Question

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 1

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.