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Year 12 Maths Advanced (2027) Sequences and series

Geometric Sequences

20 practice questions 0 video lessons Theory + worked examples
NSW · Year 12 Mathematics Advanced · Sequences and series

Geometric sequences multiply by a constant common ratio \(r\), with \(n\)th term \(T_n=ar^{\,n-1}\).

Part of the NSW Year 12 Mathematics Advanced course, in the Sequences and series focus area of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

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Theory

A geometric sequence (GP) multiplies by the same amount each step — the common ratio \(r\). This Year 12 Mathematics Advanced topic (MAV-12-03) uses \(a_n=ar^{\,n-1}\) to find terms and solve growth and decay problems.

In a geometric sequence each term is a fixed multiple of the one before: the common ratio is \(r=\dfrac{a_n}{a_{n-1}}\).

With first term \(a\), the \(n\)th term is \(a_n=ar^{\,n-1}\). \(|r|>1\) grows; \(0<|r|<1\) decays toward \(0\); a negative \(r\) makes the terms alternate in sign.

Geometric sequence 2,6,18,54The terms of a GP with ratio 3 rise exponentially. na_n 1 2 3 4
The GP \(2,6,18,54\) (\(r=3\)) rises exponentially.
\[a_n=a\,r^{\,n-1}\]
a n equals a times r to the power n minus 1

Method

  1. Find the common ratio \(r\) (divide a term by the previous one) and the first term \(a\).
  2. Substitute into \(a_n=ar^{\,n-1}\).
  3. Apply in context: a repeated percentage or fraction of the previous value.
Example 1 — Find a term
Find \(r\) and \(a_6\) for \(2,6,18,\ldots\)
Solution
\(r\)\(=\)\(3\)
\(a_6\)\(=\)\(2\times3^5=486\)
Example 2 — Rule for \(a_n\)
Find a rule for \(16,8,4,\ldots\) and hence \(a_5\).
Solution
\(a_n\)\(=\)\(16\left(\tfrac12\right)^{n-1}\)
\(a_5\)\(=\)\(16\times\tfrac{1}{16}=1\)
Example 3 — Bounce
A ball dropped from \(10\) m bounces to \(0.8\) of each height. Height after the \(3\)rd bounce?
Solution
\(h\)\(=\)\(10\times0.8^3\)
\(=\)\(5.12\text{ m}\)
Example 4 — Build the rule
For \(3,6,12,\ldots\): state \(r\), find \(a_7\), and the rule for \(a_n\).
Solution
\(r\)\(=\)\(2\)
\(a_7\)\(=\)\(3\times2^6=192\)
\(a_n\)\(=\)\(3\times2^{\,n-1}\)

Common pitfalls

The exponent is \(n-1\), not \(n\). The first term is \(ar^{0}=a\).
\(r\) can be a fraction or negative. For \(16,8,4,\ldots\), \(r=\dfrac12\).
Check the ratio is constant (not the difference).

Frequently asked questions

What is the formula for the nth term of a geometric sequence?

It is a n equals a times r to the power n minus 1, where a is the first term and r is the common ratio.

How do you find the common ratio?

Divide any term by the term before it. For 2, 6, 18 the ratio is 6 divided by 2, which is 3.

What is the difference between an arithmetic and geometric sequence?

An arithmetic sequence adds a fixed amount each step, while a geometric sequence multiplies by a fixed ratio each step.

Can the common ratio be a fraction?

Yes. A ratio between 0 and 1 makes a decaying sequence, such as 16, 8, 4 with ratio one half.