4. The fourth term of a geometric progression is 64 and the tenth term is -4096 . Find the first term and the common ratio of the progression

In a geometric progression (GP), the \( n \)-th term can be expressed as: \[ a_n = a \cdot r^{n-1} \] where \( a \) is the first term and \( r \) is the common ratio. Given: - The fourth term \( a_4 = 64 \) - The tenth term \( a_{10} = -4096 \) We can write the equations for these terms: 1. For the fourth term: \[ a \cdot r^{3} = 64 \quad \text{(1)} \] 2. For the tenth term: \[ a \cdot r^{9} = -4096 \quad \text{(2)} \] Now, we can divide equation (2) by equation (1) to eliminate \( a \): \[ \frac{a \cdot r^{9}}{a \cdot r^{3}} = \frac{-4096}{64} \] This simplifies to: \[ r^{6} = -64 \] Taking the sixth root of both sides, we find: \[ r = -2 \] Now that we have the common ratio \( r \), we can substitute it back into equation (1) to find \( a \): \[ a \cdot (-2)^{3} = 64 \] Calculating \( (-2)^{3} \): \[ -8a = 64 \] Now, solving for \( a \): \[ a = \frac{64}{-8} = -8 \] Thus, the first term \( a \) and the common ratio \( r \) of the geometric progression are: \[ \text{First term } a = -8 \] \[ \text{Common ratio } r = -2 \]