[I guess the original question has been rephrased: 4n^2-3n/4 was meant to be (4n^2-3n)/4 (=4n^2/4-3n/4).]
If a is the first term, and d the constant difference between terms, then the sum, S=a+a+d+a+2d+....
There are n a's making the sum of the a's na; and the d's form a sum d(1+2+3+...), which is d times the sum of the natural number up to n=n(n-1)/2, so:
S=na+dn(n-1)/2 has to be equal to the given expression 4n^2/4-3n/4=n^2-3n/4 and a+dn/2-d/2=n-3/4 (dividing through by n).
Equating terms: n term: d/2=1, so d=2; constant term: a-d/2=a-1=-3/4, making a=1/4.
The general term is 1/4+2n.
CHECK: series is 1/4, 1/4+2, 1/4+4, 1/4+6,... or 1/4, 9/4, 17/4, ...
Sum is n/4+2n(n-1)/2=n/4+n^2-n=n^2-3n/4
If you had any queries about the logic in the original question, I hope my expansion of it helped.
