If \(| r| > \) land \(x=a+\frac{a}{r}+\frac{a}{r^2}+\ldots\) to \(\infty\),…
If \(| r| > \) land \(x=a+\frac{a}{r}+\frac{a}{r^2}+\ldots\) to \(\infty\),
\(\mathrm{y}=\mathrm{b}-\frac{\mathrm{b}}{\mathrm{r}}+\frac{\mathrm{b}}{\mathrm{r}^2}-\ldots . \text { to } \infty\)
and \(z=c+\frac{c}{r^2}+\frac{c}{r^4}+\ldots\) to \(\infty\), then \(\frac{x y}{z}=\)
\(\frac{\mathrm{ab}}{\mathrm{c}}\)
\(\frac{\mathrm{ac}}{\mathrm{b}}\)
\(\frac{\mathrm{bc}}{\mathrm{a}}\)
1
Solution
Since $|r| > 1$, $\frac{1}{|r|} < 1$
$\therefore x=\frac{a}{1-\frac{1}{r}}=\frac{ar}{r-1}$
Similarly, $y=\frac{b}{1-\left(-\frac{1}{r}\right)}=\frac{br}{r+1}$ and
$\begin{aligned}
z&=\frac{c}{1-\frac{1}{r^2}}=\frac{cr^2}{r^2-1}\quad ...(1)\\
\therefore x y&=\frac{ar}{r-1} \times \frac{br}{r+1}=\frac{abr^2}{r^2-1} \quad ...(2)
\end{aligned}$
Dividing (2) by (1), we get
$\frac{xy}{z}=\frac{abr^2}{r^2-1} \times \frac{r^2-1}{cr^2}=\frac{ab}{c}$