Which of the following form an $AP$? Justify your answer.
$(i)$ $-1, -1, -1, -1, \ldots$
$(ii)$ $0, 2, 0, 2, \ldots$
$(iii)$ $1, 1, 2, 2, 3, 3, \ldots$

(A) $(i)$ Here,$t_{1} = -1, t_{2} = -1, t_{3} = -1$ and $t_{4} = -1$.
$t_{2} - t_{1} = -1 - (-1) = 0$
$t_{3} - t_{2} = -1 - (-1) = 0$
$t_{4} - t_{3} = -1 - (-1) = 0$
Since the common difference $d = 0$ is constant,the given list of numbers forms an $AP$.
$(ii)$ Here,$t_{1} = 0, t_{2} = 2, t_{3} = 0$ and $t_{4} = 2$.
$t_{2} - t_{1} = 2 - 0 = 2$
$t_{3} - t_{2} = 0 - 2 = -2$
Since $t_{2} - t_{1} \neq t_{3} - t_{2}$,the given list of numbers does not form an $AP$.
$(iii)$ Here,$t_{1} = 1, t_{2} = 1, t_{3} = 2$ and $t_{4} = 2$.
$t_{2} - t_{1} = 1 - 1 = 0$
$t_{3} - t_{2} = 2 - 1 = 1$
Since $t_{2} - t_{1} \neq t_{3} - t_{2}$,the given list of numbers does not form an $AP$.