Three real numbers form an arithmetic sequence, and their sum is 12, and their product is 48

Let the sequence be A-D, a, a + D. then we can get A-D + A + A + D = 12, and (A-D) a (a + D) = 48. The solution is a = 4, d = 2 & nbsp; or & nbsp; a = 4 & nbsp; d = - 2, and the sequence is 2, 4, 6 or 6, 4, 2

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