If the sum of three numbers is equal to 18 and the sum of their squares is equal to 116, then the three numbers are equal______ .
Let the sequence be A-D, a, a + D. then we can get A-D + A + A + D = 18, and (A-D) 2 + A2 + (a + D) 2 = 116. The solution is a = 6, d = 2 & nbsp; or & nbsp; a = 6 & nbsp; d = - 2, so the three numbers are 4, 6, 8. So the answer is: 4, 6, 8
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