6.5 The sum of three numbers forming an arithmetic progression is 2, and the sum of the squares of these same numbers is $\frac{14}{9}$. Find these numbers.
6.5 According to the condition, we write the system

$\left\{\begin{array}{l}a_{1}+a_{2}+a_{3}=2, \\ a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=\frac{14}{9} .\end{array}\right.$

Next, using formula (6.2), we have $\frac{a_{1}+a_{3}}{2} \cdot 3=2$, from which $a_{1}+a_{3}=\frac{4}{3}$ and, therefore, $a_{2}=\frac{2}{3}$. Thus, we arrive at the system

$\left\{\begin{array}{l}a_{1}+a_{3}=\frac{4}{3}, \\ a_{1}^{2}+a_{3}^{2}=\frac{10}{9},\end{array}\right.$

solving which we get $a_{1}=\frac{1}{3}, a_{3}=1$, or $a_{1}=1$, $a_{3}=\frac{1}{3}$. Therefore, the sought numbers are $\frac{1}{3} ; \frac{2}{3} ; 1$.

Answer: $\frac{1}{3} ; \frac{2}{3} ; 1$.