The base $4$ repeating decimal $0.\overline{12}_4$ can be expressed in the form $\frac{a}{b}$ in base 10, where $a$ and $b$ are relatively prime positive integers. Compute the sum of $a$ and $b$.

[i]2020 CCA Math Bonanza Team Round #2[/i]
1. First, let's express the repeating decimal \(0.\overline{12}_4\) as a geometric series. In base 4, the repeating decimal \(0.\overline{12}_4\) can be written as:
   \[
   0.\overline{12}_4 = 0.121212\ldots_4
   \]

2. We can express this as an infinite series:
   \[
   0.\overline{12}_4 = \frac{1}{4} + \frac{2}{4^2} + \frac{1}{4^3} + \frac{2}{4^4} + \frac{1}{4^5} + \frac{2}{4^6} + \cdots
   \]

3. Notice that this series can be split into two separate geometric series:
   \[
   S = \left(\frac{1}{4} + \frac{1}{4^3} + \frac{1}{4^5} + \cdots\right) + \left(\frac{2}{4^2} + \frac{2}{4^4} + \frac{2}{4^6} + \cdots\right)
   \]

4. Each of these series is a geometric series. The first series has the first term \(a_1 = \frac{1}{4}\) and common ratio \(r = \frac{1}{4^2} = \frac{1}{16}\):
   \[
   \sum_{n=0}^{\infty} \left(\frac{1}{4}\right) \left(\frac{1}{16}\right)^n = \frac{\frac{1}{4}}{1 - \frac{1}{16}} = \frac{\frac{1}{4}}{\frac{15}{16}} = \frac{1}{4} \cdot \frac{16}{15} = \frac{4}{15}
   \]

5. The second series has the first term \(a_2 = \frac{2}{4^2} = \frac{2}{16} = \frac{1}{8}\) and common ratio \(r = \frac{1}{16}\):
   \[
   \sum_{n=0}^{\infty} \left(\frac{1}{8}\right) \left(\frac{1}{16}\right)^n = \frac{\frac{1}{8}}{1 - \frac{1}{16}} = \frac{\frac{1}{8}}{\frac{15}{16}} = \frac{1}{8} \cdot \frac{16}{15} = \frac{2}{15}
   \]

6. Adding these two sums together, we get:
   \[
   S = \frac{4}{15} + \frac{2}{15} = \frac{6}{15}
   \]

7. Simplify the fraction \(\frac{6}{15}\):
   \[
   \frac{6}{15} = \frac{2}{5}
   \]

8. The fraction \(\frac{2}{5}\) is already in its simplest form, where \(a = 2\) and \(b = 5\). The sum of \(a\) and \(b\) is:
   \[
   a + b = 2 + 5 = 7
   \]

The final answer is \(\boxed{7}\).