Verify that each of the following is an $AP$,and then write its next three terms.
$\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, \ldots$

(N/A) Here,$a_{1}=\sqrt{3}, a_{2}=2 \sqrt{3}$ and $a_{3}=3 \sqrt{3}$.
Calculate the common difference:
$a_{2}-a_{1}=2 \sqrt{3}-\sqrt{3}=\sqrt{3}$
$a_{3}-a_{2}=3 \sqrt{3}-2 \sqrt{3}=\sqrt{3}$
Since $a_{2}-a_{1}=a_{3}-a_{2}=\sqrt{3}$,which is the common difference $(d)$,the given sequence forms an $AP$.
The next three terms are:
$a_{4}=a_{3}+d=3 \sqrt{3}+\sqrt{3}=4 \sqrt{3}$
$a_{5}=a_{4}+d=4 \sqrt{3}+\sqrt{3}=5 \sqrt{3}$
$a_{6}=a_{5}+d=5 \sqrt{3}+\sqrt{3}=6 \sqrt{3}$