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MySolution Using (1) (a+b+c+4)/4 > (a +b+c)/3 or a+b+c<12 No clue about a, b, c. Insufficient
Using (2) four possibilities: (a, b, c, 4), median (b+c)/2 < b or c < b, not possible (4, a, b, c), median (b+a)/2 < b or b > a, already known (a, 4, b, c), median (b+4)/2 < b or b > 4 (a, b, 4, c), median (b+c)/2 < b or b > 4 Considering all the above cases: b > 4 But still no clue about a & c. Insufficient.
Using (1) & (2), b_m_i_n = 5, (a, b, c are integers) assuming all are in A.P. which means, a = b - k c = b + k from (1), a+b+c < 12 or (b-k)+b+(b+k) < 12 or3b < 12 or b < 4 but b > 4 (from (2)), thus the assumption is wrong. and a, b, c cannot be in AP