Solution: First look at the functional characteristics of the sum of the first n terms of arithmetic progression:

Sn = n a 1+ n(n- 1)d/2

= n a 1 + (d/2)(n？ - n)

= (d/2)n？ +(a 1-d/2)n (quadratic function with n as unknown and constant term 0)

= Ann? +Bn (let d/2 = a; a 1 -d/2 = B)

Theorem: A necessary and sufficient condition for the sequence {an} to be arithmetic progression is that the sum of the first n terms is Sn = An? +Bn form.

Multiply a = d/2; B= a 1 -d/2 available.

d = 2A； a 1=A+B .

According to the above theorem and conclusion, in your question, A=4 and B= 1, so a 1 = A+B = 5 and d =2A =8.

Therefore, an = a1+(n-1) d.

= 5 + 8(n- 1)

= 8n -3