If a is the first term, and d the constant difference between terms, then the sum, S=a+a+d+a+2d+....
The first term is repeated n times, so that gives us na; the d terms sum to d times the sum of the natural numbers, which is n(n-1)/2 giving us dn(n-1)/2:
S=na+dn(n-1)/2 must equal the given expression 4n^2-3n/4 and so, dividing through by n, we get a+dn/2-d/2=4n-3/4.
Equating terms: (n term) d/2=4, so d=8 and (constant term) a-d/2=a-4=-3/4, making a=13/4.
The general term is 13/4+8n.
CHECK: series is 13/4, 13/4+8, 13/4+16, 13/4+24, ... or 13/4, 45/4, 77/4, ...
Sum is 13n/4+8n(n-1)/2=13n/4+4n^2-4n=4n^2-3n/4