If three numbers form an arithmetic sequence, the sum of them is equal to 12, and the product of the beginning and the end is 12, find the three numbers
The middle 12 △ 3 = 4
(4-d)(4+d)=12
16-d²=12
d²=4
d=±2
So it's 2 4 6 or 6 4 2
In the arithmetic sequence 3, 12, 21, 30, 39, 48 912 is the second______ The number of them
(912-3) △ 9 + 1, = 909 △ 9 + 1, = 101 + 1, = 102; so the answer is: 102
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