When the [mean](https://artofproblemsolving.com/wiki/index.php/Mean), [median](https://artofproblemsolving.com/wiki/index.php/Median), and [mode](https://artofproblemsolving.com/wiki/index.php/Mode) of the list
\[10,2,5,2,4,2,x\]
are arranged in increasing order, they form a non-constant [arithmetic progression](https://artofproblemsolving.com/wiki/index.php/Arithmetic_progression). What is the sum of all possible real values of $x$?
$\text {(A)}\ 3 \qquad \text {(B)}\ 6 \qquad \text {(C)}\ 9 \qquad \text {(D)}\ 17 \qquad \text {(E)}\ 20$
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 $3 + 17 = 20\ \mathrm{(E)}$.