1952 AMC 12 第 30 题

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30.

若一个等差数列前十项之和是前五项之和的四倍,则首项与公差之比为:

When the sum of the first ten terms of an arithmetic progression is four times the sum of the first five terms, the ratio of the first term to the common difference is:

1:21:2

2:12:1

1:41:4

4:14:1

1:11:1

答案:A
知识点:等差数列求和比与比例
难度评级:1590
小提示:

用首项 aa 和公差 dd 表示两个部分和

Write both partial sums in terms of the first term aa and common difference dd

大提示:

使用 Sn=n2(2a+(n1)d)S_n=\dfrac n2(2a+(n-1)d),并化简 S10=4S5S_{10}=4S_5

Use Sn=n2(2a+(n1)d)S_n=\dfrac n2(2a+(n-1)d) and simplify S10=4S5S_{10}=4S_5

解答:

求和公式给出 S10=5(2a+9d),S5=52(2a+4d) \begin{aligned} S_{10}&=5(2a+9d),\\ S_5&=\frac52(2a+4d) \end{aligned}\text{。}方程 S10=4S5S_{10}=4S_5 化为 2a+9d=2(2a+4d) 2a+9d=2(2a+4d)\text{,}所以 d=2ad=2a。因此 a:d=1:2a:d=1:2

因此,正确答案是 A

The sum formula gives S10=5(2a+9d),S5=52(2a+4d). \begin{aligned} S_{10}&=5(2a+9d),\\ S_5&=\frac52(2a+4d). \end{aligned} The equation S10=4S5S_{10}=4S_5 becomes 2a+9d=2(2a+4d), 2a+9d=2(2a+4d), so d=2a.d=2a. Hence a:d=1:2.a:d=1:2.

Thus, the correct answer is A.

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