
The value of \[0.0\overline{37}\] where \[0.0\overline{37}\] stands for $0.0373737...$ is
A. \[\dfrac{37}{1000}\]
B. \[\dfrac{37}{990}\]
C. \[\dfrac{1}{37}\]
D. \[\dfrac{1}{27}\]
Answer
164.1k+ views
Hint: In this question, we are to find the sum of the repeated terms in the given non-terminating decimal number. By writing the non-terminating decimal number in the form of a series, we are able to find the sum of the series using the appropriate formula, and then adding with the remaining non-repeating term we get the required value
Formula used: Here the non-terminating values will form a series by writing them repeatedly. The obtained series represents a geometric series. Thus, the sum of the infinite terms is calculated by
${{S}_{\infty }}=\dfrac{a}{1-r}$ where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{\infty }}$- Sum of infinite terms of the series; $a$ - The first term and $r$ - The common ratio.
Complete step by step solution: The given value is \[0.0\overline{37}\]
On simplifying the series,
\[\begin{align}
& 0.0\overline{37}=0.0373737 \\
& \text{ }=0.037+0.00037+0.0000037+...\infty \\
& \text{ }=\dfrac{37}{{{10}^{3}}}+\dfrac{37}{{{10}^{5}}}+\dfrac{37}{{{10}^{7}}}+...\infty \\
& \text{ }=\dfrac{37}{{{10}^{3}}}\left[ 1+\dfrac{1}{{{10}^{2}}}+\dfrac{1}{{{10}^{4}}}+...\infty \right]\text{ }...\text{(1)} \\
\end{align}\]
Since the series formed in equation (1) is a geometric series, the common ratio is
$r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
\[\begin{align}
& \Rightarrow r=\dfrac{{{a}_{2}}}{{{a}_{1}}} \\
& \text{ =}\dfrac{\dfrac{1}{{{10}^{2}}}}{1} \\
& \text{ =}\dfrac{1}{100} \\
\end{align}\]
Thus, the sum of the infinite terms of the obtained geometric series is calculated by the formula,
${{S}_{\infty }}=\dfrac{a}{1-r}$
On substituting $a=1;r=\dfrac{1}{{{10}^{2}}}$, we get
\[\begin{align}
& {{S}_{\infty }}=\dfrac{1}{1-\dfrac{1}{100}} \\
& \text{ }=\dfrac{1}{\dfrac{99}{100}} \\
& \text{ }=\dfrac{100}{99} \\
\end{align}\]
Then, substituting the obtained sum in equation (1) we get,
\[\begin{align}
& 0.0\overline{37}=0.0373737...\infty \\
& \text{ }=\dfrac{37}{1000}\left( \dfrac{100}{99} \right) \\
& \text{ }=\dfrac{37}{990} \\
\end{align}\]
Thus, Option (D) is correct.
Note: Here the given value is a non-terminating decimal number. So, on expanding the decimal, a geometric series is formed. So, we can find the common ratio from this easily and we can use it for finding the sum of infinite terms. On substituting these values in the expansion of the given decimal we get the required value of the decimal in the rational form.
Formula used: Here the non-terminating values will form a series by writing them repeatedly. The obtained series represents a geometric series. Thus, the sum of the infinite terms is calculated by
${{S}_{\infty }}=\dfrac{a}{1-r}$ where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{\infty }}$- Sum of infinite terms of the series; $a$ - The first term and $r$ - The common ratio.
Complete step by step solution: The given value is \[0.0\overline{37}\]
On simplifying the series,
\[\begin{align}
& 0.0\overline{37}=0.0373737 \\
& \text{ }=0.037+0.00037+0.0000037+...\infty \\
& \text{ }=\dfrac{37}{{{10}^{3}}}+\dfrac{37}{{{10}^{5}}}+\dfrac{37}{{{10}^{7}}}+...\infty \\
& \text{ }=\dfrac{37}{{{10}^{3}}}\left[ 1+\dfrac{1}{{{10}^{2}}}+\dfrac{1}{{{10}^{4}}}+...\infty \right]\text{ }...\text{(1)} \\
\end{align}\]
Since the series formed in equation (1) is a geometric series, the common ratio is
$r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
\[\begin{align}
& \Rightarrow r=\dfrac{{{a}_{2}}}{{{a}_{1}}} \\
& \text{ =}\dfrac{\dfrac{1}{{{10}^{2}}}}{1} \\
& \text{ =}\dfrac{1}{100} \\
\end{align}\]
Thus, the sum of the infinite terms of the obtained geometric series is calculated by the formula,
${{S}_{\infty }}=\dfrac{a}{1-r}$
On substituting $a=1;r=\dfrac{1}{{{10}^{2}}}$, we get
\[\begin{align}
& {{S}_{\infty }}=\dfrac{1}{1-\dfrac{1}{100}} \\
& \text{ }=\dfrac{1}{\dfrac{99}{100}} \\
& \text{ }=\dfrac{100}{99} \\
\end{align}\]
Then, substituting the obtained sum in equation (1) we get,
\[\begin{align}
& 0.0\overline{37}=0.0373737...\infty \\
& \text{ }=\dfrac{37}{1000}\left( \dfrac{100}{99} \right) \\
& \text{ }=\dfrac{37}{990} \\
\end{align}\]
Thus, Option (D) is correct.
Note: Here the given value is a non-terminating decimal number. So, on expanding the decimal, a geometric series is formed. So, we can find the common ratio from this easily and we can use it for finding the sum of infinite terms. On substituting these values in the expansion of the given decimal we get the required value of the decimal in the rational form.
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