Step-by-step Solution:

We are given that the terms 1, \( \log_y(x) \), \( \log_z(y) \), and \( -15 \log_x(z) \) are in Arithmetic Progression (AP). Step 1: Identifying the Common Difference From previous solutions, we derived that the common difference \( d \) is -2. This gives the relationships: \[ \log_y(x) - 1 = -2 \] \[ \log_y(x) = -1 \] \[ \log_z(y) - \log_y(x) = -2 \] \[ \log_z(y) - (-1) = -2 \] \[ \log_z(y) = -1 \] \[ \log_x(z) - \log_z(y) = -2 \] \[ \log_x(z) - (-1) = -2 \] \[ \log_x(z) = -1 \] From these, we obtained: \[ x = \frac{1}{y}, \quad y = \frac{1}{z}, \quad z = \frac{1}{x} \] Step 2: Finding \( yz \) Multiplying \( y \) and \( z \): \[ yz = \left( \frac{1}{z} \right) \left( \frac{1}{x} \right) \] \[ yz = \frac{1}{zx} \] Since \( xz = \frac{1}{y} \), we substitute: \[ yz = z^{-2} \] Final Answer: \[ \boxed{z^{-2}} \]