Let, the first 3 terms of the A.P. are a−d,a,a+d
⇒sum=3a=33⇒a=11
And product (a−d)a(a+d)=a(a2−d2)=1155
⇒11(112−d2)=1155
⇒121−d2=111155=105
⇒d2=121−105=16
⇒d=±4
For a=11,d=4, the A.P. is 7,11,15,…
The nth term of an A.P., with first term A and common difference d is Tn=A+(n−1)d
So, T11=7+10×4=47
For a=11,d=−4, the A.P. is 15,11,7,…
So, T11=15+10(−4)=−25.