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Year 12 Maths Advanced (2027) Sequences and series

Geometric Sequences

20 practice questions 0 video lessons Theory + worked examples

Learn geometric sequences for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-03). In a geometric progression each term is found by multiplying the previous one by the same number, the common ratio, so the terms grow or shrink by a constant factor.

You will learn to test whether a list is geometric, find the common ratio, use the nth term formula \(ar^{n-1}\) to find a term or its position, and recognise the exponential pattern behind it — the maths of doubling, halving and percentage change.

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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.