Question

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 $24$ 与 $30$ 。若末项比首项大 $10.5$ ，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 $24$ and $30,$ respectively. If the last term exceeds the first by $10.5,$ the number of terms in the A.P. is

$20$

$18$

$12$

$10$

$8$

答案：E

知识点：等差数列方程组配对与分组

难度评级：1850

小提示：

将项数写成 $2n$ ，公差写成 $d$

Write the number of terms as $2n$ and the common difference as $d$

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 $nd=6$

Pair each odd-numbered term with the following even-numbered term to get $nd=6$

解答：

设数列有 $2n$ 项，公差为 $d$ 。将每个奇数编号项与后一项配对，可得 $nd=30-24=6\text{。}$ 末项与首项之差为 $(2n-1)d=10.5\text{。}$ 因为 $2nd=12$ ，相减得到 $d=1.5$ 。于是 $n=\frac{6}{1.5}=4$ ，该数列共有 $2n=8$ 项。

所以正确答案是 E。

Let the progression have $2n$ terms and common difference $d.$ Pairing each odd-numbered term with its successor shows that $nd=30-24=6.$ The difference between the last and first terms is $(2n-1)d=10.5.$ Since $2nd=12,$ subtraction gives $d=1.5.$ Hence $n=\frac{6}{1.5}=4,$ and the progression has $2n=8$ terms.

Therefore, the correct answer is E.

1973 AMC 12 第 26 题

答案：E

难度评级：1850

小提示：

大提示：

解答：

其他年份的第 26 题