Three numbers are known to be equal difference sequence, and their sum is 12 and product is 60
Let the middle number be a, then the first number be A-D, the third number be a + D, a + (A-D) + (a + D) = 12, and a = 4
A (A-D) (a + D) = 60, D1 = 1, D2 = - 1, when d = 1, a + D = 5, A-D = 3, when d = - 1, a + D = 3, A-D = 5
To sum up, the three numbers are 3, 4 and 5 respectively