[Algebra]
Problem: The geometric sequence with $n$ terms $t_{1}, t_{2}, \ldots, t_{n-1}, t_{n}$ has $t_{1} t_{n}=3$. Also, the product of all $n$ terms equals 59049 (that is, $t_{1} t_{2} \cdots t_{n-1} t_{n}=59049$ ). Determine the value of $n$.

(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.)

Solution: Suppose that the first term in the geometric sequence is $t_{1}=a$ and the common ratio in the sequence is $r$.

Then the sequence, which has $n$ terms, is $a, a r, a r^{2}, a r^{3}, \ldots, a r^{n-1}$.

In general, the $k$ th term is $t_{k}=a r^{k-1}$; in particular, the $n$th term is $t_{n}=a r^{n-1}$.

Since $t_{1} t_{n}=3$, then $a \cdot a r^{n-1}=3$ or $a^{2} r^{n-1}=3$.

Since $t_{1} t_{2} \cdots t_{n-1} t_{n}=59049$, then

$$
\begin{aligned}
(a)(a r) \cdots\left(a r^{n-2}\right)\left(a r^{n-1}\right) & =59049 \\
a^{n} r r^{2} \cdots r^{n-2} r^{n-1} & =59049 \\
a^{n} r^{1+2+\cdots+(n-2)+(n-1)} & =59049 \\
a^{n} r^{\frac{1}{2}(n-1)(n)} & =59049
\end{aligned}
$$

$$
a^{n} r r^{2} \cdots r^{n-2} r^{n-1}=59049 \quad \text { (since there are } n \text { factors of } a \text { on the left side) }
$$

since $1+2+\cdots+(n-2)+(n-1)=\frac{1}{2}(n-1)(n)$.

Since $a^{2} r^{n-1}=3$, then $\left(a^{2} r^{n-1}\right)^{n}=3^{n}$ or $a^{2 n} r^{(n-1)(n)}=3^{n}$.

Since $a^{n} r^{\frac{1}{2}(n-1)(n)}=59049$, then $\left(a^{n} r^{\frac{1}{2}(n-1)(n)}\right)^{2}=59049^{2}$ or $a^{2 n} r^{(n-1)(n)}=59049^{2}$.

Since the left sides of these equations are the same, then $3^{n}=59049^{2}$.

Now

$$
59049=3(19683)=3^{2}(6561)=3^{3}(2187)=3^{4}(729)=3^{5}(243)=3^{6}(81)=3^{6} 3^{4}=3^{10}
$$

Since $59049=3^{10}$, then $59049^{2}=3^{20}$ and so $3^{n}=3^{20}$, which gives $n=20$.

Final answer: 20