3. Three numbers form a geometric progression. If the second term is increased by 8, the progression turns into an arithmetic one, but if then the third term of the obtained progression is increased by 64, it turns back into a geometric progression. Find these numbers.
Solution: Let $a, b, c$ be the required numbers. Then

$$
\begin{gathered}
\left\{\begin{array} { c } 
{ a c = b ^ { 2 } , } \\
{ a + c = 2 ( b + 8 ) , } \\
{ a ( c + 6 4 ) = ( b + 8 ) ^ { 2 } , }
\end{array} \Leftrightarrow \left\{\begin{array} { c } 
{ a c = b ^ { 2 } , } \\
{ a + c = 2 ( b + 8 ) , } \\
{ a c + 6 4 a = b ^ { 2 } + 1 6 b + 6 4 , }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
a c=b^{2}, \\
a+c=2(b+8), \\
a c+64 a=b^{2}+16 b+64,
\end{array}\right.\right.\right. \\
\Leftrightarrow\left\{\begin{array} { c } 
{ a c = b ^ { 2 } , } \\
{ c = 2 ( b + 8 ) - c , } \\
{ 6 4 a = 1 6 b + 6 4 , }
\end{array} \Leftrightarrow \left\{\begin{array} { c } 
{ a c = b ^ { 2 } , } \\
{ c = \frac { 7 } { 4 } b + 1 5 , } \\
{ a = \frac { 1 } { 4 } b + 1 , }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
(7 b+60)(b+4)=16 b^{2}, \\
c=\frac{7}{4} b+15, \\
a=\frac{1}{4} b+1 .
\end{array}\right.\right.\right.
\end{gathered}
$$

We arrive at the quadratic equation: $9 b^{2}-88 b-240=0, \frac{D}{4}=64^{2}$, $b_{1}=-\frac{20}{9}, \quad b_{2}=12$

The required numbers: $a_{1}=\frac{4}{9}, \quad b_{1}=-\frac{20}{9}, \quad c_{1}=\frac{100}{9}$, or $a_{2}=4, \quad b_{2}=12, \quad c_{2}=36$.

Answer: $a_{1}=\frac{4}{9}, \quad b_{1}=-\frac{20}{9}, \quad c_{1}=\frac{100}{9}, \quad$ or $a_{2}=4, \quad b_{2}=12, \quad c_{2}=36$.