Chapter 12.3, Problem 64E

To determine

Express the given repeating decimal $0.2\overline{53}\mathrm{.....}$ as a fraction

Answer to Problem 64E

The given repeating decimal $0.2\overline{53}\mathrm{.....}$ as a fraction is $\frac{251}{990}$ .

Explanation of Solution

Given information:

It is given in the question that, the repeating decimal $0.2\overline{53}\mathrm{.....}$

Formula used:

  $S=\frac{a}{1-r}\&a_n=a{r}^{n-1}$

Where,

  $\begin{array}{l}\begin{array}{l}S=sumofinfiniteterm\hfill \\ a=firstterm\hfill \\ r=commonratio\hfill \end{array}\\ a_{n}isthenthterm\\ n=numberofterm\end{array}$

Calculation:

The given repeating decimal $0.2\overline{53}\mathrm{.....}$ is written in the following sequence,

  $\frac{2}{10}+\frac{53}{1000}+\frac{53}{100000}+\mathrm{.......}$

  $\Rightarrow \frac{2}{10}\&\frac{53}{1000}+\frac{53}{100000}+\mathrm{.......}$

So from the above sequence, $\frac{53}{1000}+\frac{53}{100000}+\mathrm{......}$

  $\begin{array}{l}{a}_1=\frac{53}{1000}\Rightarrow a=\frac{53}{1000}\\ a_2=\frac{53}{100000}\Rightarrow ar=\frac{53}{100000}\Rightarrow r=\frac{\frac{53}{100000}}{\frac{53}{1000}}\Rightarrow r=\frac{53}{100000}*\frac{1000}{53}\Rightarrow r=\frac{1}{100}\end{array}$

  $a=\frac{53}{1000}\&r=\frac{1}{100}$

By using formula $S=\frac{a}{1-r}$

  $\begin{array}{l}S=\frac{\frac{53}{1000}}{1-\left(\frac{1}{100}\right)}\\ =\frac{\frac{53}{1000}}{1-\left(\frac{1}{100}\right)}\\ =\frac{\frac{53}{1000}}{\left(\frac{100-1}{100}\right)}\\ =\frac{53}{1000}*\frac{100}{99}\\ =\frac{53}{990}\end{array}$

So the final sum is

  $\begin{array}{l}\frac{2}{10}+\frac{53}{990}\\ \frac{53+198}{990}\\ \frac{251}{990}\end{array}$