The sum of three consecutive terms of a geometric progression is $a$, the sum of the squares of the terms is $b$; determine the first term, the common ratio, and form the progression.
If the first term is $x$, and the ratio is $y$, then:

$$
x\left(1+y+y^{2}\right)=a
$$

$$
x^{2}\left(1+y^{2}+y^{4}\right)=b
$$

Squaring (1) and dividing by (2), we get:

$$
\frac{\left(1+y+y^{2}\right)^{2}}{1+y^{2}+y^{4}}=\frac{a^{2}}{b}
$$

Dividing the numerator and the denominator of the left side of this equation by $1+y+y^{2}$:

$$
\frac{1+y+y^{2}}{1-y+y^{2}}=\frac{a^{2}}{b}
$$

from which

$$
\begin{gathered}
\left(a^{2}-b\right) y^{2}-\left(a^{2}+b\right) y+a^{2}-b=0 \\
y=\frac{1}{2\left(a^{2}-b\right)}\left[a^{2}+b \pm \sqrt{\left(a^{2}+b\right)^{2}-4\left(a^{2}-b\right)^{2}}\right]
\end{gathered}
$$

or

$$
y=\frac{1}{2\left(a^{2}-b\right)}\left[a^{2}+b \pm \sqrt{\left(3 a^{2}-b\right)\left(3 b-a^{2}\right)}\right]
$$

To calculate the first term of the progression, divide (2) by (1); then we get:

$$
x\left(1-y+y^{2}\right)=\frac{b}{a}
$$

subtracting this from (1):

$$
2 x y=\frac{a^{2}-b}{a}
$$

from which

$$
x=\frac{a^{2}-b}{2 a y}
$$

Substituting the value of the ratio from (4) into this:

$$
x=\frac{1}{4 a}\left[a^{2}+b \mp \sqrt{\left(3 a^{2}-b\right)\left(3 b-a^{2}\right)}\right]
$$

The second and third terms of the progression will be:

$$
\frac{a^{2}-b}{2 a}, \frac{1}{4 a}\left[a^{2}+b \pm \sqrt{\left(3 a^{2}-b\right)\left(3 b-a^{2}\right)}\right]
$$

The problem was solved by: Friedmann Bernát, senior gym. VII. grade, Satoraljaújhely; Geiszt Emil, senior VII. grade, Győr; Goldstein Zsigmond, senior gym. VII. grade, Nyíregyháza; Grünhut Béla, senior VII. grade, Pécs, Hofbauer Ervin and Kántor Nándor, senior high evang. senior gym. VII. grade, Budapest; Szabó István senior VI. grade, Debrecen; Visnya Aladár, senior VIII. grade, Pécs; Weisz Herman, senior gym. VI. grade, Satoraljaújhely.