Question 258336
Let the first term of an arithmetic sequence be a.
Given,the common difference is twice the first term.
So the common difference d = 2a
Also, the sum of the first six terms is equal to the square of the first term. 
     (6/2)(a+a+5d) = a^2
      3(2a+5*2a) = a^2
      3(12a)=a^2
      36a-a^2=0
      a(36-a) = 0
      a=0  or  a=36
Therefore the first term is 36