1961 AMC 12 第 26 题

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26.

某等差数列的前 5050 项之和为 200200,接下来的 5050 项之和为 27002700。该数列的首项为:

For a given arithmetic series the sum of the first 5050 terms is 200,200, and the sum of the next 5050 terms is 2700.2700. The first term in the series is:

1212-12\dfrac12

21.5-21.5

20.5-20.5

33

3.53.5

答案:C
知识点:等差数列求和方程组
难度评级:1280
小提示:

用首项 aa 和公差 dd 表示第一个和

Write the first sum in terms of the first term aa and common difference dd

大提示:

5050 项的首项为 a+50da+50d

The next 5050 terms have first term a+50da+50d

解答:

两个和给出 25(2a+49d)=200,25(2a+149d)=2700 \begin{aligned} 25(2a+49d)&=200,\\ 25(2a+149d)&=2700 \end{aligned}\text{。} 因此 2a+49d=82a+49d=82a+149d=1082a+149d=108。相减得 d=1d=1,继而 2a=412a=-41,所以 a=20.5a=-20.5

因此,正确答案是 C

The two sums give 25(2a+49d)=200,25(2a+149d)=2700. \begin{aligned} 25(2a+49d)&=200,\\ 25(2a+149d)&=2700. \end{aligned} Thus 2a+49d=82a+49d=8 and 2a+149d=108.2a+149d=108. Subtraction gives d=1,d=1, and then 2a=41,2a=-41, so a=20.5.a=-20.5.

Therefore, the correct answer is C.

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