## Task A-4.1.

The product of the second and fourth terms of an arithmetic sequence with difference $d$ is $-d^{2}$. Determine the product of the third and fifth terms of this sequence.
## Solution.

Let's denote the first five terms of the given arithmetic sequence as $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$.

Since the common difference is $d$, we have $a_{2}=a_{1}+d, a_{3}=a_{1}+2 d, a_{4}=a_{1}+3 d, a_{5}=a_{1}+4 d$.

According to the condition of the problem, the product of the second and fourth terms is equal to $-d^{2}$ on one side, and on the other side,

$$
a_{2} \cdot a_{4}=\left(a_{1}+d\right)\left(a_{1}+3 d\right)=a_{1}^{2}+4 d a_{1}+3 d^{2}
$$

Equating the two expressions, we get:

$$
\begin{aligned}
a_{1}^{2}+4 d a_{1}+3 d^{2} & =-d^{2} \\
a_{1}^{2}+4 d a_{1}+4 d^{2} & =0 \\
\left(a_{1}+2 d\right)^{2} & =0 \\
a_{1}+2 d & =0
\end{aligned}
$$

Substituting the obtained result into the product $a_{3} \cdot a_{5}$, we get

$$
a_{3} \cdot a_{5}=\left(a_{1}+2 d\right)\left(a_{1}+4 d\right)=0
$$

since the first parenthesis is equal to zero.