Shapovvoov A.V.

The sequence of natural numbers $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ is such that for each n, the equation $a_{n+2} x^{2} + a_{n+1} x^{x} + a_{n} = 0$ has a real root. Can the number of terms in this sequence be
a) equal to 10;
b) infinite
a) Here is an example of a sequence of 10 terms: $1, 2^{10}, 2^{18}, 2^{24}, 2^{26}, 2^{26}, 2^{24}, 2^{18}, 2^{10}, 1$. For each triplet of numbers $a_{n}, a_{n+1}, a_{n+2}$, the condition $a_{n+1}^{2}=4 a_{n} a_{n+2}$ is satisfied, which is sufficient for the existence of a real root of the equation $a_{n+2} x^{2}+a_{n+1} x^{x+a_{n}}=0$.

b) For such a sequence, for each natural $n$, $a_{n}^{2} \geq 4 a_{n-1} a_{n+1}$, which means $\frac{a_{n+1}}{a_{n}} \leq \frac{1}{4} \cdot \frac{a_{n}}{a_{n-1}}$. Therefore, $\frac{a_{n+1}}{a_{n}} \leq \frac{1}{4} \cdot \frac{a_{n}}{a_{n-1}} \leq \frac{1}{4^{2}} \cdot \frac{a_{n-1}}{a_{n-2}} \leq \ldots \leq \frac{1}{4^{n-1}} \cdot \frac{a_{2}}{a_{1}}$. Hence, $a_{n+1}<a_{n}$ for "large" $n$, meaning the sequence is decreasing. Contradiction: a sequence of natural numbers cannot decrease indefinitely.

## Answer

a) It can; b) it cannot.