# 2000 AMC 12 Problems/Problem 14

The following problem is from both the 2000 AMC 12 #14 and 2000 AMC 10 #23, so both problems redirect to this page.

## Problem

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 $x$? $\text {(A)}\ 3 \qquad \text {(B)}\ 6 \qquad \text {(C)}\ 9 \qquad \text {(D)}\ 17 \qquad \text {(E)}\ 20$

## 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 $3 + 17 = 20\ \mathrm{(E)}$.

## Video Solution

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 