
Problem 4. The real numbers $a_{1}, a_{2}, \ldots, a_{n}(n \geq 3)$ form an arithmetic progression. There exists a permutation $a_{i_{1}}, a_{i_{2}}, \ldots, a_{i_{n}}$ of the same numbers, which is a geometric progression.

Find the numbers $a_{1}, a_{2}, \ldots, a_{n}$, if they are pairwise different and the biggest among them is equal to 1996 .


Solution. Let $a_{1}1$, i.e. $q>1$ or $q1)$ and then $a_{i_{1}}0$.)

Assume now that 3 among the numbers $a_{1}, a_{2}, \ldots, a_{n}$ are positive. Then $0<a_{n-2}<a_{n-1}<$ $a_{n}$ and they form a geometric as well as an arithmetic progression. Therefore $2 a_{n-1}=a_{n-2}+a_{n}$ and $a_{n-1}^{2}=a_{n-2} a_{n}$. From here $a_{n-2}=a_{n-1}=a_{n}$, which is a contradiction.

Thus at most two among the numbers are positive. Analogously, at most two among the numbers are negative. Consequently, $n \leq 4$.

Let $n=4$. Then $a_{1}<a_{2}<0<a_{3}<a_{4}$ and $2 a_{2}=a_{1}+a_{3}, 2 a_{3}=a_{2}+a_{4}$. But $q<-1$ and the geometric progression is either $a_{3}, a_{2}, a_{4}, a_{1}$ or $a_{2}, a_{3}, a_{1}, a_{4}$. Let it be $a_{3}, a_{2}, a_{4}, a_{1}$. Then $a_{2}=a_{3} q, a_{4}=a_{3} q^{2}$ and $a_{1}=a_{3} q^{3}$. Thus, $2 a_{3} q=a_{3} q^{3}+a_{3}$ and $2 a_{3}=a_{3} q+a_{3} q^{2}$. From here $q=1$, which contradicts to $q<-1$.

So $n=3$. There are two possibilities:
I. $a_{1}<a_{2}<0<a_{3}=1996$. Then the geometric progression is $a_{2}, a_{3}=a_{2} q, a_{1}=a_{2} q^{2}$. It follows from $2 a_{2}=a_{1}+a_{3}$ that $2 a_{2}=a_{2} q^{2}+a_{2} q$, i.e. $q^{2}+q-2=0$. Thus, $q=-2,-2 a_{2}=1996$, $a_{2}=-998$ and the numbers are $(-3992 ;-998 ; 1996)$.

II. $a_{1}<0<a_{2}<a_{3}=1996$. Now the geometric progression is $a_{2}, a_{1}=a_{2} q, a_{3}=a_{2} q^{2}$. From $2 a_{2}=a_{1}+a_{3}$ we obtain $2 a_{2}=a_{2} q+a_{2} q^{2}$, i.e. again $q=-2$. Therefore, $a_{3}=4 a_{2}=1996$ and $a_{2}=499$. The numbers are ( $-998 ; 499 ; 1996$ ).
