1. Determine all real numbers $c$ for which the equation

$$
x^{2}+\frac{5}{2} x+c=0
$$

has two real roots that can be arranged with the number $c$ to form a three-term arithmetic sequence.
1. Assume that the number $c$ has the required property. Let the difference of the corresponding arithmetic sequence be $d$. We will distinguish two cases depending on whether the number $c$ lies between the roots $x_{1}$ and $x_{2}$ of the given quadratic equation, or not:

a) If $c$ is the middle term of the assumed arithmetic sequence, then $x_{1}=c-d$ and $x_{2}=c+d$. For the sum of the roots, according to Viète's formula, we get $-\frac{5}{2}=x_{1}+x_{2}=2 c$, from which $c=-\frac{5}{4}$. Moreover, for negative $c$, the discriminant of the given equation is positive, so it has two real roots. (For $c=-\frac{5}{4}$, the given equation has roots $x_{1,2}=-\frac{5}{4} \pm \frac{3}{4} \sqrt{5}$.)

b) If the coefficient $c$ is an extreme term of the assumed arithmetic sequence, let the roots of the given equation be such that $x_{1}=c+d, x_{2}=c+2 d$. For their sum, this time we get $-\frac{5}{2}=x_{1}+x_{2}=2 c+3 d$. If we express $d$ from this, we get $d=-\frac{5}{6}-\frac{2}{3} c$ and substitute it into the relations $x_{1}=c+d$ and $x_{2}=c+2 d$, we get $x_{1}=\frac{1}{6}(2 c-5), x_{2}=-\frac{1}{3}(c+5)$. If we substitute both expressions into Viète's formula $x_{1} x_{2}=c$ for the product of the roots, we get the quadratic equation $2 c^{2}+23 c-25=0$ after simplification, which has roots $1$ and $-\frac{25}{2}$. (We do not need to check the condition on the discriminant this time, as the given procedure guarantees that the real numbers $x_{1,2}$ corresponding to both found values of $c$ satisfy both Viète's formulas, so they are indeed the roots of the relevant equation. For $c=1$, the given quadratic equation has roots $x_{1}=-\frac{1}{2}, x_{2}=-2$; for $c=-\frac{25}{2}$, the equation has roots $x_{1}=-5, x_{2}=\frac{5}{2}$.)

Conclusion. The real numbers $c$ that satisfy the problem are from the set $\left\{-\frac{25}{2} ;-\frac{5}{4} ; 1\right\}$.

For a complete solution of the problem, award 6 points. For finding each of the three solutions of the problem, award 2 points each. Do not deduct points if the contestant does not determine the roots of the corresponding quadratic equation for the found values of the coefficient $c$, as the problem does not require it. However, deduct one point if the student does not mention the existence of real roots in case a).