If A=1+ra+r2a+r3a+…upto ∞and B = 1 +rb+r2b+r3b+… upto ∞, thenab is equal to

$A = \frac{1}{1 - r^{a}}, B = \frac{1}{1 - r^{b}}$

$\Rightarrow 1 - r^{a} = \frac{1}{A}, 1 - r^{b} = \frac{1}{B}$

$\Rightarrow a \log r = \log (1 - 1 / A)$ and

$b \log r = \log (1 - 1 / B)$

$∴ \frac{a}{b} = \frac{\log (1 - A^{- 1})}{\log (1 - B^{- 1})}$

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