1973 AMC 12 第 26 题

先试着解答 1973 AMC 12 第 26 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 1973 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。

26.

一个等差数列的项数为偶数。奇数编号各项之和与偶数编号各项之和分别为 2424 与 3030。若末项比首项大 10.510.5,则该等差数列的项数为
The number of terms in an A.P. (Arithmetic Progression) is even. The sums of the odd- and even-numbered terms are 2424 and 30,30, respectively. If the last term exceeds the first by 10.5,10.5, the number of terms in the A.P. is
2020
1818
1212
1010
88
答案:E
知识点:等差数列方程组配对与分组
小提示:
将项数写成 2n2n,公差写成 dd
Write the number of terms as 2n2n and the common difference as dd
大提示:
将每个奇数编号项与紧随其后的偶数编号项配对,得到 nd=6nd=6
Pair each odd-numbered term with the following even-numbered term to get nd=6nd=6
解答:
设数列有 2n2n 项,公差为 dd。将每个奇数编号项与后一项配对,可得 nd=30−24=6。 nd=30-24=6\text{。}末项与首项之差为 (2n−1)d=10.5。 (2n-1)d=10.5\text{。}因为 2nd=122nd=12,相减得到 d=1.5d=1.5。于是 n=61.5=4n=\frac{6}{1.5}=4,该数列共有 2n=82n=8 项。
所以正确答案是 E。
Let the progression have 2n2n terms and common difference d.d. Pairing each odd-numbered term with its successor shows that nd=30−24=6. nd=30-24=6. The difference between the last and first terms is (2n−1)d=10.5. (2n-1)d=10.5. Since 2nd=12,2nd=12, subtraction gives d=1.5.d=1.5. Hence n=61.5=4,n=\frac{6}{1.5}=4, and the progression has 2n=82n=8 terms.
Therefore, the correct answer is E.
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