Official solution
Let the three terms are a - d, a, a + d
Given a - d + a + a + d = 33
\[ \Rightarrow \] 3a = 33
\[ \Rightarrow \] a = 11
Also given,
(a - d)a(a + d) = 1155
\[ \Rightarrow \] (a2 - d2)a = 1155
\[ \Rightarrow \] (112 - d2)11 = 1155
\[ \Rightarrow \] (112 - d2) = 105
\[ \Rightarrow \] d = \[ \pm \] 4
When d = 4 and a = 11 then series is
7, 11, 15, ....
\[ \therefore \] T11 = a + 10d = 7 + 10\[ \times \] 4 = 47
When d = -4 and a = 11 then series is
15, 11, 7, ....
\[ \therefore \] T11 = a + 10d = 15 + 10\[ \times \] -4 = -25