[Algebra]
Problem: The sequence $a_{1}, a_{2}, a_{3}, \ldots$ is a geometric sequence with $a_{20}=8$ and $a_{14}=2^{21}$. Compute $a_{21}$.

Solution: Let $r$ be the common ratio of the sequence. Then $a_{20}=r^{20-14} \cdot a_{14}$, hence $8=r^{6} \cdot 2^{21} \Rightarrow r^{6}=$ $\frac{2^{3}}{2^{21}}=2^{-18}$, so $r=2^{-3}=\frac{1}{8}$. Thus $a_{21}=r \cdot a_{20}=\frac{1}{8} \cdot 8=\mathbf{1}$.

Final answer: 1