Question

If $x=a+\frac{a}{r}+\frac{a}{r^2}+....\infty ,y=b-\frac{b}{r}+\frac{b}{r^2}-....\infty ,andz=c+\frac{c}{r^2}+\frac{c}{r^4}+....\infty ,\left(r\right|>1,c\ne 0then\frac{xy}{z}=$
.

• A. $\frac{ab}{c}$
• B. $\frac{ac}{b}$
• C. abc
• D. $\frac{c}{ab}$

Answer 1

The correct option is A $\frac{ab}{c}$

$x=a+\frac{a}{r}+\frac{a}{r^2}+....\infty ,\Rightarrow x=\frac{a}{\left(}=\frac{ar}{r-1}y=b-\frac{b}{r}+\frac{b}{r^2}-....\infty \Rightarrow y=\frac{b}{\left(}=\frac{br}{r+1}z=c+\frac{c}{r^2}+\frac{c}{r^4}+....\infty ,\Rightarrow z=\frac{c}{\left(}=\frac{c{r}^2}{r^2-1}\Rightarrow \frac{z}{c}=\frac{r^2}{r^2-1}.....\left(1\right]xy=\frac{ar}{r-1}\times \frac{br}{r+1}=ab\frac{r^2}{r^2-1}\Rightarrow xy=ab\times \frac{z}{c}\Rightarrow \frac{xy}{z}=\frac{ab}{c}$