series topic of Unit 1.
You will learn to find any term, the first term \(a\), the common ratio \(r\) or the position \(n\), add the first \(n\) terms with the partial sum \(S_n=\dfrac{a\big(r^{\,n}-1\big)}{r-1}\), and — when \(|r|<1\) — find the sum to infinity \(S_\infty=\dfrac{a}{1-r}\) — skills that underpin recurrence relations and financial modelling.
Theory
A geometric sequence multiplies by the same common ratio \(r\) at every step. In Year 11 Specialist Mathematics you use the \(n\)th-term rule \(t_n = a\,r^{\,n-1}\) to find any term, the first term \(a\), the common ratio \(r\) or the position \(n\); the partial sum \(S_n = \dfrac{a\big(r^{\,n}-1\big)}{r-1}\) to add the first \(n\) terms; and, when \(|r|<1\), the sum to infinity \(S_\infty = \dfrac{a}{1-r}\). This page shows how, with fully worked examples.
A sequence is an ordered list of numbers called terms, written \(t_1, t_2, t_3, \ldots\) In a geometric sequence each term is found by multiplying the term before it by a fixed number, the common ratio \(r\). The first term is \(a\) (also written \(t_1\)).
Because the same \(r\) multiplies each time, the ratio of any term to the one before it is constant: \(r = \dfrac{t_2}{t_1} = \dfrac{t_3}{t_2} = \cdots\). For example \(3,\ 6,\ 12,\ 24,\ \ldots\) is geometric with \(a=3\) and \(r=2\), while \(24,\ 12,\ 6,\ 3,\ \ldots\) is geometric with \(a=24\) and \(r=\tfrac{1}{2}\).
Multiplying by \(r\) each time means the \(n\)th term is \(a\) times \((n-1)\) factors of \(r\): \(t_n = a\,r^{\,n-1}\). A series is the sum of the terms; the partial sum \(S_n = t_1 + t_2 + \cdots + t_n\) adds the first \(n\) terms and equals \(\dfrac{a\big(r^{\,n}-1\big)}{r-1}\) (for \(r\neq 1\)).
The limiting behaviour of the terms depends entirely on \(|r|\). When \(|r|<1\) each term is a fraction of the one before, so the terms shrink toward \(0\) as \(n\to\infty\); when \(|r|>1\) the terms grow without bound. Only when \(|r|<1\) do the partial sums settle on a finite limit, the sum to infinity \(S_\infty = \dfrac{a}{1-r}\).
The \(n\)th term of a geometric sequence is the first term \(a\) multiplied by \((n-1)\) factors of the common ratio \(r\):
The partial sum of the first \(n\) terms (for \(r\neq 1\)):
When \(|r|<1\), the terms tend to \(0\) and the partial sums approach the sum to infinity:
Choosing and applying the rule
- Identify \(a\) and \(r\). The first term is \(a\); the common ratio is any term divided by the one before it, \(r = \dfrac{t_2}{t_1}\).
- Finding a term: substitute \(a\), \(r\) and the position \(n\) into \(t_n = a\,r^{\,n-1}\).
- Finding \(a\), \(r\) or \(n\): substitute what you know into \(t_n = a\,r^{\,n-1}\) and solve. Two given terms give two equations — divide them so \(a\) cancels and a power of \(r\) is left, then take the root.
- Finding a partial sum: use \(S_n = \dfrac{a\big(r^{\,n}-1\big)}{r-1}\) with \(a\), \(r\) and \(n\).
- Sum to infinity: first check \(|r|<1\). If so, \(S_\infty = \dfrac{a}{1-r}\); if \(|r|\ge 1\) the sum to infinity does not exist.
For a worded problem, name \(a\) and \(r\) from the context first (for example a first swing of \(80\) cm followed by swings \(\tfrac{3}{4}\) as long gives \(a=80\), \(r=\tfrac{3}{4}\)), then apply the matching rule.
