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2000 AMC 12 Problems/Problem 14

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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, which results in a contradiction.
  • If the median is 4 (x \ge 4), then 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.
  • If the median is x (2 \le x \le 4), then the mean can either be 1, 5/2, 4 to form an arithmetic progression. Solving for x yields -14,-7.5, 3 respectively, of which only 3 works.

The answer is 3 + 17 = 20\ \mathrm{(E)}.

See also

2000 AMC 12 (ProblemsResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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