x = 1 + a + {a^2} + ....infty ,(a

Question

(a) Since the series are $G.P.$, therefore

$x = \frac{1}{{1 - a}}$

$\Rightarrow a = \frac{{x - 1}}{x}$ and $y = \frac{1}{{1 - b}}$

$\Rightarr

Vedclass · Accepted Answer

$\frac{{xy}}{{x + y - 1}}$

Vedclass · Answer

(a) Since the series are $G.P.$, therefore

$x = \frac{1}{{1 - a}}$

$\Rightarrow a = \frac{{x - 1}}{x}$ and $y = \frac{1}{{1 - b}}$

$\Rightarrow b = \frac{{y - 1}}{y}$

$	herefore $$1 + ab + {a^2}{b^2} + ..........\infty = \frac{1}{{1 - ab}}$

$ = \frac{1}{{1 - \frac{{x - 1}}{x}.\frac{{y - 1}}{y}}} = \frac{{xy}}{{x + y - 1}}$.

$x = 1 + a + {a^2} + ....\infty \,(a < 1)$ $y = 1 + b + {b^2}.......\infty \,(b < 1)$ Then the value of $1 + ab + {a^2}{b^2} + ..........\infty $ is

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