2000 AMC 10 Problem 23

Question

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?

3

6

9

17

20

The mode is

2

and the mean is

\dfrac{25 + x}{7};

the median depends on where

x

falls

Split into cases by the size of

x,

and require the three ordered values to be equally spaced

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.

Problem 23 in other years