Question

三个实数组成一个等差数列，第一项为 $9$ 。若第二项加 $2$ ，第三项加 $20$ ，所得三个数形成等比数列。该等比数列第三项的最小可能值是多少？

A sequence of three real numbers forms an arithmetic progression with a first term of $9.$ If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?

$1$

$4$

$36$

$49$

$81$

答案：A

知识点：等差数列等比数列二次方程

难度评级：1630

解答：

等差数列为 $9$ 、 $9 + d$ 、 $9 + 2d$ ，所以新的三个数为 $9$ 、 $11 + d$ 、 $29 + 2d$ 。

等比条件给出化简得 $d^2 + 4d - 140 = 0$ ，所以 $d = 10$ 或 $d = -14$ 。 $(11 + d)^2 = 9(29 + 2d),$

第三项 $29 + 2d$ 分别为 $49$ 和 $1$ 。最小是 $1$ 。

所以正确答案是 A。

The arithmetic progression is $9,$ $9 + d,$ $9 + 2d,$ so the geometric progression is $9,$ $11 + d,$ $29 + 2d.$

The geometric condition gives $(11 + d)^2 = 9(29 + 2d),$ which simplifies to $d^2 + 4d - 140 = 0,$ so $d = 10$ or $d = -14.$

The third terms $29 + 2d$ are $49$ and $1.$ The smallest is $1.$

Thus, the correct answer is A.

2004 AMC 10A 第 18 题

答案：A

难度评级：1630

解答：

其他年份的第 18 题