# Problem 2. (2 points)

In a positive non-constant geometric progression, the arithmetic mean of the third, fourth, and eighth terms is equal to some term of this progression. What is the minimum possible number of this term?
Answer: 4

Solution:

The third element is either the larger or the smaller of the three specified, so it cannot be equal to the arithmetic mean of all three. The first and second elements are even less suitable.

To prove that the answer "4" is possible, let's introduce the notation: let $b_{n}=$ $b q^{n-1}$. Then we need to solve the equation $3 b q^{3}=b q^{2}+b q^{3}+b q^{7}$. Simplifying it, we get $q^{5}-$ $2 q+1=0$.

This equation has a root $q=1$, but it does not suit us because the progression is not constant. $q^{5}-2 q+1=(q-1)\left(q^{4}+q^{3}+q^{2}+q-1\right)$. The second factor is negative at $q=0$ and positive at $q=1$, so this expression has a root between zero and one. Thus, we can choose the common ratio of the progression and take any positive number as the first term for the answer 4.