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oxed{\left( rac5011, rac711, rac1711 - Silent Sales Machine

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 The square of the sum is $ (4a)^2 = 16a^2 $.

📌 2a^2 + \frac{9d^2}{2} = 16a^2

📌 \Rightarrow \frac{9d^2}{2} = 14a^2

📌 \Rightarrow d^2 = \frac{28a^2}{9}.

📌 The third term is $ a + \frac{d}{2} $, and the first term is $ a - \frac{3d}{2} $. We want the ratio:

📌 \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})}.

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