Question

设 $S_n$ 与 $T_n$ 分别为两个等差数列前 $n$ 项的和。若对所有 $n$ 都有 $S_n:T_n=(7n+1):(4n+27)$ ，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 $S_n$ and $T_n$ be the respective sums of the first $n$ terms of two arithmetic series. If $S_n:T_n=(7n+1):(4n+27)$ for all $n,$ the ratio of the eleventh term of the first series to the eleventh term of the second series is:

$4:3$

$3:2$

$7:4$

$78:71$

无法确定

undetermined

答案：A

知识点：等差数列比与比例

难度评级：1840

小提示：

在等差数列中，前 $21$ 项的中间项等于这些项的平均数

In an arithmetic sequence, the middle term of the first $21$ terms equals their average

大提示：

将每个第十一项表示为相应的前 $21$ 项之和除以 $21$

Express each eleventh term as its corresponding $21$ -term sum divided by $21$

解答：

对等差数列而言，第十一项是前 $21$ 项的平均数。因此两个数列的第十一项分别为 $\frac{S_{21}}{21}$ 与 $\frac{T_{21}}{21}$ 。它们的比为 $\frac{S_{21}}{T_{21}} =\frac{7(21)+1}{4(21)+27} =\frac{148}{111} =\frac43\text{。}$

所以正确答案是 A。

For an arithmetic sequence, the eleventh term is the average of the first $21$ terms. Thus the respective eleventh terms are $\frac{S_{21}}{21}$ and $\frac{T_{21}}{21}.$ Their ratio is $\frac{S_{21}}{T_{21}} =\frac{7(21)+1}{4(21)+27} =\frac{148}{111} =\frac43.$

Therefore, the correct answer is A.

1969 AMC 12 第 33 题

答案：A

难度评级：1840

小提示：

大提示：

解答：

其他年份的第 33 题