If r - 1 and x-a-ar-ar2- to - y-b-br-br2- to - and z-c-cr2-cr4- to - then xyz-

The correct option is B abc Given: x=a+ar+ar^2+.....+∞ #160; #160; =a1 1/r #160; #160; =arr 1 y=b br+br^2 .....+∞ #160; ...

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Standard XII

Mathematics

Geometric Progression

If r > 1 an...

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 $\frac{x y}{z} =$

\frac{a b}{c}

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\frac{a c}{b}

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\frac{b c}{a}

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None of these

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Solution

The correct option is B $\frac{a b}{c}$
Given:
$x = a + \frac{a}{r} + \frac{a}{r^{2}} + . . . . . + \infty$
$= \frac{a}{1 - \frac{1}{r}}$
$= \frac{a r}{r - 1}$

$y = b - \frac{b}{r} + \frac{b}{r^{2}} - . . . . . + \infty$
$= \frac{b}{1 + \frac{1}{r}}$
$= \frac{b r}{r + 1}$

$z = c + \frac{c}{r^{2}} + \frac{c}{r^{4}} + . . . . . + \infty$
$= \frac{c}{1 - \frac{1}{r^{2}}}$
$= \frac{c r^{2}}{r^{2} - 1}$

$∴ \frac{x y}{z} = \frac{a r}{r - 1} . \frac{b r}{r + 1} . \frac{r^{2} - 1}{c r^{2}}$
$= \frac{a b}{c}$
Hence, the answer is $\frac{a b}{c}$ .

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If r>1 and x=a+ar+ar2+...... to ∞, y=b−br+br2−... to ∞ and z=c+cr2+cr4+.... to ∞, then xyz=

The correct option is B abcGiven:x=a+ar+ar2+.....+∞ =a1−1r =arr−1y=b−br+br2−.....+∞ =b1+1r =brr+1z=c+cr2+cr4+.....+∞ =c1−1r2 =cr2r2−1∴xyz=arr−1.brr+1.r2−1cr2 =abcHence, the answer is abc.

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 $\frac{x y}{z} =$