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 \( a \) (the first term) and \( r \) (the common ratio) of the geometric progression (GP), follow these steps: 1. **Set Up the Equations:** The general formula for the \( n \)-th term of a GP is: \[ T_n = a \cdot r^{n-1} \] Given: - The 4th term (\( T_4 \)) is 7: \[ a \cdot r^{3} = 7 \quad \text{(Equation 1)} \] - The 8th term (\( T_8 \)) is 567: \[ a \cdot r^{7} = 567 \quad \text{(Equation 2)} \] 2. **Divide Equation 2 by Equation 1:** \[ \frac{a \cdot r^{7}}{a \cdot r^{3}} = \frac{567}{7} \] Simplifying: \[ r^{4} = 81 \] Taking the fourth root (and considering only the positive value since \( r \) is positive): \[ r = 3 \] 3. **Find \( a \) Using \( r = 3 \):** Substitute \( r = 3 \) back into Equation 1: \[ a \cdot 3^{3} = 7 \Rightarrow a \cdot 27 = 7 \Rightarrow a = \frac{7}{27} \] **Final Answer:** \[ r = 3 \quad \text{and} \quad a = \frac{7}{27} \]