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

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

1. A

$\frac{\mathrm{log}\left(1+A\right)}{\mathrm{log}\left(1+B\right)}$

2. B

$\frac{\mathrm{log}\left(A-1\right)}{\mathrm{log}\left(B-1\right)}$

3. C

${\mathrm{log}}_{B}A$

4. D

none of these

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### Solution:

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

$⇒a\mathrm{log}r=\mathrm{log}\left(1-1/A\right)$ and

$b\mathrm{log}r=\mathrm{log}\left(1-1/B\right)$

$\therefore \frac{a}{b}=\frac{\mathrm{log}\left(1-{A}^{-1}\right)}{\mathrm{log}\left(1-{B}^{-1}\right)}$

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