In a geometric progression (G.P.), the \(n\)-th term can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
where \(a\) is the first term and \(r\) is the common ratio.
Given:
- Third term \(T_3 = 24\)
- Common ratio \(r = 2\)
The third term can be expressed as:
\[ T_3 = a \cdot r^{3-1} = a \cdot r^2 \]
Substituting the values we have:
\[ 24 = a \cdot 2^2 \]
This simplifies to:
\[ 24 = a \cdot 4 \]
Now, solving for \(a\):
\[ a = \frac{24}{4} = 6 \]
Now that we have the first term \(a\), we can find the second term \(T_2\):
\[ T_2 = a \cdot r^{2-1} = a \cdot r = a \cdot 2 \]
Substituting \(a = 6\):
\[ T_2 = 6 \cdot 2 = 12 \]
Thus, the second term is equal to \( \boxed{12} \).