4.047. Find three numbers that form a geometric progression, given that their product is 64, and their arithmetic mean is $14 / 3$.
## Solution.

From the condition we have:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ b _ { 1 } \cdot b _ { 2 } \cdot b _ { 3 } = 6 4 } \\
{ \frac { b _ { 1 } + b _ { 2 } + b _ { 3 } } { 3 } \frac { 1 4 } { 3 } } \\
{ \{ \begin{array} { l } 
{ b _ { 1 } \cdot b _ { 1 } q \cdot b _ { 1 } q ^ { 2 } = 6 4 , } \\
{ b _ { 1 } + b _ { 1 } q + b _ { 1 } q ^ { 2 } = 1 4 , }
\end{array} } \\
{ \Rightarrow b _ { 1 } = \frac { 4 } { q } , \frac { 4 ( 1 + q + q ^ { 2 } ) } { q } = 1 4 , 2 q ^ { 2 } - 5 q + 2 = 0 , \text { hence } q _ { 1 } = 2 , q _ { 2 } = \frac { 1 } { 2 } } \\
{ b _ { 1 } ( 1 + q + q ^ { 2 } ) = 1 4 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\left.b_{1} q\right)^{3}=64, \\
b_{1}\left(1+q+q^{2}\right)=14,
\end{array} \Rightarrow\right.\right.
\end{aligned}
$$

Then $b_{1}^{\prime}=\frac{4}{2}=2, b_{2}^{\prime}=2 \cdot 2=4, b_{3}^{\prime}=2 \cdot 4=8$ or $b_{1}^{\prime \prime}=\frac{4}{\frac{1}{2}}=8$,

$$
b_{2}^{n}=8 \cdot \frac{1}{2}=4, b_{3}^{\prime \prime}=8 \cdot \frac{1}{4}=2
$$

Answer: 1) $2,4,8 ; 2) 8,4,2$.