If (|r| ) land (x=a+ (a)/(r)+(a)/(r^2)+ to ( ∞ , ( y=b- b r + b r^2 - . to ∞ and (z=c+ (c)/(r^2)+(c)/(r^4)+ to ( ∞ , then ( (x y)/(z)=

Correct answer: A. ( ab c ) Since (|r| >1, 1 |r| <1) ( x= a 1- 1 r = ar r-1 Similarly, \ y=(b)/(1-(-(1)/(r)))=(b r)/(r+1) and gathered z=(c)/(1-(1)/(r^2))=(cr^2)/(r^2-1) ...(1) x y=(a r)/(r-1) × (b r)/(r+1)=(a b r²)/(r^2-1) ...(2) gathered ) Dividing (2) by (1), we get ( xy z = abr^2 r^2-1 × r^2-1 cr^2 = ab c )

BITSAT 2009

Exam: BITSAT
Year: 2009
Subject: Mathematics
Chapter: Circle

BITSAT 2009 · Mathematics · Circle

If (|r| ) land (x=a+ (a)/(r)+(a)/(r^2)+ to ( ∞ , ( y=b- b r + b r^2...

If (|r| \gt) land (x=a+

\frac{a}{r}+\frac{a}{r^2}+\ldots

to (

\infty

, (

\mathrm{y}=\mathrm{b}-\frac{\mathrm{b}}{\mathrm{r}}+\frac{\mathrm{b}}{\mathrm{r}^2}-\ldots . \text{

} \infty

and (z=c+

\frac{c}{r^2}+\frac{c}{r^4}+\ldots

to (

\infty

, then (

\frac{x y}{z}=

Options

A( $\frac{\mathrm{ab}}{\mathrm{c}}$ )
B( $\frac{\mathrm{ac}}{\mathrm{b}}$ )
C( $\frac{\mathrm{bc}}{\mathrm{a}}$ )
D1

Correct answer

A. ( $\frac{\mathrm{ab}}{\mathrm{c}}$ )

Step-by-step solution

Since (|\mathrm{r}| >1,

\frac{1}{|\mathrm{r}|}

<1) (\therefore \mathrm{x}=\frac{\mathrm{a}}{1-\frac{1}{\mathrm{r}}}=\frac{\mathrm{ar}}{\mathrm{r}-1}

Similarly, \$y=\frac{b}{1-\left(-\frac{1}{r}\right)}=\frac{b r}{r+1}

and

\begin{gathered}z=\frac{c}{1-\frac{1}{r^2}}=\frac{cr^2}{r^2-1} \quad ...(1)\\ \therefore \quad x y=\frac{a r}{r-1} \times \frac{b r}{r+1}=\frac{a b r²}{r^2-1} \quad ...(2) \end{gathered}) Dividing (2) by (1), we get (\frac{\mathrm{xy}}{\mathrm{z}}=\frac{\mathrm{abr}^2}{\mathrm{r}^2-1} \times \frac{\mathrm{r}^2-1}{\mathrm{cr}^2}=\frac{\mathrm{ab}}{\mathrm{c}})

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