Let the ( n^{th} ) term be a linear expression in ( n ): [ a_n = pn + q ] where ( p, q ) are constants.

Then [ a_{n+1} = p(n+1) + q = pn + p + q ]

Common difference: [ a_{n+1} - a_n = (pn + p + q) - (pn + q) = p ]

which is constant (independent of ( n )).

Therefore the progression is an arithmetic progression.

Proved.