In a GP, the ratio between consecutive terms is constant. So \(\frac{x-3}{x+1} = \frac{x-1}{x-3}\). Cross-multiplying: \((x-3)^2 = (x+1)(x-1)\). Expanding: \(x^2-6x+9 = x^2-1\). Solving: \(-6x+9 = -1\), so \(-6x = -10\), \(x = \frac{5}{3}\). The terms are \(\frac{8}{3}, -\frac{4}{3}, \frac{2}{3}\). Common ratio = \(\frac{-4/3}{8/3} = -\frac{1}{2}\).

Break down complex topics into smaller, manageable chunks for easier recall.

Zambia

• Naomi Miselo – 31 pts

• Distance of A Straight Line
• Mid-point of A Line
• Equation of a Straight Line
• Parallel and Perpendicular Lines
• COMMUNICATION SYSTEMS