Sequences的終極難題

Let d be common difference of the arithmetic sequence.
Let the three integers be a - d, a and a+d
(a-d)(a)(a+d) = 11[a-d+a+a+d]
a(a^2 - d^2) = 11(3a)
a(a^2 - d^2) = 33a
a = 0 (rejected, since the three integers are all positive) or a^2 - d^2 = 33
a^2 - d^2 = 33
As the three terms are all integers, by trial and error, we get a = 7, and d = 4
Hence the three terms are 3, 7 and 11


2007-11-10 20:28:12 補充：
Note that the value of d can also be -4, and the answer are term same...but the answer of a cannot be -7, since the three terms must be positive integers