Task A-4.4. (8 points)

Three different real numbers, different from zero, form an arithmetic sequence, and their squares in the same order form a geometric sequence. Determine all possible values of the quotient of this geometric sequence.
## Solution.

First method.

Let $a-d, a$, and $a+d$ be the numbers forming an arithmetic sequence. The terms of the geometric sequence are $(a-d)^{2}$, $a^{2}$, and $(a+d)^{2}$. This means that $(a-d)^{2}(a+d)^{2}=\left(a^{2}\right)^{2}$, or $\left(a^{2}-d^{2}\right)^{2}=a^{4}$, from which it follows that $d=0$ or $d= \pm a \sqrt{2}$.

The quotient of the geometric sequence can be $q_{1}=1$ (for $d=0$), or (for $d= \pm a \sqrt{2}$):

$$
q_{2,3}=\frac{(a+d)^{2}}{a^{2}}=1+2 \cdot \frac{d}{a}+\left(\frac{d}{a}\right)^{2}=1 \pm 2 \sqrt{2}+(\sqrt{2})^{2}=3 \pm \sqrt{2}
$$

Second method.

Let these real numbers be denoted by $a, b$, and $c$. Since they form an arithmetic sequence, we have $\left.a+c=2 b . \quad{ }^{*}\right)$ Since $a^{2}, b^{2}, c^{2}$ is a geometric sequence, we have $a^{2} c^{2}=b^{4}$.

The desired quotient is $q=\frac{b^{2}}{a^{2}}=\frac{c^{2}}{b^{2}}$.

By squaring $(*)$ we get $a^{2}+2 a c+c^{2}=4 b^{2}$, or after dividing by $b^{2}$:

$$
\frac{a^{2}}{b^{2}}+2 \cdot \frac{a c}{b^{2}}+\frac{c^{2}}{b^{2}}=4
$$

From $\left({ }^{* *}\right)$ it follows that $b^{2}=|a c|$, so $\frac{a c}{b^{2}}= \pm 1$. Therefore, from $(* * *)$ we get two equations:

$$
\frac{1}{q}+2+q=4 \quad \text { and } \quad \frac{1}{q}-2+q=4
$$

or their equivalent quadratic equations: $q^{2}-2 q+1=0$ and $q^{2}-6 q+1=0$, with solutions $q_{1}=1$ and $q_{2,3}=3 \pm \sqrt{2}$.

For 4 points: at least one of the solutions $3 \pm \sqrt{2}$.

For 8 points: the correct result and method.