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If the three terms a² * b³ * c⁴, a³ * b⁴ * c⁵, and a⁴ * b⁵ * c⁶ (where a, b, c > 0) are in arithmetic progression, find the minimum value of (a + b + c).

1. 1
2. 3
3. 4
4. 8

Correct answer: 3

Solution

For three terms in AP, the middle term equals the average of the other two: 2(a³ b⁴ c⁵) = a² b³ c⁴ + a⁴ b⁵ c⁶. Dividing both sides by a² b³ c⁴ gives 2abc = 1 + (abc)². This means (abc - 1)² = 0, so abc = 1. By AM-GM, a + b + c >= 3*(abc)^(1/3) = 3*1 = 3, with equality when a = b = c = 1.

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