To solve for \( A \) and \( R \) given the information about the two geometric progressions (GPs), we can set up equations for each GP based on their definitions.
For the first GP:
The first term is \( A \) and the common ratio is \( R \). The formula for the \( n \)-th term of a GP is:
\[ T_n = A \cdot R^{(n-1)} \]
For the 6th term:
\[ T_6 = A \cdot R^{5} = 768 \quad (1) \]
For the second GP:
The first term is \( A \) and the common ratio is \( 6R \). The third term is:
\[ T_3 = A \cdot (6R)^{(3-1)} = A \cdot (6R)^2 = A \cdot 36R^2 = 3456 \quad (2) \]
Now we have two equations:
- \( A \cdot R^{5} = 768 \)
- \( A \cdot 36R^{2} = 3456 \)
Solve for \( A \):
From equation (1):
\[ A = \frac{768}{R^5} \]
Substituting \( A \) into equation (2):
\[ \frac{768}{R^5} \cdot 36R^{2} = 3456 \]
This simplifies to:
\[ \frac{768 \cdot 36R^2}{R^5} = 3456 \]
\[ 768 \cdot 36R^{-3} = 3456 \]
To eliminate \( R^{-3} \), multiply both sides by \( R^3 \):
\[ 768 \cdot 36 = 3456R^3 \]
Calculating \( 768 \cdot 36 \):
\[ 768 \cdot 36 = 27648 \]
Thus:
\[ 27648 = 3456R^3 \]
Now solve for \( R^3 \):
\[ R^3 = \frac{27648}{3456} \]
Calculating \( \frac{27648}{3456} \):
\[ R^3 = 8 \] So,
\[ R = 2 \]
Now find \( A \) using \( R \):
Substituting \( R = 2 \) back into the equation for \( A \):
\[ A = \frac{768}{R^5} = \frac{768}{2^5} = \frac{768}{32} = 24 \]
Final values:
\[ A = 24, \quad R = 2 \]
Thus, the values are \( \boxed{24} \) for \( A \) and \( \boxed{2} \) for \( R \).