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Home » Solution: Use identity $\sin(2z) = 2\sin z \cos z$. So: - Nelissen Grade advocaten

Solution: Use identity $\sin(2z) = 2\sin z \cos z$. So: - Nelissen Grade advocaten

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 \frac{\text{third term}}{\text{first term}} = \frac{a + \frac{d}{2}}{a - \frac{3d}{2}}.

📌 Substitute $ d^2 = \frac{28a^2}{9} \Rightarrow d = a\sqrt{\frac{28}{9}} = \frac{2a\sqrt{7}}{3} $. Then:

📌 \frac{a + \frac{1}{2} \cdot \frac{2a\sqrt{7}}{3}}{a - \frac{3}{2} \cdot \frac{2a\sqrt{7}}{3}} = \frac{a + \frac{a\sqrt{7}}{3}}{a - a\sqrt{7}} = \frac{1 + \frac{\sqrt{7}}{3}}{1 - \sqrt{7}} = \frac{\frac{3 + \sqrt{7}}{3}}{1 - \sqrt{7}} = \frac{3 + \sqrt{7}}{3(1 - \sqrt{7})}.

📌 Multiply numerator and denominator by $ 1 + \sqrt{7} $:

📌 \frac{(3 + \sqrt{7})(1 + \sqrt{7})}{3(1 - 7)} = \frac{3(1) + 3\sqrt{7} + \sqrt{7} + 7}{3(-6)} = \frac{10 + 4\sqrt{7}}{-18} = -\frac{10 + 4\sqrt{7}}{18} = -\frac{5 + 2\sqrt{7}}{9}.

📌 But since the progression is real and the context suggests positive terms, we reconsider the symmetry. Instead, let the terms be $ a - \frac{3d}{2},\ a - \frac{d}{2},\ a + \frac{d}{2},\ a + \frac{3d}{2} $, and we already have $ d^2 = \frac{28}{9}a^2 $.

📌 Let $ r = \frac{a + \frac{d}{2}}{a - \frac{3d}{2}} $. Let $ x = \frac{d}{a} $, so $ x^2 = \frac{28}{9} $. Then:

📌 r = \frac{1 + \frac{x}{2}}{1 - \frac{3x}{2}}.

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