Three numbers form an arithmetic progression, while another three form a geometric progression. If we add the corresponding terms of the two progressions, we get the following sums: 27, 39, 87. The sum of the terms of the arithmetic progression is \( s=26 \). What are the two progressions?
Let the first term of the arithmetic progression be $a$, the common difference be $d$; the first term of the geometric progression be $b$, and the common ratio be $q$.

According to the problem, we have:

$$
\begin{gathered}
a+b=27 \\
a+d+b q=39 \\
a+2 d+b q^{2}=87 \\
3 a+3 d=26
\end{gathered}
$$

If we subtract the sum of the terms of the arithmetic progression from the sum of the numbers on the right side of the first three equations, we get the sum of the terms of the geometric progression. Therefore,

$$
b+b q+b q^{2}=127
$$

From equations 2) and 4), by eliminating $a+d$, we get

$$
3 b q=91
$$

If we multiply both sides of equation 5) by $3 q$ and substitute the value of $3 b q$ from equation 6), we get:

$$
91 q^{2}-290 q+91=0
$$

From this equation, the values of $q$ are:

$$
\frac{290 \pm \sqrt{50976}}{182}
$$

or:

$$
q=\frac{145 \pm 6 \sqrt{354}}{91}
$$

Thus:

$$
\begin{gathered}
b=\frac{91^{2}}{3(145 \pm 6 \sqrt{354})} \\
b=\frac{91^{2}(145 \mp 6 \sqrt{354})}{3\left(145^{2}-36 \cdot 854\right)} \\
b=\frac{8281(145 \mp 6 \sqrt{354})}{3 \cdot 8281} \\
b=\frac{145 \mp 6 \sqrt{354}}{3}
\end{gathered}
$$

From equations 1) and 9), after eliminating $b$:

$$
\begin{gathered}
a=27-\frac{145 \mp 6 \sqrt{354}}{3} \\
a=\frac{-64 \pm 6 \sqrt{354}}{3}
\end{gathered}
$$

Finally, from equations 4) and 10):

$$
\begin{gathered}
d=\frac{26}{3}-\frac{-64 \pm 6 \sqrt{354}}{3} \\
d=\frac{90 \mp 6 \sqrt{354}}{3} \\
d=30 \mp 2 \sqrt{354}
\end{gathered}
$$

The two progressions are:

$$
\begin{aligned}
& \frac{-64 \pm 6 \sqrt{354}}{3}, \frac{26}{3}, \frac{116 \mp 6 \sqrt{354}}{3} \\
& \frac{145 \mp 6 \sqrt{354}}{3}, \frac{91}{3}, \frac{145 \pm 6 \sqrt{354}}{3}
\end{aligned}
$$

(Sándor Pollák, 7th grade student, Chief Gymnasium, Győr).