Question

Show that a sequence is an AP if its nth term is a linear expression in n and in sucha a case the common difference is equal to the coefficient of n.

Answer 1

Let there be a sequence whose nth term is a linear expression in n.

$a_n=An+B$, where A, B are constants.

$a_{n+1}=A(n+1)+B$

Therefore,

$a_{n+1}-a_n=A$

It is independent of n and therefore a constant.

So, the sequence is an AP with common difference A.