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

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(B) Given $x = \sum\limits_{n = 0}^\infty {{a^n}} = \frac{1}{{1 - a}}$.$\Rightarrow 1 - a = \frac{1}{x}$ $\Rightarrow a = 1 - \frac{1}{x} = \frac{x -

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$xz + yz = xy + z$

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(B) Given $x = \sum\limits_{n = 0}^\infty {{a^n}} = \frac{1}{{1 - a}}$.$\Rightarrow 1 - a = \frac{1}{x}$ $\Rightarrow a = 1 - \frac{1}{x} = \frac{x - 1}{x}$.Similarly,$y = \sum\limits_{n = 0}^\infty {{b^n}} = \frac{1}{{1 - b}} \Rightarrow b = \frac{y - 1}{y}$.And $z = \sum\limits_{n = 0}^\infty {{{(ab)}^n}} = \frac{1}{{1 - ab}}$ $\Rightarrow 1 - ab = \frac{1}{z}$ $\Rightarrow ab = 1 - \frac{1}{z} = \frac{z - 1}{z}$.Substituting the values of $a$ and $b$ in the expression for $ab$:$\left( \frac{x - 1}{x} \right) \left( \frac{y - 1}{y} \right) = \frac{z - 1}{z}$.$\frac{xy - x - y + 1}{xy} = \frac{z - 1}{z}$.$z(xy - x - y + 1) = xy(z - 1)$.$xyz - xz - yz + z = xyz - xy$.$-xz - yz + z = -xy$.$xy + z = xz + yz$.

If $x = \sum\limits_{n = 0}^\infty {{a^n}} ,\;y = \sum\limits_{n = 0}^\infty {{b^n},\;z = \sum\limits_{n = 0}^\infty {{{(ab)}^n}} } $,where $a, b < 1$,then

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