Justify whether it is true to say that the following are the $n^{\text{th}}$ terms of an $AP.$
$(i)$ $2n-3$
$(ii)$ $3n^{2}+5$
$(iii)$ $1+n+n^{2}$

(N/A) $(i)$ Yes,here $a_{n}=2n-3$.
Put $n=1, a_{1}=2(1)-3=-1$.
Put $n=2, a_{2}=2(2)-3=1$.
Put $n=3, a_{3}=2(3)-3=3$.
Put $n=4, a_{4}=2(4)-3=5$.
The list of numbers is $-1, 1, 3, 5, \ldots$.
Here,$a_{2}-a_{1}=1-(-1)=2$,$a_{3}-a_{2}=3-1=2$,and $a_{4}-a_{3}=5-3=2$.
Since the common difference $d=2$ is constant,$2n-3$ is the $n^{\text{th}}$ term of an $AP$.
$(ii)$ No,here $a_{n}=3n^{2}+5$.
Put $n=1, a_{1}=3(1)^{2}+5=8$.
Put $n=2, a_{2}=3(2)^{2}+5=17$.
Put $n=3, a_{3}=3(3)^{2}+5=32$.
The list of numbers is $8, 17, 32, \ldots$.
Here,$a_{2}-a_{1}=17-8=9$ and $a_{3}-a_{2}=32-17=15$.
Since $a_{2}-a_{1} \neq a_{3}-a_{2}$,it does not form an $AP$.
$(iii)$ No,here $a_{n}=1+n+n^{2}$.
Put $n=1, a_{1}=1+1+(1)^{2}=3$.
Put $n=2, a_{2}=1+2+(2)^{2}=7$.
Put $n=3, a_{3}=1+3+(3)^{2}=13$.
The list of numbers is $3, 7, 13, \ldots$.
Here,$a_{2}-a_{1}=7-3=4$ and $a_{3}-a_{2}=13-7=6$.
Since $a_{2}-a_{1} \neq a_{3}-a_{2}$,it does not form an $AP$.