When given two terms of a geometric sequence, use the relationship between them. Method 1 (Direct approach): From 3rd to 6th term, we multiply by r three times (3rd→4th→5th→6th). So \(T_6 = T_3 \times r^3\). Calculation: \(96 = 12 \times r^3\). Divide: \(r^3 = \frac{96}{12} = 8\). Take cube root: \(r = \sqrt[3]{8} = 2\). Method 2 (Using formula): \(T_3 = ar^2 = 12\) ... (1) and \(T_6 = ar^5 = 96\) ... (2). Divide equation (2) by equation (1): \(\frac{ar^5}{ar^2} = \frac{96}{12}\), which gives \(r^3 = 8\), so r = 2. Both methods give r = 2. Verification: If \(T_3 = 12\) and r = 2, then \(T_4 = 24\), \(T_5 = 48\), \(T_6 = 96\) ✓. This problem type is extremely common in ECZ Paper 1, testing both GP knowledge and algebraic reasoning. The key insight: the number of ratio multiplications equals the difference in term positions. Practice this thoroughly as it appears in various forms across multiple ECZ past papers.

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Zambia

• ivwananji siwale – 29 pts

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