The next term of the A.P. $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}$, … is:

1. $\sqrt{72}$

2. $\sqrt{68}$

3. $\sqrt{75}$

4. $\sqrt{56}$

In the above A.P. next terms is the 5th term.

a = $\sqrt{3}$

n = 5

$\Rightarrow d = \sqrt{12} - \sqrt{3} \\[1em] \Rightarrow d = \sqrt{4 \times 3} - \sqrt3 \\[1em] \Rightarrow d = 2\sqrt{3} - \sqrt3 \\[1em] \Rightarrow d = \sqrt3.$

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

$\Rightarrow T_5 = \sqrt3 + (5 - 1)\sqrt3 \\[1em] = \sqrt3 + 4\sqrt3 \\[1em] = 5\sqrt3 \\[1em] = \sqrt{5^2 \times 3} \\[1em] = \sqrt{75}.$

Hence, option 3 is the correct option.