
The value of \[0.2\dot{3}\dot{4}\] is
A. $\frac{232}{990}$
B. $\frac{232}{9990}$
C. $\frac{232}{90}$
D. $\frac{232}{9909}$
Answer
163.8k+ 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. So, the series represents a geometric series since the next term in the series is multiplied by the previous term.
Thus, the sum of the infinite terms is calculated by
${{S}_{n}}=\frac{a}{1-r}$ where $r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$ is the sum of the n terms of the series; $n$ is the number of terms; $a$ is the first term in the series, and $r$ is the common ratio of the terms in the series.
Complete Step-by-step Solution:
The given value is \[0.2\dot{3}\dot{4}\]
Expanding the given non-terminating decimal, we get
\[0.2\dot{3}\dot{4}=0.2343434...\]
On simplifying the series,
\[\begin{align}
& 0.2\dot{3}\dot{4}=0.2343434... \\
& \text{ }=0.2+0.034+0.00034+0.0000034+... \\
& \text{ }=\frac{2}{10}+\frac{34}{1000}+\frac{34}{100000}+\frac{34}{10000000}+... \\
& \text{ }=\frac{2}{10}+34\left[ \frac{1}{{{10}^{3}}}+\frac{1}{{{10}^{5}}}+\frac{1}{{{10}^{7}}}+... \right]\text{ }...\text{(1)} \\
\end{align}\]
Since the series formed in equation (1) is a geometric series, the common ratio is
$r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
\[\begin{align}
& \Rightarrow r=\frac{{{a}_{2}}}{{{a}_{1}}} \\
& \text{ =}\frac{\frac{1}{{{10}^{5}}}}{\frac{1}{{{10}^{3}}}} \\
& \text{ =}\frac{1}{{{10}^{5}}}\times \frac{{{10}^{3}}}{1} \\
& \text{ =}\frac{1}{{{10}^{2}}} \\
\end{align}\]
Thus, the sum of the infinite terms of the obtained geometric series is calculated by the formula,
${{S}_{n}}=\frac{a}{1-r}$
On substituting $a=\frac{1}{{{10}^{3}}};r=\frac{1}{{{10}^{2}}}$, we get
\[\begin{align}
& {{S}_{\infty }}=\frac{\frac{1}{1000}}{1-\frac{1}{100}} \\
& \text{ }=\frac{\frac{1}{1000}}{\frac{100-1}{100}} \\
& \text{ }=\frac{1}{1000}\times \frac{100}{99} \\
& \text{ }=\frac{1}{990} \\
\end{align}\]
Then, substituting the obtained sum in equation (1) we get,
\[\begin{align}
& 0.2\dot{3}\dot{4}=\frac{2}{10}+34\left[ \frac{1}{{{10}^{3}}}+\frac{1}{{{10}^{5}}}+\frac{1}{{{10}^{7}}}+... \right] \\
& \text{ }=\frac{2}{10}+34\left( \frac{1}{990} \right) \\
& \text{ }=\frac{2}{10}+\frac{34}{990} \\
& \text{ }=\frac{232}{990} \\
\end{align}\]
Thus, Option (A) is the correct value.
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. So, the series represents a geometric series since the next term in the series is multiplied by the previous term.
Thus, the sum of the infinite terms is calculated by
${{S}_{n}}=\frac{a}{1-r}$ where $r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$ is the sum of the n terms of the series; $n$ is the number of terms; $a$ is the first term in the series, and $r$ is the common ratio of the terms in the series.
Complete Step-by-step Solution:
The given value is \[0.2\dot{3}\dot{4}\]
Expanding the given non-terminating decimal, we get
\[0.2\dot{3}\dot{4}=0.2343434...\]
On simplifying the series,
\[\begin{align}
& 0.2\dot{3}\dot{4}=0.2343434... \\
& \text{ }=0.2+0.034+0.00034+0.0000034+... \\
& \text{ }=\frac{2}{10}+\frac{34}{1000}+\frac{34}{100000}+\frac{34}{10000000}+... \\
& \text{ }=\frac{2}{10}+34\left[ \frac{1}{{{10}^{3}}}+\frac{1}{{{10}^{5}}}+\frac{1}{{{10}^{7}}}+... \right]\text{ }...\text{(1)} \\
\end{align}\]
Since the series formed in equation (1) is a geometric series, the common ratio is
$r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
\[\begin{align}
& \Rightarrow r=\frac{{{a}_{2}}}{{{a}_{1}}} \\
& \text{ =}\frac{\frac{1}{{{10}^{5}}}}{\frac{1}{{{10}^{3}}}} \\
& \text{ =}\frac{1}{{{10}^{5}}}\times \frac{{{10}^{3}}}{1} \\
& \text{ =}\frac{1}{{{10}^{2}}} \\
\end{align}\]
Thus, the sum of the infinite terms of the obtained geometric series is calculated by the formula,
${{S}_{n}}=\frac{a}{1-r}$
On substituting $a=\frac{1}{{{10}^{3}}};r=\frac{1}{{{10}^{2}}}$, we get
\[\begin{align}
& {{S}_{\infty }}=\frac{\frac{1}{1000}}{1-\frac{1}{100}} \\
& \text{ }=\frac{\frac{1}{1000}}{\frac{100-1}{100}} \\
& \text{ }=\frac{1}{1000}\times \frac{100}{99} \\
& \text{ }=\frac{1}{990} \\
\end{align}\]
Then, substituting the obtained sum in equation (1) we get,
\[\begin{align}
& 0.2\dot{3}\dot{4}=\frac{2}{10}+34\left[ \frac{1}{{{10}^{3}}}+\frac{1}{{{10}^{5}}}+\frac{1}{{{10}^{7}}}+... \right] \\
& \text{ }=\frac{2}{10}+34\left( \frac{1}{990} \right) \\
& \text{ }=\frac{2}{10}+\frac{34}{990} \\
& \text{ }=\frac{232}{990} \\
\end{align}\]
Thus, Option (A) is the correct value.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main Chemistry Question Paper with Answer Keys and Solutions

GFTI Colleges in India - List, Ranking & Admission Details

Other Pages
NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

JEE Advanced 2025 Notes

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks
