4.033. In an infinite geometric progression with positive terms and a common ratio $|q|<1$, the sum of the first three terms is 10.5, and the sum of the progression is 12. Find the progression.
Solution.

From formulas (4.6) and (4.12), we get:

$$
\left\{\begin{array} { l } 
{ b _ { 1 } + b _ { 1 } q + b _ { 1 } q ^ { 2 } = 10.5 , } \\
{ b _ { 1 } = 12 ( 1 - q ) }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}\left(1+q+q^{2}\right)=10.5 \\
b_{1}=12(1-q)
\end{array}\right.\right.
$$

From this, $12(1-q)\left(1+q+q^{2}\right)=10.5, \quad 12\left(1-q^{3}\right)=10.5, \quad q=0.5$.

Then $b_{1}=12(1-0.5)=6, b_{2}=3, b_{3}=\frac{3}{2}$.

Answer: $6,3, \frac{3}{2}, \ldots$.