AMC12 2004 A

AMC12 2004 A · Q14

AMC12 2004 A · Q14. 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 in the geometric progression?

三个实数构成一个等差数列，第一项为 $9$。如果把第二项加 $2$，把第三项加 $20$，则得到的三个数构成一个等比数列。求该等比数列第三项的最小可能值。

(A) 1 1

(B) 4 4

(D) 49 49

(E) 81 81

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $d$ be the common difference. Then $9$, $9+d+2=11+d$, $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$. The smallest possible value occurs when $d = -14$, and the third term is $2(-14) + 29 = 1\Rightarrow\boxed{\mathrm{(A)}\ 1}$.

设公差为 $d$。则等比数列的三项为 $9$，$9+d+2=11+d$，$9+2d+20=29+2d$。由于中项是两端项的几何平均数，$(11+d)^2 = 9(2d+29)$ $\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$。最小可能值在 $d = -14$ 时取得，此时第三项为 $2(-14) + 29 = 1\Rightarrow\boxed{\mathrm{(A)}\ 1}$。

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