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Assume the sequence of annual retreats is quadratic: aₙ = an² + bn + c
a₁ = 1.2 → a(1)² + b(1) + c = a + b + c = 1.2
a₂ = 1.8 → 4a + 2b + c = 1.8
a₃ = 2.5 → 9a + 3b + c = 2.5
Subtract first from second: (4a + 2b + c) − (a + b + c) = 1.8 − 1.2 → 3a + b = 0.6
Subtract second from third: (9a + 3b + c) − (4a + 2b + c) = 2.5 − 1.8 → 5a + b = 0.7
Now subtract: (5a + b) − (3a + b) = 0.7 − 0.6 → 2a = 0.1 → a = 0.05
Then 3(0.05) + b = 0.6 → 0.15 + b = 0.6 → b = 0.45
Then a + b + c = 1.2 → 0.05 + 0.45 + c = 1.2 → c = 0.7
So aₙ = 0.05n² + 0.45n + 0.7