If r 1,x = a + frac{a}{r} + frac{a}{r^2} + | vedclass

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(A) Given the infinite geometric series:$x = \frac{a}{1 - 1/r} = \frac{ar}{r - 1}$$y = \frac{b}{1 - (-1/r)} = \frac{b}{1 + 1/r} = \frac{br}{r + 1}$$z

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$ab/c$

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(A) Given the infinite geometric series:$x = \frac{a}{1 - 1/r} = \frac{ar}{r - 1}$$y = \frac{b}{1 - (-1/r)} = \frac{b}{1 + 1/r} = \frac{br}{r + 1}$$z = \frac{c}{1 - 1/r^2} = \frac{cr^2}{r^2 - 1}$Now,calculate the product $xy$:$xy = \left( \frac{ar}{r - 1} ight) \left( \frac{br}{r + 1} ight) = \frac{abr^2}{r^2 - 1}$Now,divide by $z$:$\frac{xy}{z} = \frac{abr^2}{r^2 - 1} \div \frac{cr^2}{r^2 - 1} = \frac{abr^2}{r^2 - 1} 	imes \frac{r^2 - 1}{cr^2} = \frac{ab}{c}$

If $r > 1$,$x = a + \frac{a}{r} + \frac{a}{r^2} + \dots \infty$,$y = b - \frac{b}{r} + \frac{b}{r^2} - \dots \infty$,and $z = c + \frac{c}{r^2} + \frac{c}{r^4} + \dots \infty$,then $\frac{xy}{z} = \dots$

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