# Problem 3.

The number $b$ is the arithmetic mean of the numbers $a$ and $c$. Find all ordered triples $(a, b, c)$ of such numbers for which at least one of the numbers $1 / a, 1 / b, 1 / c$ is the arithmetic mean of the other two.
# Solution.

Obviously, $a, b, c$ should not be equal to 0. As indicated,

$$
2 b=a+c
$$

and one of the following conditions must be satisfied:

$$
\frac{2}{b}=\frac{1}{a}+\frac{1}{c} \quad \text { or } \quad \frac{2}{a}=\frac{1}{b}+\frac{1}{c} \quad \text { or } \quad \frac{2}{c}=\frac{1}{a}+\frac{1}{b} .
$$

Solving the system of equations

$$
\left\{\begin{array}{l}
2 b=a+c \\
\frac{2}{b}=\frac{1}{a}+\frac{1}{c}
\end{array}\right.
$$

we get that $a=b=c(=x)$.

Solving the system of equations

$$
\left\{\begin{array}{l}
2 b=a+c \\
\frac{2}{a}=\frac{1}{b}+\frac{1}{c}
\end{array}\right.
$$

we get either $a=b=c$ (this case has already been discussed), or $a=4 b, c=-2 b(=$ $2 x)$, which gives the triplet $(-4 x,-x, 2 x)$.

Answer: This is either the triplet $(x, x, x)$, where $x \neq 0$, or the triplet $(-4 x,-x, 2 x)$, where $x \neq 0$.

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