# 2000 AMC 10 Problems/Problem 23

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

## Solution

As $2$ occurs three times and each of the three other values just once, regardless of what $x$ we choose the mode will always be $2$.

The sum of all numbers is $25+x$, therefore the mean is $\frac {25+x}7$.

The six known values, in sorted order, are $2,2,2,4,5,10$. From this sequence we conclude: If $x\leq 2$, the median will be $2$. If $2, the median will be $x$. Finally, if $x\geq 4$, the median will be $4$.

We will now examine each of these three cases separately.

In the case $x\leq 2$, both the median and the mode are 2, therefore we can not get any non-constant arithmetic progression.

In the case $2 we have $x < \frac {25+x}7$, because $\frac {25+x}7 - x = \frac{25-6x}7 > \frac{25-6\cdot 4}7 > 0$. Therefore our three values in order are $2,x,\frac {25+x}7$. We want this to be an arithmetic progression. From the first two terms the difference must be $x-2$. Therefore the third term must be $x+(x-2)=2x-2$.

Solving $2x-2 = \frac {25+x}7$ we get the only solution for this case: $x=3$.

The case $x\geq 4$ remains. Once again, we have $\frac {25+x}7 \geq \frac{25+4}7 > 4$, therefore the order is $2,4,\frac {25+x}7$. The only solution is when $6=\frac {25+x}7$, i. e., $x=17$.

The sum of all solutions is therefore $3+17=\boxed{20}$.