2000 AMC 10 Problem 23

The mode is always $2,$ and the mean is $\mu = \dfrac{25+x}{7}.$ If $x \leq 2,$ the median is also $2.$ Two equal values cannot belong to a non-constant three-term arithmetic progression, so this case gives no solutions. If $2 < x < 4,$ the values occur in the order $2 < x < \mu.$ They form an arithmetic progression exactly when $x-2 = \mu-x.$ Substituting for $\mu$ gives $$x-2 = \dfrac{25+x}{7}-x,$$ whose solution is $x=3.$ If $x \geq 4,$ the median is $4$ and $\mu>4.$ Thus the condition is $$4-2 = \mu-4,$$ so $\mu=6$ and $x=17.$ The two possible values are therefore $3$ and $17,$ whose sum is $20.$ Thus, the correct answer is E.