In a geometric sequence, the ratio between consecutive terms is constant. This means: \(\frac{\text{2nd term}}{\text{1st term}} = \frac{\text{3rd term}}{\text{2nd term}}\). Set up equation: \(\frac{k}{k+4} = \frac{2k-15}{k}\). Cross-multiply: \(k \times k = (k+4)(2k-15)\). Expand left side: \(k^2\). Expand right side: \(k \times 2k - k \times 15 + 4 \times 2k - 4 \times 15 = 2k^2 - 15k + 8k - 60 = 2k^2 - 7k - 60\). Equation: \(k^2 = 2k^2 - 7k - 60\). Rearrange: \(0 = 2k^2 - k^2 - 7k - 60\), so \(k^2 - 7k - 60 = 0\). Factor: Find factors of -60 that add to -7. These are -12 and +5. \(k^2 - 12k + 5k - 60 = 0\). \(k(k-12) + 5(k-12) = 0\). \((k+5)(k-12) = 0\). Solutions: k = -5 or k = 12. Since k is positive, k = 12. Verification: Terms are 16, 12, 9. Ratio: 12÷16 = 0.75 and 9÷12 = 0.75 ✓. This exact question type appears in ECZ 2018 Internal exam. It combines GP understanding with algebraic manipulation.

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Zambia

• Joseph Soko – 101 pts

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