Find the rational number whose decimal expansion is given below :

Question:

Find the rational number whose decimal expansion is given below :

(i) $0 . \overline{3}$

(ii) $0 . \overline{231}$

(iii) $3 . \overline{52}$

Solution:

(i) Let, x=0.3333…

$\Rightarrow x=0.3+0.03+0.003+\ldots$

$\Rightarrow x=3(0.1+0.01+0.001+0.0001+\ldots \infty)$

$\Rightarrow x=3\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}+\cdots \infty\right)$

This is an infinite geometric series.

Here, a=1/10 and r=1/10

$\therefore \operatorname{Sum}=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{\frac{1}{10}}{1-\frac{1}{10}}=\frac{1 \times 10}{9 \times 10}=\frac{1}{9}$

$\therefore \mathrm{x}^{=3 \times \frac{1}{9}=\frac{1}{3}}$

$0 . \overline{3}=\frac{1}{-}$

(ii) Let, x=0.231231231….

$\Rightarrow \mathrm{x}=0.231+0.000231+0.000000231+\ldots \infty$

$\Rightarrow \mathrm{x}=231(0.001+0.000001+0.000000001+\ldots \infty)$

$\Rightarrow x=231\left(\frac{1}{10^{3}}+\frac{1}{10^{6}}+\frac{1}{10^{9}}+\frac{1}{10^{12}}+\ldots^{\infty}\right)$

This is an infinite geometric series.

Here, $a^{=\frac{1}{10^{3}}}$ and $r=\frac{1}{10^{3}}$

$\therefore$ Sum $=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{\frac{1}{10^{3}}}{1-\frac{1}{10^{3}}}=\frac{1 \times 1000}{999 \times 1000}=\frac{1}{999}$

$\Rightarrow \mathrm{X}=231 \times \frac{1}{999}=\frac{231}{999}$

$0 . \overline{231}_{=} \frac{231}{999}$

(iii) Let, x=3.525252552…

⇒ x=3+0.52+0.0052+0.000052+…∞

$\Rightarrow x=3+52(0.01+0.0001+\ldots \infty)$

$\Rightarrow x=3+52\left(\frac{1}{10^{2}}+\frac{1}{10^{4}}+\frac{1}{10^{6}}+\frac{1}{10^{8}}+\ldots \infty\right)$

Here, $a=\frac{1}{10^{2}}$ and $r=\frac{1}{10^{2}}$

$\therefore$ Sum $=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{\frac{1}{10^{2}}}{1-\frac{1}{10^{2}}}=\frac{1 \times 100}{99 \times 100}=\frac{1}{99}$

$\Rightarrow \mathrm{X}=3+\left(52 \times \frac{1}{99}\right)=\frac{297+52}{999}=\frac{349}{999}$

$3 . \overline{52}=\frac{349}{999}$

 

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