$a_n = 4n - 11$, finding $d$ in sequence

Question

I was given the information that the last member of an arithmetic progression is: $$a_n = 4n - 11$$ I need to prove that this is an arithmetic progression and then find the equation for $S_n$ (sum). Therefore I guess my problem is to find $d$, that represents the common difference of the successive members.

Answer 1

Isn't $a_{n-1}=4(n-1)-11$ and then you can check the difference between $a_n - a_{n-1}$?

So $d=a_{n}-a_{n-1}=4n-11-4n+11+4=4$. So the difference is without $n$, so this is an arithmetic progression with $d=4$.

And now you can just put all the information to the known formula for the sum.