If 1-2-1-4-1-8-1-16-x and 1-3-1-9-1-27-1-81-y- then

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Standard XII

Mathematics

Method of Difference

If 1/2+1/4+...

Question

If $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . = x$ and $\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + . . . = y$ , then

x = y

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2 x + 4 y = 4

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x^{2} = y

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x + y = 0

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Solution

The correct option is D $2 x + 4 y = 4$
Sum of infinite series is $S = a / (1 - r)$
So $x = 1 / 2 * 1 / (1 - 1 / 2) = 1$
$x = 1$
$y = 1 / 3 * 1 / (1 - 1 / 3) = 1 / 2$
$y = 1 / 2$
$2 x + 4 y = 4$ is the only appropriate option

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If 12+14+18+116+...=x and 13+19+127+181+...=y, then

The correct option is D 2x+4y=4Sum of infinite series is S=a/(1−r)So x=1/2∗1/(1−1/2)=1x=1y=1/3∗1/(1−1/3)=1/2y=1/22x+4y=4 is the only appropriate option

If $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . = x$ and $\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + . . . = y$ , then

The correct option is D $2 x + 4 y = 4$
Sum of infinite series is $S = a / (1 - r)$
So $x = 1 / 2 * 1 / (1 - 1 / 2) = 1$
$x = 1$
$y = 1 / 3 * 1 / (1 - 1 / 3) = 1 / 2$
$y = 1 / 2$
$2 x + 4 y = 4$ is the only appropriate option