[Algebra]
Problem: Positive integers $a_{1}, a_{2}, a_{3}, \ldots$ form an arithmetic sequence. If $a_{1}=10$ and $a_{a_{2}}=100$, compute $a_{a_{a_{3}}}$.

Solution: Let $d$ be the common difference of the sequence. Then $a_{a_{2}}=a_{1}+\left(a_{2}-1\right) d=100 \Rightarrow\left(a_{2}-1\right) d=$ 90. But $a_{2}=a_{1}+d=10+d$, so $(9+d) d=90$. Solving the quadratic yields $d=-15$ or $d=6$, but the requirement that $a_{i}$ be positive for all $i$ rules out the negative value, so $d=6$ and $a_{n}=10+(n-1) \cdot 6$. Thus $a_{3}=10+2(6)=22$, and $a_{a_{3}}=a_{22}=10+21(6)=136$. Finally, $a_{a_{a_{3}}}=a_{136}=10+135(6)=\mathbf{8 2 0}$.

Final answer: 820