AMC12 2000 A

AMC12 2000 A · Q14

AMC12 2000 A · Q14. It mainly tests Averages (mean), Patterns & sequences (misc).

When the mean, median, and mode of the list \[10,2,5,2,4,2,x\] are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of $x$?

当列表 \[10,2,5,2,4,2,x\] 的均值、中位数和众数按从小到大排列时，它们构成一个非恒定的等差数列。所有可能的实数 $x$ 的和是多少？

(A) 3 3

(B) 6 6

(D) 17 17

(E) 20 20

Answer

Correct choice: (E)

正确答案：（E）

Solution

- The mean is $\frac{10+2+5+2+4+2+x}{7} = \frac{25+x}{7}$. - Arranged in increasing order, the list is $2,2,2,4,5,10$, so the median is either $2,4$ or $x$ depending upon the value of $x$. - The mode is $2$, since it appears three times. We apply casework upon the median: - If the median is $2$ ($x \le 2$), then the arithmetic progression must be constant. - If the median is $4$ ($x \ge 4$), because the mode is $2$, the mean can either be $0,3,6$ to form an arithmetic progression. Solving for $x$ yields $-25,-4,17$ respectively, of which only $17$ works because it is larger than $4$. - If the median is $x$ ($2 \le x \le 4$), we must have the arithmetic progression $2, x, \frac{25+x}{7}$. Thus, we find that $2x=2+\frac{25+x}{7}$ so $x=3$. The answer is $3 + 17 = \boxed{20\ \mathrm{(E)}}$.

- 均值为 $\frac{10+2+5+2+4+2+x}{7} = \frac{25+x}{7}$. - 按从小到大排列，该列表为 $2,2,2,4,5,10$，因此中位数根据 $x$ 的取值可能是 $2,4$ 或 $x$。 - 众数为 $2$，因为它出现了三次。对中位数分类讨论： - 若中位数为 $2$（$x \le 2$），则等差数列必须是常数列。 - 若中位数为 $4$（$x \ge 4$），由于众数为 $2$，均值可以为 $0,3,6$ 以构成等差数列。分别解得 $x$ 为 $-25,-4,17$，其中只有 $17$ 符合因为它大于 $4$。 - 若中位数为 $x$（$2 \le x \le 4$），则等差数列必须为 $2, x, \frac{25+x}{7}$。因此有 $2x=2+\frac{25+x}{7}$，解得 $x=3$。答案为 $3 + 17 = \boxed{20\ \mathrm{(E)}}$.

Topics