Year 12 Maths Advanced (2027) Sequences and series

Recurring Decimals as Infinite GPs

20 practice questions 0 video lessons Theory + worked examples

NSW · Year 12 Mathematics Advanced · Sequences and series

Recurring decimals can be written as infinite geometric series and converted to exact fractions using the limiting sum \(S_\infty=\dfrac{a}{1-r}\).

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

Every recurring decimal is an infinite geometric series, so the limiting sum turns it into an exact fraction. This Year 12 Mathematics Advanced topic (MAV-12-03) converts recurring decimals to fractions.

A recurring decimal repeats a block of digits forever, which is a geometric series with \(|r|<1\). Summing it gives an exact fraction.

Shortcut: one repeating digit is over \(9\) (\(0.\dot{d}=\dfrac{d}{9}\)); a two-digit block is over \(99\); a three-digit block over \(999\). A useful check is \(0.\dot{9}=\dfrac{9}{9}=1\).

\(0.\dot{6}=0.6+0.06+0.006+\cdots\), a geometric series.

\[0.\dot{d}=\dfrac{d}{9},\qquad 0.\dot{a}\dot{b}=\dfrac{ab}{99}\]

0 point d recurring equals d over 9; 0 point a b recurring equals a b over 99

Method

Identify the repeating block and its length.
Write it over \(9\), \(99\), \(999\), \(\ldots\) (matching the block length), or sum it as a GP.
Simplify the fraction.

Example 1 — One digit

Write \(0.\dot{4}\) as a fraction.

Solution

\(0.\dot{4}\)

\(=\)

\(\dfrac{4}{9}\)

Example 2 — Two digits

Write \(0.\dot{2}\dot{7}\) in simplest form.

Solution

\(0.\dot{2}\dot{7}\)

\(=\)

\(\dfrac{27}{99}=\dfrac{3}{11}\)

Example 3 — Via the GP sum

Use \(S_\infty\) to write \(0.\dot{6}\) as a fraction.

Solution

\(0.\dot{6}\)	\(=\)	\(\dfrac{0.6}{1-0.1}\)
	\(=\)	\(\dfrac{0.6}{0.9}=\dfrac{2}{3}\)

Example 4 — Two conversions

Write \(0.\dot{1}\dot{2}\) and \(0.\dot{9}\) as fractions.

Solution

\(0.\dot{1}\dot{2}\)	\(=\)	\(\dfrac{12}{99}=\dfrac{4}{33}\)
\(0.\dot{9}\)	\(=\)	\(\dfrac{9}{9}=1\)

Common pitfalls

Match block length to the denominator. One digit over \(9\); two digits over \(99\).

Only the repeating part is the GP. Handle any non-repeating start separately.

Always simplify (e.g. \(\dfrac{27}{99}=\dfrac{3}{11}\)).

Frequently asked questions

How do you turn a recurring decimal into a fraction?

A recurring decimal is an infinite geometric series. Put the repeating block over as many nines as its length: one digit over 9, two digits over 99, then simplify.

What is 0.6 recurring as a fraction?

It is a geometric series with first term 0.6 and ratio 0.1, so the sum is 0.6 divided by 0.9, which simplifies to two thirds.

Is 0.9 recurring equal to 1?

Yes. As a fraction 0.9 recurring is 9 over 9, which equals exactly 1.

What is 0.27 recurring as a fraction?

The two-digit block 27 goes over 99, giving 27 over 99, which simplifies to 3 over 11.

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