2003 AMC 12B 第 6 题

先试着解答 2003 AMC 12B 第 6 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2003 AMC 12B 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

6.

一个等比数列的第二项和第四项分别是 22 和 66。下列哪一项可能是第一项?
The second and fourth terms of a geometric sequence are 22 and 6.6. Which of the following is a possible first term?
−3-\sqrt{3}
−233-\dfrac{2\sqrt{3}}{3}
−33-\dfrac{\sqrt{3}}{3}
3\sqrt{3}
33
答案:B
知识点:等比数列根式
小提示:
若第一项为 aa,公比为 rr,则 ar=2ar = 2 且 ar3=6ar^3 = 6
If the first term is aa and the ratio is r,r, then ar=2ar = 2 and ar3=6ar^3 = 6
大提示:
两式相除得到 r2=3r^2 = 3,所以 r=±3r = \pm\sqrt{3}
Dividing the two equations gives r2=3,r^2 = 3, so r=±3r = \pm\sqrt{3}
解答:
设第一项为 aa,公比为 rr,则 ar=2ar = 2,且 ar3=6ar^3 = 6,因此 r2=3r^2 = 3,r=±3r = \pm\sqrt{3}。
第一项为 a=2r=±23=±233。 a = \frac{2}{r} = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}\text{。} 选项中出现的是 −233-\dfrac{2\sqrt{3}}{3}。
因此,正确答案是 B。
Let the first term be aa and the common ratio r.r. Then ar=2ar = 2 and ar3=6,ar^3 = 6, so r2=3r^2 = 3 and r=±3.r = \pm\sqrt{3}.
The first term is a=2r=±23=±233. a = \frac{2}{r} = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}. The choice −233-\dfrac{2\sqrt{3}}{3} appears among the options.
Thus, the correct answer is B.
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