MathematicsChapter 4: Sequences and Series

Infinite Gp and Recurring Decimals

If |r|<1, the infinite sum S = a / (1-r).

Learning objectives

State the main ideas and notation used in infinite gp and recurring decimals.
Apply the rules of infinite gp and recurring decimals in a clear, step-by-step solution.
Check signs, conditions, and the final answer for mathematical accuracy.

Definition

If |r|<1, the infinite sum S = a1-r. If |r|>=1 the infinite GP does not settle to a finite school sum (except the special a=0). A recurring decimal is a GP in disguise: 0.333... = 310 + 3100 + ... with r=110. Meaning: each extra term is a smaller bite, so the tail vanishes when |r|<1.

Mathematical form

a = 210, r = 110

S = 2101-110 = 15910 = 29

S = a1-r. If

a=0). A recurring decimal is a GP in disgu

Where / symbols

INFINITE GP AND RECURRING DECIMALS: If |r|<1, the infinite sum S = a / (1-r). If |r|>=1 the infinite GP does not settle to a finite school sum (except the special a=0). A recurring decimal is a GP in disguise: 0.333... = 3/10 + 3/100 + ... with r=1/10. Meaning: each extra term is a smaller bite, so the tail vanishes when |r|<1.
1. INFINITE SUM: Need |r|<1. 1+1/2+1/4+... = 2. If |r|>1 terms blow up. If r=-1/2, still |r|<1, S = a/(1-r).
2. 0.abc repeating: 0.\dot{3} = 3/10 / (1-1/10) = 1/3. Split non-repeating prefix if 0.12\dot{3} (school method: two infinites or 10^k tricks). Show the first repeating block as a.

Relationship

|r|<1: S=a1-r

Overview

When it converges, then decimals:

Infinite Sum

Need |r|<1. 1+12+14+... = 2. If |r|>1 terms blow up. If r=-12, still |r|<1, S = a1-r.

Examples a/(1-r) · First term a · Not for |r|>=1

0.abc repeating

0.3 = 3/101-110 = 13. Split non-repeating prefix if 0.12\dot{3} (school method: two infinites or 10^k tricks). Show the first repeating block as a.

Examples 0.555...=5/9 · 0.2727...=27/99 · Prefix then GP

WHAT — INFINITE GP AND RECURRING DECIMALS

RULE — 1. INFINITE SUM

Need |r|<1. 1+12+14+... = 2. If |r|>1 terms blow up. If r=-12, still |r|<1, S = a1-r.

METHOD — 2. 0.abc repeating

0.3 = 3/101-110 = 13. Split non-repeating prefix if 0.12\dot{3} (school method: two infinites or 10^k tricks). Show the first repeating block as a.

WORKED EXAMPLE

Worked: 0.222... a = 210, r = 110 S = 2101-110 = 15910 = 29 check: 29 = 0.222...

How to answer a infinite gp and recurring decimals question

Identify the given quantities, notation, and required result.
Choose the matching rule and write it before substitution.
Work one algebraic step at a time, keeping brackets and signs clear.
Check restrictions, simplify the result, and verify it where possible.

Worked example

0.222...
a = 210, r = 110
S = 2101-110 = 15910 = 29
check: 29 = 0.222...

In simple words

Infinite formula needs |r|<1. 1+2+4+... is not 11-2.

Key points

|r|<1: S=a1-r
repeating decimal -> infinite GP
|r|<1 only
geometric tail

Exam preparation

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