Winter Mathematical Competition

Writing the equation in the form $$(x-1)(x^2+(1-a)x-a+a^2)=0,$$ we obtain $x_1=1$. Let $x_2$ and $x_3$ be the roots of the quadratic equation. If $1$ is the second term of the progression then $x_2+x_3=2$, giving $a-1=2$, i.e. $a=3$. When $a=3$ the roots of the quadratic equation are not real. If $x_1=1$ is not the second term we may assume that $1+x_2=2x_3$, which together with $x_2+x_3=a-1$ implies $3x_3=a$. Therefore $\frac{a}{3}$ is a root of the quadratic equation, i.e. $${\left(\frac{a}{3}\right)}^2+(1-a)\frac{a}{3}-a+a^2=0.$$ Thus, $a=0$ or $a=\frac{6}{7}$. When $a=0$ we obtain $x_2=-1$, $x_3=0$ and when $a=\frac{6}{7}$ we find $x_2=-\frac{3}{7}$ and $x_3=\frac{2}{7}$. Hence the desired values are $a=0$ and $a=\frac{6}{7}$.