# Prove that the coefficient of $n$ is the common difference.

Suppose the sequence <$a_n$> is an Arithmetic Progression if its $n^{th}$ term is a linear expression in $n$ then show that common difference is equal to the coefficient of $n$.

• I can’t proceed on any result except ... Isn't that precisely what a "linear expression in $n$" means? – dxiv Jan 16 '17 at 3:27

## 1 Answer

Let $$a_n=An+B$$then$$a_{n+1}=A(n+1)+B$$now$$a_{n+1}-a_n=An+A+B-An-B$$ $$=A$$ Hope it helps!!!