Q83 / 125 · BITSAT 2009 · Mathematics
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 \(\frac{x y}{z}=\)
- Correct answerA\(\frac{\mathrm{ab}}{\mathrm{c}}\)
- B\(\frac{\mathrm{ac}}{\mathrm{b}}\)
- C\(\frac{\mathrm{bc}}{\mathrm{a}}\)
- D1
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
\(\begin{gathered}
z=\frac{c}{1-\frac{1}{r^2}}=\frac{cr^2}{r^2-1} \quad ...(1)\\
\therefore \quad x y=\frac{a r}{r-1} \times \frac{b r}{r+1}=\frac{a b r^2}{r^2-1} \quad ...(2)
\end{gathered}\)
Dividing (2) by (1), we get
\(\frac{\mathrm{xy}}{\mathrm{z}}=\frac{\mathrm{abr}^2}{\mathrm{r}^2-1} \times \frac{\mathrm{r}^2-1}{\mathrm{cr}^2}=\frac{\mathrm{ab}}{\mathrm{c}}\)
Hence, the correct answer is option: 'A'.