3. Three numbers, the sum of which is 114, are, on the one hand, three consecutive terms of a geometric progression, and on the other - the first, fourth, and twenty-fifth terms of an arithmetic progression, respectively. Find these numbers.
Solution: Let $a, b, c$ be the required numbers, and $d$ be the common difference of the arithmetic progression. Then

$$
\begin{gathered}
\left\{\begin{array} { c } 
{ a + b + c = 114 , } \\
{ a c = b ^ { 2 } , } \\
{ b = a + 3 d , } \\
{ c = a + 24 d , }
\end{array} \Leftrightarrow \left\{\begin{array} { c } 
{ 3 a + 27 d = 114 , } \\
{ a ( a + 24 d ) = ( a + 3 d ) ^ { 2 } } \\
{ b = a + 3 d , } \\
{ c = a + 24 d , }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
a=38-9 d \\
24 a d=6 a d+9 d^{2}, \\
b=a+3 d \\
c=a+24 d
\end{array}\right.\right.\right. \\
\Leftrightarrow\left\{\begin{array} { c } 
{ a = 38 - 9 d , } \\
{ d ( 2 a - d ) = 0 , } \\
{ b = a + 3 d , } \\
{ c = a + 24 d , }
\end{array} \Leftrightarrow \left\{\begin{array} { c } 
{ a = 38 - 9 d , } \\
{ [ \begin{array} { c } 
{ d = 0 , } \\
{ d = 2 a , } \\
{ b = a + 3 d , } \\
{ c = a + 24 d }
\end{array} }
\end{array} \Leftrightarrow \left[\begin{array}{c}
d=0, a=b=c=38 \\
d=4, a=2, b=14, c=98
\end{array}\right.\right.\right.
\end{gathered}
$$

Answer: $a_{1}=38, b_{1}=38, c_{1}=38$, or $a_{2}=2, b_{2}=14, \quad c_{2}=98$.