An arithmetic sequence has first term a a\,a and common difference ddd. The sum of the first 25 terms of the sequence is equal to the square of the sum of the first 10 terms.
Show that 4a2+36ad+81d2=a+12d4a^2 + 36ad + 81d^2 = a + 12d4a2+36ad+81d2=a+12d.
Given that the 10th term of the sequence is 2, find the possible values of aaa.