A GP has a first term of A,a common ratio of R and it's 6th term is 768. Another GP has a first term of A, a common ratio of 6R and its 3rd term is 3456. Evaluate A and R

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:
1. \( A \cdot R^{5} = 768 \)
2. \( A \cdot 36R^{2} = 3456 \)

### 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 \( R \).