Chapter 8, Problem 21CR

To determine

Find the simple fraction for the given repeating decimal.

Answer to Problem 21CR

The simple fraction of the given decimal is $\frac{4}{33}$ .

Explanation of Solution

Given:

The given repeating decimal is $0.1212$ .

Calculation:

Rewrite the given decimal.

  $\begin{array}{l}0.1212\\ =0.12+0.0012+0.000012---\end{array}$

In an geometric sequence, the ratio of any term to the previous term, called the common ratio, is constant.

Find the common ratio of any term to the previous term of the sequence $0.12+0.0012+0.000012---$ .

  $\begin{array}{l}\frac{0.0012}{0.12}=0.01\\ \frac{0.000012}{0.0012}=0.01\end{array}$

The common ratio $\left(0.01\right)$ is constant.

The sum of the infinite geometric series is $s_n=\frac{a_1}{1-r}$ , where $a_1$ is the first number and $r$ is the common ratio.

In the given series $a_1=0.12$ and $r=0.01$ .

  $\begin{array}{l}{s}_n=\frac{0.12}{1-\left(0.01\right)}\\ =\frac{0.12}{1-\left(0.01\right)}\\ =\frac{0.12}{0.99}\\ =\frac{4}{33}\end{array}$

Hence the simple fraction of the given decimal is $\frac{4}{33}$ .