4 th and 8 th term of a GP are 7 and 567 respectively. find the positive value of \( r \) and a

To solve for the positive values of \( r \) (common ratio) and \( a \) (first term) of the geometric progression (GP), we can use the information provided about the 4th and 8th terms. The nth term of a GP is given by: \[ T_n = a \cdot r^{(n-1)} \] From the question: - The 4th term \( T_4 = a \cdot r^3 = 7 \) - The 8th term \( T_8 = a \cdot r^7 = 567 \) Now, we can set up our equations: 1. \( a \cdot r^3 = 7 \) (Equation 1) 2. \( a \cdot r^7 = 567 \) (Equation 2) To eliminate \( a \), we can divide Equation 2 by Equation 1: \[ \frac{a \cdot r^7}{a \cdot r^3} = \frac{567}{7} \] This simplifies to: \[ r^4 = 81 \] Taking the fourth root: \[ r = 3 \] Now substitute \( r \) back into Equation 1 to find \( a \): \[ a \cdot (3)^3 = 7 \implies a \cdot 27 = 7 \implies a = \frac{7}{27} \] Therefore, the positive values are: - \( a = \frac{7}{27} \) - \( r = 3 \)