If x = sum_{n=0}^{infty} a^{n},y = sum_{n=0}^ | vedclass

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(B) Given $x = \sum_{n=0}^{\infty} a^{n} = \frac{1}{1-a}$,$y = \frac{1}{1-b}$,and $z = \frac{1}{1-c}$.From these,we get $a = 1 - \frac{1}{x}$,$b = 1 -

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$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$

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(B) Given $x = \sum_{n=0}^{\infty} a^{n} = \frac{1}{1-a}$,$y = \frac{1}{1-b}$,and $z = \frac{1}{1-c}$.From these,we get $a = 1 - \frac{1}{x}$,$b = 1 - \frac{1}{y}$,and $c = 1 - \frac{1}{z}$.Since $a, b, c$ are in $A.P.$,we have $2b = a + c$.Substituting the expressions in terms of $x, y, z$:$2(1 - \frac{1}{y}) = (1 - \frac{1}{x}) + (1 - \frac{1}{z})$$2 - \frac{2}{y} = 2 - (\frac{1}{x} + \frac{1}{z})$$\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$.This condition implies that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$

If $x = \sum_{n=0}^{\infty} a^{n}$,$y = \sum_{n=0}^{\infty} b^{n}$,$z = \sum_{n=0}^{\infty} c^{n}$,where $a, b, c$ are in $A.P.$ and $|a| < 1, |b| < 1, |c| < 1$,$abc \neq 0$,then:

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