4. In the Cartesian coordinate system $x O y$, line $l$ passes through the origin, $\bar{n}=(3,1)$ is a normal vector of $l$. It is known that the sequence $\left\{a_{n}\right\}$ satisfies: for any positive integer $n$, the point $\left(a_{n+1}, a_{n}\right)$ is on $l$. If $a_{2}=6$, then the value of $a_{1} a_{2} a_{3} a_{4} a_{5}$ is $\qquad$
Answer: -32.
Solution: It is easy to know that the equation of line $l$ is $3x + y = 0$. Therefore, for any positive integer $n$, we have $3a_{n+1} + a_n = 0$, which means $a_{n+1} = -\frac{1}{3}a_n$, so $\{a_n\}$ is a geometric sequence with a common ratio of $-\frac{1}{3}$. Thus, $a_3 = -\frac{1}{3}a_2 = -2$. By the property of geometric sequences, we get $a_1 a_2 a_3 a_4 a_5 = a_3^5 = (-2)^5 = -32$.