# Thread: Sequence and Series (guide me, thanks!)

1. ## Sequence and Series (guide me, thanks!)

In an arithmetic progression, the sum of the first n even number terms exceeds the sum of the first n odd number terms by 3n. Find the common difference of this series.

my working:

Sigma a(2r) - Sigma a(2r-1) = 2 . upper limit of sigma - n , lower limit - 1
Note that 2r and 2r-1 are both subscripts.

I have no idea what to do next, help me please, thank you!

2. 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?

Thanks.