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Q.

If x=1+a+a2+…(|a|<1)and y=1+b+b2+…(|b|<1) then some of the series 1+ab+a2b2+…is

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a

xyx−y−1

b

xyx−y+1

c

xyx+y+1

d

xyx+y−1

answer is D.

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Detailed Solution

x=1+a+a2+⋯=11−ay=11−b⇒a=1−1/x,b=1−1/yNow 1+ab+a2b2+…=11−ab=11−(1−1/x)(1−1/y)=xyx+y−1
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