Recurring Ratios

Since all the ratios are equal, we can find a value for each of the ratios in terms of a and k: $\begin{aligned}\frac{r_0}{a+\frac{k}{9}}&=\frac{a}{k}&\implies&&r_0&=\frac{a^2}{k}+\frac{a}{9}&\implies&r_0=\frac{a^2}{k}+\frac{a}{9}\left(1\right)\\ \frac{r_1}{r_0+\frac{k}{9}}=\frac{r_1}{\frac{a^2}{k}+\frac{a}{9}+\frac{k}{9}}&=\frac{a}{k}&\implies&&r_1&=\frac{a^3}{k^2}+\frac{a^2}{9k}+\frac{a}{9}&\implies&r_1=\frac{a^3}{k^2}+\frac{a}{9}\left(1+\frac{a}{k}\right)\\ \frac{r_2}{r_1+\frac{k}{9}}=\frac{r_2}{\frac{a^3}{k^2}+\frac{a^2}{9k}+\frac{a}{9}+\frac{k}{9}}&=\frac{a}{k}&\implies&&r_2&=\times\frac{a^4}{k^3}+\frac{a^3}{9k^2}+\frac{a^2}{9k}+\frac{a}{9}&\implies&r_2=\frac{a^4}{k^3}+\frac{a}{9}\left(1+\frac{a}{k}+\frac{a^2}{k^2}\right)\end{aligned}$ Now we can see a pattern appearing, and write a general rule: $r_n=\frac{a^{n+2}}{k^{n+1}}+\frac{a}{9}\left(1+\frac{a}{k}+\frac{a^2}{k^2}+\dots+\frac{a^n}{k^n}\right)$ We know that as n approaches infinity, r approaches 0.148148148..., which can be written as $\frac{148}{999}$ or $\frac{4}{27}$ . Now we know that when n approaches infinity, the above formula approaches this value: $\lim_{n\to\infty}\frac{a^{n+2}}{k^{n+1}}+\frac{a}{9}\left(1+\frac{a}{k}+\frac{a^2}{k^2}+\dots+\frac{a^n}{k^n}\right)=\frac{4}{27}$ The infinite sum can be solved using the formula for the sum of an infinite geometric series, $S_\infty=\frac{A}{1-r}$ where A is the first term and r is the common ratio. We know that this converges, as $a>k$ , so $0<\frac{a}{k}<1$ $\lim_{n\to\infty}\frac{a^{n+2}}{k^{n+1}}+\frac{a}{9}\left(\frac{1}{1-\frac{a}{k}}\right)=\frac{4}{27}\\ \lim_{n\to\infty}a\times\left(\frac{a}{k}\right)^{n+1}+\left(\frac{ak}{9k-9a}\right)=\frac{4}{27}$ Since $0<\frac{a}{k}<1, \lim_{n\to\infty}\left(\frac{a}{k}\right)^{n+1}=0$ , so the other part of the limit disappears, leaving us with $\frac{ak}{9k-9a}=\frac{4}{27}$ Now we find that $k=4$ and $a=1$ satisfy this equation, so the answer is $\boxed4$