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

1 answer

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 \( R \).