Question

Can the following expression be $n^{th}$ term of an AP?
If true then enter $1$ and if false then enter $0$.

$1+n+n^2$

Answer 1

We have
$a_n=1+n+n^2$

Changing to n to n+1 in $a_n$, we get
$a_{n+1}=1+(n+1)+(n+1{)}^2$
$\Rightarrow a_{n+1}=3+3n+n^2$

Common difference = $a_{n+1}-a_n=(3+3n+n^2)-(1+n+n^2)=2n+2$

since $a_{n+1}-a_n$ is depends upon $n$ and therefore not a constant.

So, the given sequence is not an AP