1969 AMC 12 第 33 题

先试着解答 1969 AMC 12 第 33 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 1969 AMC 12 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。

33.

设 SnS_n 与 TnT_n 分别为两个等差数列前 nn 项的和。若对所有 nn 都有 Sn:Tn=(7n+1):(4n+27)S_n:T_n=(7n+1):(4n+27),则第一个数列的第十一项与第二个数列的第十一项之比为:
Let SnS_n and TnT_n be the respective sums of the first nn terms of two arithmetic series. If Sn:Tn=(7n+1):(4n+27)S_n:T_n=(7n+1):(4n+27) for all n,n, the ratio of the eleventh term of the first series to the eleventh term of the second series is:
4:34:3
3:23:2
7:47:4
78:7178:71
无法确定
undetermined
答案:A
知识点:等差数列比与比例
小提示:
在等差数列中,前 2121 项的中间项等于这些项的平均数
In an arithmetic sequence, the middle term of the first 2121 terms equals their average
大提示:
将每个第十一项表示为相应的前 2121 项之和除以 2121
Express each eleventh term as its corresponding 2121-term sum divided by 2121
解答:
对等差数列而言,第十一项是前 2121 项的平均数。因此两个数列的第十一项分别为 S2121\frac{S_{21}}{21} 与 T2121\frac{T_{21}}{21}。它们的比为 S21T21=7(21)+14(21)+27=148111=43。 \frac{S_{21}}{T_{21}} =\frac{7(21)+1}{4(21)+27} =\frac{148}{111} =\frac43\text{。}
所以正确答案是 A。
For an arithmetic sequence, the eleventh term is the average of the first 2121 terms. Thus the respective eleventh terms are S2121\frac{S_{21}}{21} and T2121.\frac{T_{21}}{21}. Their ratio is S21T21=7(21)+14(21)+27=148111=43. \frac{S_{21}}{T_{21}} =\frac{7(21)+1}{4(21)+27} =\frac{148}{111} =\frac43.
Therefore, the correct answer is A.
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