Apply the \(n\)th-term rule \(t_n = a\,r^{\,n-1}\) with \(n=6\):
| \(t_6\) | \(=\) | \(a\,r^{6-1}\) |
| \(=\) | \(3\times 2^{5}\) | |
| \(=\) | \(3\times 32\) | |
| \(=\) | \(96\) |
The \(6\)th term is \(t_6 = 96\).
Read off \(a=3\) and \(r=4\), then apply \(S_n = \dfrac{a\big(r^{\,n}-1\big)}{r-1}\) with \(n=4\):
| \(S_4\) | \(=\) | \(\dfrac{3\big(4^{4}-1\big)}{4-1}\) |
| \(=\) | \(\dfrac{3\,(256-1)}{3}\) | |
| \(=\) | \(\dfrac{3\times 255}{3}\) | |
| \(=\) | \(255\) |
The sum of the first \(4\) terms is \(S_4 = 255\).
Find the common ratio \(r = \dfrac{t_2}{t_1}\) and check its size:
| \(r\) | \(=\) | \(\dfrac{12}{24} = \dfrac{1}{2}\) |
| \(|r|\) | \(=\) | \(\tfrac{1}{2} < 1\) |
Since \(|r|<1\), each term is half the one before, so the terms shrink toward \(0\); the sum to infinity exists, \(S_\infty = \dfrac{a}{1-r}\):
| \(S_\infty\) | \(=\) | \(\dfrac{24}{1-\tfrac{1}{2}}\) |
| \(=\) | \(\dfrac{24}{\tfrac{1}{2}}\) | |
| \(=\) | \(48\) |
As \(n\to\infty\) the terms \(t_n\to 0\), and the sum to infinity is \(S_\infty = 48\).
The swing lengths are geometric with \(a=80\) and \(r=\tfrac{3}{4}\). Third swing, using \(t_n = a\,r^{\,n-1}\):
| \(t_3\) | \(=\) | \(80\times\left(\tfrac{3}{4}\right)^{2}\) |
| \(=\) | \(80\times\tfrac{9}{16}\) | |
| \(=\) | \(45\) |
Total distance: since \(|r|=\tfrac{3}{4}<1\), use \(S_\infty = \dfrac{a}{1-r}\):
| \(S_\infty\) | \(=\) | \(\dfrac{80}{1-\tfrac{3}{4}}\) |
| \(=\) | \(\dfrac{80}{\tfrac{1}{4}}\) | |
| \(=\) | \(320\) |
The \(3\)rd swing is \(45\) cm, and the total distance travelled over all swings is \(320\) cm.
Common pitfalls
Frequently asked questions
What is a geometric sequence?
It is a sequence in which each term is a fixed multiple of the one before it. That multiplier is the common ratio \(r\). Starting from the first term \(a\), the terms are \(a,\ ar,\ ar^2,\ ar^3,\ \ldots\)
What is the formula for the nth term?
The \(n\)th term is \(t_n = a\,r^{\,n-1}\), where \(a\) is the first term, \(r\) is the common ratio and \(n\) is the position of the term.
How do I find the sum of a geometric series?
For the first \(n\) terms use \(S_n = \dfrac{a\big(r^{\,n}-1\big)}{r-1}\) (equivalently \(\dfrac{a\big(1-r^{\,n}\big)}{1-r}\)), where \(a\) is the first term, \(r\) the common ratio and \(n\) the number of terms.
What happens to the terms as n gets large?
It depends on \(|r|\). When \(|r|<1\) each term is a fraction of the previous one, so the terms \(t_n\) shrink toward \(0\). When \(|r|>1\) the terms grow without bound, and when \(|r|=1\) they stay the same size.
When does the sum to infinity exist, and what is it?
The sum to infinity exists exactly when \(|r|<1\), because only then do the terms tend to \(0\) and the partial sums settle on a finite limit. Its value is \(S_\infty = \dfrac{a}{1-r}\).
How do I find the common ratio from two terms?
Write \(t_n = a\,r^{\,n-1}\) for each given term and divide one by the other so \(a\) cancels, leaving a power of \(r\). For example \(\dfrac{t_5}{t_2} = r^{3}\), so if \(t_5=48\) and \(t_2=6\) then \(r^{3}=8\) and \(r=2\).