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If A=1+ra+r2a+r3a+ and If  B=1+rb+r2b+r3b+,then ab is equal to 

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a
logB⁡A
b
log1−B⁡(1−A)
c
logB−1B⁡A−1A
d
None of these

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detailed solution

Correct option is C

A=11−ra⇒1−ra=1A⇒ra=1−1A=A−1AB=11−rb⇒1−rb=1B⇒rb=1−1B=B−1B∴ alog⁡r=log⁡A−1A and blog⁡r=log⁡B−1Bab=log⁡A−1Alog⁡B−1B=logB−1B⁡A−1A


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