A) \[\frac{xy}{x+y-1}\]
B) \[\frac{xy}{x+y+1}\]
C) \[\frac{xy}{x-y-1}\]
D) \[\frac{xy}{x-y+1}\]
Correct Answer: A
Solution :
Since the series are G.P., therefore \[x=\frac{1}{1-a}\Rightarrow a=\frac{x-1}{x}\] and \[y=\frac{1}{1-b}\Rightarrow b=\frac{y-1}{y}\] \[\therefore \]\[1+ab+{{a}^{2}}{{b}^{2}}+..........\infty =\frac{1}{1-ab}\] \[=\frac{1}{1-\frac{x-1}{x}.\frac{y-1}{y}}=\frac{xy}{x+y-1}\].
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