(G) Sum to Infinity of Geometric Progressions (Part 2)

Posted on May 24, 2020 by Myhometuition

(H) Recurring Decimal

Example of recurring decimal:

$\begin{array}{l} \frac{2}{9} = 0.2222222222222..... \\ \frac{8}{33} = 0.242424242424..... \\ \frac{41}{333} = 0.123123123123..... \end{array}$

Recurring decimal can be changed to fraction using the sum to infinity formula:

$S \infty = \frac{a}{1 - r}$

Example (Change recurring decimal to fraction)

Express each of the following recurring decimals as a fraction in its lowest terms.

(a) 0.8888 ...

(b) 0.171717...

(c) 0.513513513 ….

Solution:

(a)

0.8888 = 0.8 + 0.08 + 0.008 +0.0008 + ….. (recurring decimal)

$\begin{array}{l} G P, a = 0.8, r = \frac{0.08}{0.8} = 0.1 \\ S_{\infty} = \frac{a}{1 - r} \\ S_{\infty} = \frac{0.8}{1 - 0.1} \\ S_{\infty} = \frac{0.8}{0.9} \\ S_{\infty} = \frac{8}{9} \to \begin{array}{l} check using calculator \\ \frac{8}{9} = 0.888888.... \end{array} \end{array}$

(b)

0.17171717 …..

= 0.17 + 0.0017 + 0.000017 + 0.00000017 + …..

$\begin{array}{l} G P, a = 0.17, r = \frac{0.0017}{0.17} = 0.01 \\ S_{\infty} = \frac{a}{1 - r} \\ S_{\infty} = \frac{0.17}{1 - 0.01} = \frac{0.17}{0.99} = \frac{17}{99} \to \begin{array}{l} remember to check the \\ answer using calculator \end{array} \end{array}$

(c)

0.513513513…..

= 0.513 + 0.000513 + 0.000000513 + …..

$\begin{array}{l} G P, a = 0.513, r = \frac{0.00513}{0.513} = 0.001 \\ S_{\infty} = \frac{a}{1 - r} \\ S_{\infty} = \frac{0.513}{1 - 0.001} = \frac{0.513}{0.999} = \frac{513}{999} = \frac{19}{37} \end{array}$