
Express the value of \[0.14189189189.....\] as a rational number
A. \[\frac{7}{3700}\]
B. \[\frac{7}{50}\]
C. \[\frac{525}{111}\]
D. \[\frac{21}{148}\]
Answer
161.1k+ 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. 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
${{S}_{\infty }}=\frac{a}{1-r}$ where $r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{\infty }}$ is the sum of infinite terms of the series; $a$ 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 \[0.14189189189.....\]
On simplifying the series,
\[\begin{align}
& 0.14189189189...=0.14+0.00189+0.00000189+...\infty \\
& \text{ }=\frac{14}{100}+\frac{189}{{{10}^{5}}}+\frac{189}{{{10}^{8}}}+...\infty \\
& \text{ }=\frac{7}{50}+189\left[ \frac{1}{{{10}^{5}}}+\frac{1}{{{10}^{8}}}+...\infty \right]\text{ }...\text{(1)} \\
\end{align}\]
Since the series formed in equation (1) is a geometric series, the common ratio is
$r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
\[\begin{align}
& \Rightarrow r=\frac{{{a}_{2}}}{{{a}_{1}}} \\
& \text{ =}\frac{\frac{1}{{{10}^{8}}}}{\frac{1}{{{10}^{5}}}} \\
& \text{ =}\frac{1}{{{10}^{8}}}\times \frac{{{10}^{5}}}{1} \\
& \text{ =}\frac{1}{{{10}^{3}}} \\
\end{align}\]
Thus, the sum of the infinite terms of the obtained geometric series is calculated by the formula,
${{S}_{\infty }}=\frac{a}{1-r}$
On substituting $a=\frac{1}{{{10}^{5}}};r=\frac{1}{{{10}^{3}}}$, we get
\[\begin{align}
& {{S}_{\infty }}=\frac{\frac{1}{{{10}^{5}}}}{1-\frac{1}{{{10}^{3}}}} \\
& \text{ }=\frac{\frac{1}{{{10}^{5}}}}{\frac{999}{{{10}^{3}}}} \\
& \text{ }=\frac{1}{999\times 100} \\
\end{align}\]
Then, substituting the obtained sum in equation (1) we get,
\[\begin{align}
& 0.14189189...=\frac{7}{50}+189\left[ \frac{1}{999\times 100} \right] \\
& \text{ }=\frac{7}{50}+\frac{189}{999\times 100} \\
& \text{ }=\frac{7}{50}+\frac{7}{3700} \\
& \text{ }=\frac{525}{3700}=\frac{21}{148} \\
\end{align}\]
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. 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
${{S}_{\infty }}=\frac{a}{1-r}$ where $r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{\infty }}$ is the sum of infinite terms of the series; $a$ 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 \[0.14189189189.....\]
On simplifying the series,
\[\begin{align}
& 0.14189189189...=0.14+0.00189+0.00000189+...\infty \\
& \text{ }=\frac{14}{100}+\frac{189}{{{10}^{5}}}+\frac{189}{{{10}^{8}}}+...\infty \\
& \text{ }=\frac{7}{50}+189\left[ \frac{1}{{{10}^{5}}}+\frac{1}{{{10}^{8}}}+...\infty \right]\text{ }...\text{(1)} \\
\end{align}\]
Since the series formed in equation (1) is a geometric series, the common ratio is
$r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
\[\begin{align}
& \Rightarrow r=\frac{{{a}_{2}}}{{{a}_{1}}} \\
& \text{ =}\frac{\frac{1}{{{10}^{8}}}}{\frac{1}{{{10}^{5}}}} \\
& \text{ =}\frac{1}{{{10}^{8}}}\times \frac{{{10}^{5}}}{1} \\
& \text{ =}\frac{1}{{{10}^{3}}} \\
\end{align}\]
Thus, the sum of the infinite terms of the obtained geometric series is calculated by the formula,
${{S}_{\infty }}=\frac{a}{1-r}$
On substituting $a=\frac{1}{{{10}^{5}}};r=\frac{1}{{{10}^{3}}}$, we get
\[\begin{align}
& {{S}_{\infty }}=\frac{\frac{1}{{{10}^{5}}}}{1-\frac{1}{{{10}^{3}}}} \\
& \text{ }=\frac{\frac{1}{{{10}^{5}}}}{\frac{999}{{{10}^{3}}}} \\
& \text{ }=\frac{1}{999\times 100} \\
\end{align}\]
Then, substituting the obtained sum in equation (1) we get,
\[\begin{align}
& 0.14189189...=\frac{7}{50}+189\left[ \frac{1}{999\times 100} \right] \\
& \text{ }=\frac{7}{50}+\frac{189}{999\times 100} \\
& \text{ }=\frac{7}{50}+\frac{7}{3700} \\
& \text{ }=\frac{525}{3700}=\frac{21}{148} \\
\end{align}\]
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.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
