1. It is known that the sequence of numbers $a_{1}, a_{2}, \ldots$, is an arithmetic progression, and the sequence of numbers $a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4} \ldots$, is a geometric progression. It is known that $a_{1}=1$. Find $a_{2017}$.
Answer: $a_{2017}=1$;

Trunov K.V.

## Solution:

Since the sequence of numbers $a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4} \ldots$, is a geometric progression, then $\left(a_{n} a_{n+1}\right)^{2}=\left(a_{n-1} a_{n}\right)\left(a_{n+1} a_{n+2}\right)$ for $n \geq 2$. From this, we obtain that $a_{n} a_{n+1}=a_{n-1} a_{n+2}$ (since $a_{n} \neq 0$ for any $n$, otherwise the geometric progression would contain an infinite number of zeros, which would mean that the arithmetic progression $a_{1}, a_{2}, \ldots$, contains an infinite number of zeros, which is impossible). Since $a_{1}, a_{2}, \ldots$, is an arithmetic progression, we can write $a_{n} a_{n+1}=a_{n-1} a_{n+2}$ as $a_{n}\left(a_{n}+d\right)=\left(a_{n}-d\right)\left(a_{n}+2 d\right) \Leftrightarrow$

$\Leftrightarrow a_{n}^{2}+a_{n} d=a_{n}^{2}+a_{n} d-2 d^{2} \Leftrightarrow-2 d^{2}=0 \Leftrightarrow d=0 \Rightarrow a_{2017}=1$.

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