Given an arithmetic sequence \ with a common difference of \, if \, \, and \ form a geometric sequence, then \a_2=\ \_\_\_\_\_\_.
Problem 14
Official solution
Since the common difference of the arithmetic sequence \ is \, and \, \, \ form a geometric sequence,
we have \(a_1+6)^2=a_1(a_1+9)\.
Thus, \a_1=-12\,
and therefore \a_2=-9\.
So, the answer is: \\boxed{-9}\.
From the condition \(a_1+6)^2=a_1(a_1+9)\, we deduce that \a_1=-12\, which leads to the conclusion.
This problem tests the general term of an arithmetic sequence and involves the application of the geometric mean, classified as a medium-level question.