If $a$, $6$, and $b$, in that order, form an arithmetic sequence, compute $a+b$.
1. Let the common difference of the arithmetic sequence be \( d \).
2. In an arithmetic sequence, the terms are given by:
   - The first term: \( a \)
   - The second term: \( a + d \)
   - The third term: \( a + 2d \)
3. Given that the second term is 6, we have:
   \[
   a + d = 6
   \]
4. The third term is given by:
   \[
   b = a + 2d
   \]
5. We need to find \( a + b \). Substitute \( b \) from the above equation:
   \[
   a + b = a + (a + 2d) = 2a + 2d
   \]
6. From step 3, we know \( a + d = 6 \). Therefore, we can express \( d \) in terms of \( a \):
   \[
   d = 6 - a
   \]
7. Substitute \( d \) back into the expression for \( a + b \):
   \[
   a + b = 2a + 2(6 - a) = 2a + 12 - 2a = 12
   \]
8. Thus, we have:
   \[
   a + b = 12
   \]

The final answer is \( \boxed{12} \).