If A=1+ra+r2a+r3a+……∞ and B=1+rb+r2b+……∞ then ab=

A=11-ra ⇒ 1-ra=1A ⇒ ra=1-1A=A-1A B=11-rb ⇒ rb=B-1B a=logrA-1A b=logrB-1B ab=logB-1BA-1A

If A=1+ra+r2a+r3a+……∞ and B=1+rb+r2b+……∞ then ab=

A=11-ra ⇒ 1-ra=1A ⇒ ra=1-1A=A-1A B=11-rb ⇒ rb=B-1B a=logrA-1A b=logrB-1B ab=logB-1BA-1A

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If $A = 1 + r^{a} + r^{2 a} + r^{3 a} + \dots \dots \infty$ and $B = 1 + r^{b} + r^{2 b} + \dots \dots \infty then \frac{a}{b} =$

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$\log_{(1 - B)}^{(1 - A)}$

$\log_{(\frac{B - 1}{B})}^{(\frac{A - 1}{A})}$

$\log_{B}^{A}$

$\log_{A}^{B}$

answer is B.

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Detailed Solution

$A = \frac{1}{1 - r^{a}} \Rightarrow 1 - r^{a} = \frac{1}{A} \Rightarrow r^{a} = 1 - \frac{1}{A} = \frac{A - 1}{A} B = \frac{1}{1 - r^{b}} \Rightarrow r^{b} = \frac{B - 1}{B} a = \log_{r} (\frac{A - 1}{A}) b = \log_{r} (\frac{B - 1}{B}) \frac{a}{b} = \log_{(\frac{B - 1}{B})}^{(\frac{A - 1}{A})}$