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Bunuel
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1) c-a=2(c-b)
(c-a)/2 = b , which means b is average mean of a and c, and also is median.
2) first case: a+b+c=3a (Not possible because for that to happen b and c either need to be equal to a or less than a)
second case: a+b+c=3b
a+c=2b (again, it means b is average mean of a and c)
third case: a+b+c=3c (Not possible since for that to happen a and b needs to be at least equal to c, which is not true)
Hence either of the statement is sufficient.
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Hi Bunuel! THank you for the question, it's a really interesting one! My only concern is that (2) in 2 cases contradicts the initial statement in the question, because if a < b < c, then a + b + c != 3a or 3c. Is there a way to rephrase the question to eliminate this confusion?
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Hi Bunuel! THank you for the question, it's a really interesting one! My only concern is that (2) in 2 cases contradicts the initial statement in the question, because if a < b < c, then a + b + c != 3a or 3c. Is there a way to rephrase the question to eliminate this confusion?

Yes, because 0 < a < b < c, then from (2) we get that a + b + c = 3b only but where is the contradiction here?
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