AP again!

Let $T_n$ be the n th term, a be the first term and d be the common difference.
So we get 2 equations, $T_{11}=a+10d$ and $T_{19}=a+18d.$
Now let $S_n$ be the sum of the first $n$ terms.
We get $S_{11}=11a+55d$ and $S_{19}=19a+171d.$ [According to the $A.P$ sum formula]
Since $S_{11}=S_{19}\Rightarrow 11a+55d=19a+171d\Rightarrow 2a+29d=0.$
$T_{30}=a+29d.$ Therefore $S_{30}=15(2a+29d)=15(0)=\boxed0.$

Let a = first term and d the difference between 2 consecutive terms. Then the 11th term is a + 10d, the 19th term is a + 18d. The sum of the first 11 terms = (11/2)*(a + a + 10d) = 11(a + 5d) = 11a + 55d. The sum of the first 19 terms is (19/2)(a + a + 18d) =19(a + 9d) = 19a + 171d. If these sums are equal, 11a + 55d = 19a + 171d, or 29d = -2a. The 30th term is a + 29d = a - 2a = -a. The sum of the first 30 terms = (30/2)( a + 29d) = 15(a + (-a)) = 0.