Which of the following form an $AP$? Justify your answer.
$(i)$ $11, 22, 33, \ldots$
$(ii)$ $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots$
$(iii)$ $2, 2^2, 2^3, 2^4, \ldots$
$(iv)$ $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots$

(I, IV) $(i)$ Here,$t_1 = 11, t_2 = 22, t_3 = 33$.
$t_2 - t_1 = 22 - 11 = 11$.
$t_3 - t_2 = 33 - 22 = 11$.
Since the common difference $d = 11$ is constant,this forms an $AP$.
$(ii)$ Here,$t_1 = \frac{1}{2}, t_2 = \frac{1}{3}, t_3 = \frac{1}{4}$.
$t_2 - t_1 = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$.
$t_3 - t_2 = \frac{1}{4} - \frac{1}{3} = -\frac{1}{12}$.
Since the differences are not equal,this does not form an $AP$.
$(iii)$ The sequence is $2, 4, 8, 16, \ldots$.
$t_2 - t_1 = 4 - 2 = 2$.
$t_3 - t_2 = 8 - 4 = 4$.
Since the differences are not equal,this does not form an $AP$.
$(iv)$ The sequence is $\sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3}, \ldots$.
$t_2 - t_1 = 2\sqrt{3} - \sqrt{3} = \sqrt{3}$.
$t_3 - t_2 = 3\sqrt{3} - 2\sqrt{3} = \sqrt{3}$.
$t_4 - t_3 = 4\sqrt{3} - 3\sqrt{3} = \sqrt{3}$.
Since the common difference $d = \sqrt{3}$ is constant,this forms an $AP$.