Question

在一个由四个递增正整数组成的数列中，前三项成等差数列，后三项成等比数列，且第一项与第四项相差 $30$ 。求这四项之和。

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.

答案

答案：129

知识点：等差数列等比数列分类讨论

难度评级：2210

解答：

将四项写成 $a$ 、 $a + d$ 、 $a + 2d$ 和 $a + 30$ ，其中 $a$ 和 $d$ 是正整数。后三项成等比数列给出 $(a + 30)(a + d) = (a + 2d)^2$ 。展开并化简得即 $30a + 30d = 3ad + 4d^2,$ $3a(10 - d) = 2d(2d - 15).$

因为 $a, d \gt 0$ ，因子 $10 - d$ 和 $2d - 15$ 必须同号，故 $7.5 \lt d \lt 10$ ，所以 $d = 8$ 或 $d = 9$ 。当 $d = 8$ 时，得到 $6a = 16$ ，没有整数解。当 $d = 9$ 时，得到 $3a = 54$ ，所以 $a = 18$ 。

数列为 $18, 27, 36, 48$ （确实 $27, 36, 48$ 的公比为 $\frac{4}{3}$ ），其和为 $18 + 27 + 36 + 48 = 129$ 。

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, $30a + 30d = 3ad + 4d^2,$ that is $3a(10 - d) = 2d(2d - 15).$

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 第 8 题

答案：129

难度评级：2210

解答：

其他年份的第 8 题