I seem to have done it but not with the working shown in the first post.
Sum of first n even number terms = (a+d) + (a+3d) + (a+5d) + ... + a + ( 2n-1 ) d -----------1
Sum of first n odd number terms = (a) + ( a+2d) + (a+4d) +... +a + ( 2n-2)d ---------------- 2
Notice that the series has n terms, therefore , the number a of in both sum are equal.
However, the number of d in both sum are different such that 1 is more than 2 for nd ( since a2-a1 = d a4-a3 =d a6-a5=d .... an - an-1 = d and there is a total of n terms, therefore , the sum of first n even number terms exceeds the sum of first n odd number terms by nd. )
Given that 1 -2 = 3n and what I found , 1 - 2 = nd ,
Therefore, 3n = nd
d = 3
Any better way or any correction in my working here?