Question

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 $30.$ Find the sum of the four terms.

Answer

Solution:

Write the terms as $a,$ $a + d,$ $a + 2d,$ and $a + 30,$ where $a$ and $d$ are positive integers. The geometric condition on the last three terms says $(a + 30)(a + d) = (a + 2d)^2.$ Expanding both sides and simplifying, $30a + 30d = 3ad + 4d^2, \qquad \text{that is} \qquad 3a(10 - d) = 2d(2d - 15).$

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

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

2003 AIME I Problem 8

Solution:

Problem 8 in Other Years