2003 AIME I Problem 8

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8.

In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by 30.30. Find the sum of the four terms.

Answer: 129
Concepts:arithmetic sequencegeometric sequencecasework
Difficulty rating: 2210
Solution:

Write the terms as a,a, a+d,a + d, a+2d,a + 2d, and a+30,a + 30, where aa and dd are positive integers. The geometric condition on the last three terms says (a+30)(a+d)=(a+2d)2.(a + 30)(a + d) = (a + 2d)^2. Expanding both sides and simplifying, 30a+30d=3ad+4d2,30a + 30d = 3ad + 4d^2, that is 3a(10d)=2d(2d15).3a(10 - d) = 2d(2d - 15).

Since a,d>0,a, d \gt 0, the factors 10d10 - d and 2d152d - 15 must have the same sign, forcing 7.5<d<10,7.5 \lt d \lt 10, so d=8d = 8 or d=9.d = 9. For d=8,d = 8, we get 6a=16,6a = 16, which has no integer solution. For d=9,d = 9, we get 3a=54,3a = 54, so a=18.a = 18.

The sequence is 18,27,36,4818, 27, 36, 48 (indeed 27,36,4827, 36, 48 has ratio 43\frac{4}{3}), and the sum is 18+27+36+48=129.18 + 27 + 36 + 48 = 129.

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