4.034. Find the first three terms of an arithmetic progression, for which the sum of any number of terms is equal to three times the square of this number.
Solution.

Let $n=2$ and $n=3$. From the condition we have

$$
\left\{\begin{array}{l}
a_{1}+a_{2}=3 \cdot 2^{2} \\
a_{1}+a_{2}+a_{3}=3 \cdot 3^{2}
\end{array}\right.
$$

By formula (4.1) we get

$$
\left\{\begin{array} { l } 
{ a _ { 1 } + a _ { 1 } + d = 1 2 , } \\
{ a _ { 1 } + a _ { 1 } + d + a _ { 1 } + 2 d = 2 7 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 2 a _ { 1 } + d = 1 2 , } \\
{ a _ { 1 } + d = 9 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
d=6 \\
a_{1}=3
\end{array}\right.\right.\right.
$$

Then $a_{2}=9, a_{3}=15$.

Answer: $3,9,15$.