AMC10 2004 A

AMC10 2004 A · Q18

AMC10 2004 A · Q18. It mainly tests Quadratic equations, Sequences & recursion (algebra).

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?

三个实数的等差数列，第一项为 9。如果向第二项加 2，向第三项加 20，则所得三个数形成等比数列。几何数列第三项的最小可能值为多少？

(A) 1 1

(B) 4 4

(D) 49 49

(E) 81 81

Answer

Correct choice: (A)

正确答案：（A）

Solution

The terms of the arithmetic progression are 9, $9 + d$, and $9 + 2d$ for some real number $d$. The terms of the geometric progression are 9, $11+d$, and $29+2d$. Therefore $(11 + d)^2 = 9(29 + 2d)$ so $d^2 + 4d -140 = 0$. Thus $d = 10$ or $d = -14$. The corresponding geometric progressions are 9, 21, 49 and 9, −3, 1, so the smallest possible value for the third term of the geometric progression is 1.

等差数列的项为 9、$9 + d$ 和 $9 + 2d$，其中 $d$ 为实数。等比数列的项为 9、$11+d$ 和 $29+2d$。因此 $(11 + d)^2 = 9(29 + 2d)$，即 $d^2 + 4d -140 = 0$。解得 $d = 10$ 或 $d = -14$。对应的等比数列为 9, 21, 49 和 9, −3, 1，因此几何数列第三项的最小可能值为 1。

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