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Home » Substitute the values: \( R = \frac50^2 \times 0.8669.8 \). - Redraw

Substitute the values: \( R = \frac50^2 \times 0.8669.8 \). - Redraw

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 Let the annual retreat amounts be: Year 1–4: ?, ?, ?, ?; Year 5: A, Year 6: B, Year 7: C.

📌 But since the acceleration is linear, model the *annual retreat* as a sequence with constant second difference.

📌 Let the retreats (in meters/year) for the 5 years be: $ a, a+d, a+2d, a+3d, a+4d $.

📌 But the first 4 years total 120: $ a + (a+d) + (a+2d) + (a+3d) = 4a + 6d = 120 $

📌 The next 3 years: $ (a+3d) + (a+4d) + (a+5d) = 3a + 12d = 210 $

📌 We now solve the system:

📌 From (2): $ a = 70 - 4d $

📌 Substitute into (1): $ 2(70 - 4d) + 3d = 60 $ → $ 140 - 8d + 3d = 60 $ → $ -5d = -80 $ → $ d = 16 $

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