This problem requires solving for n in the sum formula. Given: a = 5, d = 8 - 5 = 3, \(S_n = 155\). Using \(S_n = \frac{n}{2}[2a + (n-1)d]\): \(155 = \frac{n}{2}[2(5) + (n-1)3]\). Simplify: \(155 = \frac{n}{2}[10 + 3n - 3]\). \(155 = \frac{n}{2}[7 + 3n]\). Multiply both sides by 2: \(310 = n[7 + 3n]\). Expand: \(310 = 7n + 3n^2\). Rearrange: \(3n^2 + 7n - 310 = 0\). Factor or use quadratic formula. For factoring: Find factors of \(3 \times (-310) = -930\) that add to 7. Factors are 31 and -30. Rewrite: \(3n^2 + 31n - 30n - 310 = 0\). Group: \(n(3n + 31) - 10(3n + 31) = 0\). Factor: \((n - 10)(3n + 31) = 0\). Solutions: n = 10 or n = -31/3. Since n must be positive, n = 10. Answer: 10 terms are needed. This type of problem is challenging but appears regularly in ECZ Paper 2. It combines arithmetic sequences with quadratic equations, testing multiple skills simultaneously.

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

Zambia

• Joseph Soko – 101 pts

• Distance of A Straight Line
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• Equation of a Straight Line
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• COMMUNICATION SYSTEMS