4.067. The sum of three numbers is $11 / 18$, and the sum of their reciprocals, which form an arithmetic progression, is 18. Find these numbers.
## Solution.

From the condition, we have $a_{1}+a_{2}+a_{3}=\frac{11}{18}, \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}=18$, where $\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}$ are terms of an arithmetic progression.

Thus,

$$
\begin{aligned}
& \left\{\begin{array}{l}
a_{1}+a_{2}+a_{3}=\frac{11}{18}, \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}=18 \\
\frac{2}{a_{2}}=\frac{1}{a_{1}}+\frac{1}{a_{3}},
\end{array},\left\{\begin{array}{l}
a_{1}+a_{2}+a_{3}=\frac{11}{18}, \\
a_{2} a_{3}+a_{1} a_{3}+a_{1} a_{2}=18 a_{1} a_{2} a_{3}, \Leftrightarrow \\
2 a_{1} a_{3}=a_{2} a_{3}+a_{1} a_{2},
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
a_{1}+a_{2}+a_{3}=\frac{11}{18}, \\
\left(18 a_{1} a_{2}-a_{1}-a_{2}\right) a_{3}=a_{1} a_{2}, \Rightarrow 18 a_{1} a_{2}-a_{1}-a_{2}=2 a_{1}-a_{2}, \\
\left(2 a_{1}-a_{2}\right) a_{3}=a_{1} a_{2} .
\end{array}\right.
\end{aligned}
$$

$$
18 a_{1} a_{2}=3 a_{1}, a_{2}=\frac{1}{6}
$$

From the second equation of the system, we get

$$
\left(2 a_{1}-\frac{1}{6}\right) a_{3}=\frac{a_{1}}{6}, a_{3}=\frac{a_{1}}{12 a_{1}-1} .
$$

Then, substituting $a_{2}, a_{3}$ into the first equation of the system, after simplifications, we get $27 a_{1}^{2}-12 a_{1}+1=0$, from which $a_{1}^{\prime}=\frac{1}{9}, a_{1}^{\prime \prime}=\frac{1}{3}$. Then $a_{3}^{\prime}=\frac{1}{3}, a_{3}^{\prime \prime}=\frac{1}{9}$. We have $a_{1}^{\prime}=\frac{1}{9}, a_{2}^{\prime}=\frac{1}{6}, a_{3}^{\prime}=\frac{1}{3}$ or $a_{1}^{\prime \prime}=\frac{1}{3}, a_{2}^{\prime \prime}=\frac{1}{6}, a_{3}^{\prime \prime}=\frac{1}{9}$. According to the condition, the values $a_{1}=\frac{1}{9}, a_{2}=\frac{1}{6}, a_{3}=\frac{1}{3}$ are suitable.

Answer: $\frac{1}{9}, \frac{1}{6}, \frac{1}{3}$.