## Task B-4.4.

A sequence of real numbers $\left(a_{n}\right)$ is defined recursively by

$$
a_{1}=1, \quad a_{2}=\frac{3}{5}, \quad a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}} \quad \text { for } n \geqslant 3
$$

Determine $a_{2020}$.
## First Solution.

The terms of the sequence $\left(a_{n}\right)$ are $1, \frac{3}{5}, \frac{3}{7}, \frac{3}{9}, \ldots$, and the sequence of their reciprocals is $1, \frac{5}{3}, \frac{7}{3}, \frac{9}{3}, \ldots$ We will show that the sequence $\left(\frac{1}{a_{n}}\right)$ is an arithmetic sequence. The general term of the sequence of reciprocals is

$$
\frac{1}{a_{n}}=\frac{2 a_{n-2}-a_{n-1}}{a_{n-2} a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}}
$$

It follows that

$$
\frac{1}{2}\left(\frac{1}{a_{n-2}}+\frac{1}{a_{n}}\right)=\frac{1}{a_{n-1}}
$$

so each term of the sequence $\left(\frac{1}{a_{n}}\right)$, starting from the second term, is the arithmetic mean of its neighboring terms.

The first term of this sequence is 1, and the common difference $d=\frac{2}{3}$. Then $\frac{1}{a_{2020}}=1+2019 \cdot \frac{2}{3}=1347$, and the desired term $a_{2020}=\frac{1}{1347}$.