
If \[\left| r \right|>1\] and \[x=a+\frac{a}{r}+\frac{a}{{{r}^{2}}}+...\infty \], \[y=b-\frac{b}{r}+\frac{b}{{{r}^{2}}}-...\infty \] and \[z=c+\frac{c}{{{r}^{2}}}+\frac{c}{{{r}^{4}}}+...\infty \] then \[\frac{xy}{z}=\]
A. \[\frac{ab}{c}\]
B. \[\frac{ac}{b}\]
C. \[\frac{bc}{a}\]
D. \[1\]
Answer
163.2k+ views
Hint: In this question, we have to find the expression \[\frac{xy}{z}\] for the given geometric series $x,y,z$. Since the series are given to the infinity, we can use the sum of the infinite G.P formula. By these sums, we can calculate the given expression.
Formula Used: If the series is a geometric series, then the sum of the $n$ terms of the series is calculated by
${{S}_{n}}=\frac{a({{r}^{n}}-1)}{r-1}$ where $r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$- Sum of the $n$ terms of the series; $n$ - Number of terms; $a$ First term; $r$ - Common ratio.
The sum of the infinite G.P is calculated by
${{S}_{\infty }}=\frac{a}{1-r}$
Here ${{S}_{\infty }}$ - Sum of the infinite terms of the series.
Complete step by step solution: Given that, for \[\left| r \right|>1\], there are three geometric series.
They are:
\[x=a+\frac{a}{r}+\frac{a}{{{r}^{2}}}+...\infty \]
\[y=b-\frac{b}{r}+\frac{b}{{{r}^{2}}}-...\infty \]
And
\[z=c+\frac{c}{{{r}^{2}}}+\frac{c}{{{r}^{4}}}+...\infty \]
For finding the sum of the infinite G.P, the common ratio must be $\left| r \right|<1$.
So, for the given series, we have
\[\left| r \right|>1\Rightarrow\dfrac{1}{\left| r \right|}<1\]
Then, the sums of the given infinite geometric series are:
\[x=\frac{a}{1-\frac{1}{r}}=\frac{ar}{r-1}\]
\[y=\frac{b}{1-\left(\dfrac{-1}{r} \right)}=\frac{br}{r+1}\]
And
\[z=\frac{c}{1-\frac{1}{{{r}^{2}}}}=\frac{c{{r}^{2}}}{{{r}^{2}}-1}\text{ }...(1)\]
The product of $x$ and $y$ is
\[xy=\frac{ar}{r-1}\times\dfrac{br}{r+1}=\frac{ab{{r}^{2}}}{{{r}^{2}}-1}\text{ }...(2)\]
From (1) and (2), we get
\[\frac{xy}{z}=\frac{ab{{r}^{2}}}{{{r}^{2}}-1}\times\dfrac{{{r}^{2}}-1}{c{{r}^{2}}}=\frac{ab}{c}\]
Option ‘B’ is correct
Note: Here we need to remember that, the common ratio should be reciprocal. If the given series is the infinite geometric series, then to find their sums, the common ratio should be $\left| r \right|<1$. But for the given series, it is given that \[\left| r \right|>1\]. So, we need to change this by writing it in the reciprocal, we get the required condition. So, that we can use the sum to infinite G.P formula. By substituting these sums in the given expression, we can get the required values.
Formula Used: If the series is a geometric series, then the sum of the $n$ terms of the series is calculated by
${{S}_{n}}=\frac{a({{r}^{n}}-1)}{r-1}$ where $r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$- Sum of the $n$ terms of the series; $n$ - Number of terms; $a$ First term; $r$ - Common ratio.
The sum of the infinite G.P is calculated by
${{S}_{\infty }}=\frac{a}{1-r}$
Here ${{S}_{\infty }}$ - Sum of the infinite terms of the series.
Complete step by step solution: Given that, for \[\left| r \right|>1\], there are three geometric series.
They are:
\[x=a+\frac{a}{r}+\frac{a}{{{r}^{2}}}+...\infty \]
\[y=b-\frac{b}{r}+\frac{b}{{{r}^{2}}}-...\infty \]
And
\[z=c+\frac{c}{{{r}^{2}}}+\frac{c}{{{r}^{4}}}+...\infty \]
For finding the sum of the infinite G.P, the common ratio must be $\left| r \right|<1$.
So, for the given series, we have
\[\left| r \right|>1\Rightarrow\dfrac{1}{\left| r \right|}<1\]
Then, the sums of the given infinite geometric series are:
\[x=\frac{a}{1-\frac{1}{r}}=\frac{ar}{r-1}\]
\[y=\frac{b}{1-\left(\dfrac{-1}{r} \right)}=\frac{br}{r+1}\]
And
\[z=\frac{c}{1-\frac{1}{{{r}^{2}}}}=\frac{c{{r}^{2}}}{{{r}^{2}}-1}\text{ }...(1)\]
The product of $x$ and $y$ is
\[xy=\frac{ar}{r-1}\times\dfrac{br}{r+1}=\frac{ab{{r}^{2}}}{{{r}^{2}}-1}\text{ }...(2)\]
From (1) and (2), we get
\[\frac{xy}{z}=\frac{ab{{r}^{2}}}{{{r}^{2}}-1}\times\dfrac{{{r}^{2}}-1}{c{{r}^{2}}}=\frac{ab}{c}\]
Option ‘B’ is correct
Note: Here we need to remember that, the common ratio should be reciprocal. If the given series is the infinite geometric series, then to find their sums, the common ratio should be $\left| r \right|<1$. But for the given series, it is given that \[\left| r \right|>1\]. So, we need to change this by writing it in the reciprocal, we get the required condition. So, that we can use the sum to infinite G.P formula. By substituting these sums in the given expression, we can get the required values.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

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

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

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

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 Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NEET 2025 – Every New Update You Need to Know
