1. Consider any arithmetic sequences of real numbers $\left(x_{i}\right)_{i=1}^{\infty}$ and $\left(y_{i}\right)_{i=1}^{\infty}$, which have the same first term and satisfy for some $k>1$ the equalities

$$
x_{k-1} y_{k-1}=42, \quad x_{k} y_{k}=30 \quad \text { and } \quad x_{k+1} y_{k+1}=16 .
$$

Find all such sequences for which the index $k$ is the largest possible.

(J. Šimša)
Solution. Let $c$, resp. $d$ be the differences of the sought sequences, so that from the expressions $x_{i}=$ $=x_{1}+(i-1) c$ and $y_{i}=x_{1}+(i-1) d$ we then get for each $i$ the equality

$$
x_{i} y_{i}=x_{1}^{2}+(i-1) x_{1}(c+d)+(i-1)^{2} c d \text {. }
$$

We will thus address the question of when for some index $k>1$ the system of equations holds

$$
\begin{aligned}
x_{1}^{2}+(k-2) x_{1}(c+d)+(k-2)^{2} c d & =42, \\
x_{1}^{2}+(k-1) x_{1}(c+d)+(k-1)^{2} c d & =30, \\
x_{1}^{2}+k x_{1}(c+d)+k^{2} c d & =16 .
\end{aligned}
$$

By subtracting the sum of equations (1) and (3) from twice equation (2), we get the equality $c d=-1$ after simplification. By subtracting equation (2) from equation (3), we get the relation

$$
x_{1}(c+d)+(2 k-1) c d=14 \text {, }
$$

from which, after substituting the value $c d=-1$, we arrive at the equality

$$
x_{1}(c+d)=2 k-15 \text {. }
$$

Substituting this result into equation (3) gives us the relation

$$
x_{1}^{2}+k(2 k-15)-k^{2}=16 \text {, }
$$

from which we express $x_{1}^{2}$ as a quadratic function of the index $k$ :

$$
x_{1}^{2}=16-k(2 k-15)+k^{2}=16+15 k-k^{2}=(k+1)(16-k) .
$$

Since $x_{1}^{2} \geqq 0$ and $k>1$, it follows from the last formula that $k \leqq 16$. However, in the case $k=16$, we get $x_{1}=0$ and equation (4) then becomes $0(c+d)=2$, which is impossible. For $k=15$ we get $x_{1}^{2}=16$, so $x_{1}= \pm 4$. For $x_{1}=4($ and $k=15)$, from (4) we get $c+d=\frac{15}{4}$, which together with the equality $c d=-1$ leads to the conclusion that $\{c, d\}=\left\{4,-\frac{1}{4}\right\}$. This means that both sequences are (up to order) determined by the formulas:

$$
x_{i}=4+(i-1) 4 \quad \text { and } \quad y_{i}=4-\frac{i-1}{4} \quad \text { for each } i
$$

For such a pair of sequences, it is indeed true that

$$
x_{14} y_{14}=56 \cdot \frac{3}{4}=42, \quad x_{15} y_{15}=60 \cdot \frac{1}{2}=30 \quad \text { and } \quad x_{16} y_{16}=64 \cdot \frac{1}{4}=16 .
$$

Similarly, for the other possible value $x_{1}=-4$, we get sequences whose terms are the negatives of the terms of the sequences (5), i.e., the sequences

$$
x_{i}=-4-(i-1) 4 \quad \text { and } \quad y_{i}=-4+\frac{i-1}{4} \quad \text { for each } i .
$$

Answer. The largest value of the index $k$ is 15 and all the sequences that satisfy the conditions are (up to a possible swap of the order in the pair) determined by the relations (5) and (6).