Dear experts, please solve this :
If A = 1+r^a +r^2a +r^3a + ..... infinity and B = 1 +r^b +r^2b + r^3b + ..... infinity, then find a/b (where |r|<1).

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{If} \left|\mathrm{Common} \mathrm{ratio}\right|<1, \mathrm{then} \mathrm{sum} \mathrm{of} \mathrm{infinite} \mathrm{terms} \mathrm{of} \mathrm{G}.\mathrm{P}. \mathrm{is} \\ \mathrm{S}=\frac{\mathrm{First} \mathrm{Term}}{1-\mathrm{Common} \mathrm{Ratio}} \\ \mathrm{A}=1+{\mathrm{r}}^{\mathrm{a}}+{\mathrm{r}}^{2\mathrm{a}}+{\mathrm{r}}^{3\mathrm{a}}+... \\ \mathrm{Common} \mathrm{ratio}=\frac{{\mathrm{r}}^{\mathrm{a}}}{1}={\mathrm{r}}^{\mathrm{b}} \\ \mathrm{A}=\frac{1}{1-{\mathrm{r}}^{\mathrm{a}}} \\ \Rightarrow 1-{\mathrm{r}}^{\mathrm{a}}=\frac{1}{\mathrm{A}} \\ \Rightarrow {\mathrm{r}}^{\mathrm{a}}=1-\frac{1}{\mathrm{A}} \\ \Rightarrow {\mathrm{r}}^{\mathrm{a}}=\frac{\mathrm{A}-1}{\mathrm{A}} \\ \mathrm{Take} \log \mathrm{on} \mathrm{both} \\ \log \left({\mathrm{r}}^{\mathrm{a}}\right)=\log \left(\frac{\mathrm{A}-1}{\mathrm{A}}\right) \\ \mathrm{As} \log \left({\mathrm{m}}^{\mathrm{n}}\right)=\mathrm{nlog}\left(\mathrm{m}\right) \\ \Rightarrow \mathrm{alog}\left(\mathrm{r}\right)=\log \left(\frac{\mathrm{A}-1}{\mathrm{A}}\right)....\left(\mathrm{i}\right) \\ \mathrm{Similarly} \\ \mathrm{blog}\left(\mathrm{r}\right)=\log \left(\frac{\mathrm{B}-1}{\mathrm{B}}\right)....\left(\mathrm{ii}\right) \\ \frac{\left(\mathrm{i}\right)}{\left(\mathrm{ii}\right)} \\ \Rightarrow \frac{\mathrm{alog}\left(\mathrm{r}\right)}{\mathrm{blog}\left(\mathrm{r}\right)}=\frac{\log \left(\frac{\mathrm{A}-1}{\mathrm{A}}\right)}{\log \left(\frac{\mathrm{B}-1}{\mathrm{B}}\right)} \\ \mathbf{\Rightarrow }\frac{\mathbf{a}}{\mathbf{b}}\mathbf{=}\frac{\mathbf{log}\left(\frac{\mathbf{A}\mathbf{-}\mathbf{1}}{\mathbf{A}}\right)}{\mathbf{log}\left(\frac{\mathbf{B}\mathbf{-}\mathbf{1}}{\mathbf{B}}\right)} \end{gathered}$$

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