Let a (n) denote the n-th term in the progression. Consecutive terms in the sequence differ by a fixed constant - call it c - such that
a (n) = a (n - 1) + c
Let a = a (1) be the first term in the sequence. We can solve for a (n) in terms of a alone:
a (n) = a (n - 1) + c
a (n) = (a (n - 2) + c) + c = a (n - 2) + 2c
a (n) = (a (n - 3) + c) + c = a (n - 3) + 3c
and so on, down to
a (n) = a + (n - 1) c
The sum of the first n terms is then
[tex]\displaystyle\sum_{k=1}^n a(k)=\sum_{k=1}^n(a+(k-1)c)=a\sum_{k=1}^n1+c\sum_{k=1}^n(k-1)=an+\frac{cn(n-1)}2=\frac c2n^2+\left(a-\frac c2\right)n[/tex]
Since this is equal to 3n ^2 + 5n, it follows that
c/2 = 3 => c = 6
a - c/2 = 5 => a = 8
So the sequence is
a (n) = 8 + (n - 1) 6 = 6n + 2
