The first term of a sequence is equal to $3^{2012}$. Each subsequent term is equal to the sum of the digits of the previous term. Find the 10th term.
From the correction of the previous exercise, we know that all terms are divisible by 9. On the other hand, $a_{1}=3^{2012}=9^{1006}<10^{1006}$, so the number of digits of $a_{1}$ is less than 1005, and since each digit is less than 10, we have $a_{2} \leq 9 \cdot 1005 < 10000$. Similarly, $a_{3}<40$, $a_{4}<13$, and $a_{5}<10$. Therefore, from $a_{5}$ onwards, the sequence is constant and equal to 9.