## Task 1 - 281231

Determine all sequences of three elements $\left(a_{1}, a_{2}, a_{3}\right)$ and $\left(b_{1}, b_{2}, b_{3}\right)$ that, with two suitable non-zero real numbers $p, r$ and with $q=5$, satisfy the following conditions:

(1) It holds that $a_{1}=\frac{1}{p}, a_{2}=\frac{2}{q}, a_{3}=\frac{1}{r}$.

(2) It holds that $b_{1}=\frac{1}{a_{1}}, b_{2}=\frac{1}{a_{1} \cdot a_{2}}, b_{3}=\frac{1}{a_{2} \cdot a_{3}}$.

(3) The sequence $\left(a_{1}, a_{2}, a_{3}\right)$ is an arithmetic sequence.

(4) The sequence $\left(b_{1}, b_{2}, b_{3}\right)$ is an arithmetic sequence.
If the sequence is to be arithmetic, it must hold that $a_{2}-a_{1}=a_{3}-a_{2}$ or $2 a_{2}=a_{1}+a_{3}$. This gives the first condition as:

$$
\begin{aligned}
\frac{4}{q} & =\frac{1}{p}+\frac{1}{r} \\
4 p r & =q r+p q
\end{aligned}
$$

Similarly, $2 b_{2}=b_{1}+b_{3}$, or

$$
\begin{gathered}
p q=p+\frac{1}{2} q r \\
(q-1) p=\frac{1}{2} q r \\
r=2 \frac{q-1}{q} p
\end{gathered}
$$

We substitute this into (1):

$$
8 \frac{q-1}{q} p^{2}=2(q-1) p+q p
$$

One solution to this equation would be $p=0$, but this would lead to division by zero in the sequence $a_{i}$. Therefore, the solution is only:

$$
\begin{gathered}
8 \frac{q-1}{q} p=3 q-2 \\
p=\frac{q(3 q-2)}{8(q-1)}
\end{gathered}
$$

and

$$
r=\frac{3 q-2}{4}
$$

With $q=5$, we get $p=\frac{65}{32}$ and $r=\frac{13}{4}$.