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Given are the first term, last term, and the number of terms of a geometric sequence. What is the product of its terms?
I. Solution: Let the sought product be denoted by $P_{n}$.

$$
P_{n}=a_{1} a_{2} \ldots a_{n}=a_{1} \cdot a_{1} q \cdot a_{1} q^{2} \ldots a_{1} q^{n-1}=a_{1}^{n} \cdot q^{1+2+\ldots+(n-1)}
$$

The sum of the arithmetic sequence in the exponent of $q$ is $\frac{(n-1) n}{2}$, and thus

$$
P_{n}=a_{1}^{n} q^{\frac{(n-1) n}{2}}=a_{1}^{n}\left(q^{n-1}\right)^{\frac{n}{2}}
$$

Since $a_{1} q^{n-1}=a_{n}$, we have

$$
q^{n-1}=\frac{a_{n}}{a_{1}}
$$

Substituting this value into (1) gives

$$
P_{n}=a_{1}^{n} \frac{a_{n}^{\frac{n}{2}}}{a_{1}^{\frac{n}{2}}}=a_{1}^{\frac{n}{2}} a_{n}^{\frac{n}{2}}=\left(a_{1} a_{n}\right)^{\frac{n}{2}}
$$

Remark: The expression under (1) can also be transformed as follows

$$
a_{1}^{n}\left(q^{n-1}\right)^{\frac{n}{2}}=\left(a_{1}^{2} q^{n-1}\right)^{\frac{n}{2}}=\left(a_{1} \cdot a_{1} q^{n-1}\right)^{\frac{n}{2}}=\left(a_{1} a_{n}\right)^{\frac{n}{2}}
$$

# II. Solution: 

$$
P_{n}=a_{1} \cdot a_{1} q \cdot a_{1} q^{2} \ldots a_{1} q^{k-1} \ldots a_{1} q^{n-1}
$$

writing the terms of the geometric sequence as factors in reverse order starting from $a_{n}$

$$
P_{n}=a_{n} \cdot \frac{a_{n}}{q} \cdot \frac{a_{n}}{q^{2}} \ldots \frac{a_{n}}{q^{k-1}} \ldots \frac{a_{n}}{q^{n-1}}
$$

The product of (1) and (2) can be formed by multiplying the factors that are directly below each other in pairs. Thus,

$$
P_{n}^{2}=\left(a_{1} a_{n}\right)\left(a_{1} a_{n}\right) \ldots\left(a_{1} a_{n}\right) \ldots\left(a_{1} a_{n}\right)=\left(a_{1} a_{n}\right)^{n}
$$

from which

$$
P_{n}=\sqrt{\left(a_{1} a_{n}\right)^{n}}=\left(a_{1} a_{n}\right)^{\frac{n}{2}}
$$