Task B-4.1. (20 points) In an increasing arithmetic sequence, the product of the second and third term is 3, and the product of the third and fifth term is -3. How many of the first terms of the sequence need to be summed to achieve the minimum sum? What is that sum?
Solution. We will represent the members of the sequence using the first term $a_{1}$ and the difference of the sequence $d$.

According to the condition of the problem, we have these two equations:

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

If we add these two equations, after rearranging, we get the homogeneous quadratic equation $2 a_{1}^{2}+9 a_{1} d+10 d^{2}=0$. Dividing by $d^{2}$, we get the quadratic equation

\[
2\left(\frac{a_{1}}{d}\right)^{2}+9 \cdot \frac{a_{1}}{d}+10=0
\]

Its solutions are $\frac{a_{1}}{d}=-2, \frac{a_{1}}{d}=-\frac{5}{2}$. This gives us $a_{1}=-2 d, a_{1}=-\frac{5}{2} d$.

The first solution $a_{1}=-2 d$ does not make sense because substituting it into the first equation gives $-d \cdot 0=3$!

For $a_{1}=-\frac{5}{2} d$, from the first equation we get $d= \pm 2$. However, $d=-2$ does not satisfy the condition because the sequence must be increasing. For $d=2$, we get $a_{1}=-5$.

The sum of the first $n$ terms of the sequence is

\[
S_{n}=\frac{n}{2}\left(a_{1}+a_{n}\right)=\frac{n}{2}\left(a_{1}+a_{1}+(n-1) d\right)=n^{2}-6 n=(n-3)^{2}-9
\]

The minimum for $S_{n}$ is obtained for $n=3$ and it is $S_{\min }=-9$.

Note. If we divide the second equation by the first, we get $\frac{a_{1}+4 d}{a_{1}+d}=-1$, from which we obtain $a_{1}=-\frac{5}{2} d$, and then proceed as above.