4.011. The sum of three numbers forming an arithmetic progression is 2, and the sum of the squares of these same numbers is 14/9. Find these numbers.
Solution.

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

Using formula (4.1), we get

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ a _ { 1 } + a _ { 1 } + d + a _ { 1 } + 2 d = 2 , } \\
{ a _ { 1 } ^ { 2 } + ( a _ { 1 } + d ) ^ { 2 } + ( a _ { 1 } + 2 d ) ^ { 2 } = \frac { 1 4 } { 9 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1}+d=\frac{2}{3}, \\
a_{1}^{2}+\frac{4}{9}+\left(\frac{2}{3}+d\right)^{2}=\frac{14}{9}
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
d=\frac{2-3 a_{1}}{3}, \\
3 a_{1}^{2}-4 a_{1}+1=0 .
\end{array} \text { From this we have: } 1\right)\left\{\begin{array} { l } 
{ a _ { 1 } ^ { \prime } = 1 , } \\
{ d ^ { \prime } = - \frac { 1 } { 3 } ; }
\end{array} \text { 2) } \left\{\begin{array}{l}
a_{\mathrm{i}}^{\prime \prime}=\frac{1}{3} \\
d^{\prime \prime}=\frac{1}{3}
\end{array}\right.\right.
\end{aligned}
$$

Then

1) $a_{1}^{\prime}=1, a_{2}^{\prime}=\frac{2}{3}, a_{3}^{\prime}=\frac{1}{3}$; 2) $a_{1}^{\prime \prime}=\frac{1}{3}, a_{2^{\prime}}^{\prime \prime}=\frac{2}{3}, a_{3}^{\prime \prime}=1$.

Answer: 1) $\left.1, \frac{2}{3}, \frac{1}{3} ; 2\right) \frac{1}{3}, \frac{2}{3}, 1$.