2 Bulgarian Winter Tournament

Let $a_1$ and $d$ be the first term and the difference of the arithmetic progression, respectively. From the condition $a_1$, $a_1+6d$ and $a_1+16d$ are consecutive members of a geometric progression, i.e. $$(a_1+6d{)}^2=a_1\cdot (a_1+16d)⟺d\cdot (a_1-9d)=0.$$ Since $d\ne 0$, we get that $a_1=9d$. Furthermore, we have $(x-9)(x+\sqrt{12-x})=0$ and $x\le 12$. Then $x=9$ or $\sqrt{12-x}=-x$, i.e. $x^2+x-12=0$ and $x\le 0$, whence $x=-4$. Then $a_1=9$ and $a_1=-4$, as $d=1$ and $d=-\frac{4}{9}$, respectively. $\square$