Sequence And Series Question 329
Question: If $ | ,r, |>1 $ and $ x=a+\frac{a}{r}+\frac{a}{r^{2}}+….to\infty $ , $ y=b-\frac{b}{r}+\frac{b}{r^{2}}-….,to\infty $ and $ z=c+\frac{c}{r^{2}}+\frac{c}{r^{4}}+….to\infty $ then $ \frac{xy}{z}= $
Options:
A) $ \frac{ab}{c} $
B) $ \frac{ac}{b} $
C) $ \frac{bc}{a} $
D) 1
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Since $ | r |>1,\frac{1}{| r |}<1 $
$ \therefore ,x=\frac{a}{1-\frac{1}{r}}=\frac{ar}{r-1} $ Similarly, $ y=\frac{b}{1-( -\frac{1}{r} )}=\frac{br}{r+1} $ and $ z=\frac{c}{1-\frac{1}{r^{2}}}=\frac{cr^{2}}{r^{2}-1} $ ?. (1)
$ \therefore xy=\frac{ar}{r-1}\times \frac{br}{r+1}=\frac{abr^{2}}{r^{2}-1} $ ?. (2) 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} $
BETA
