Given A.P. is 3,15,27,39,…a=3d=a2​−a1​=15−3=12a54​=a+(54−1)d=3+(53)(12)=3+636=639Then, 132 more than its 54th term is 132+639=771We have to find the term of this A.P. which is 771.Let nth term be 771.⇒an​=a+(n−1)d⇒771=3+(n−1)12⇒768=(n−1)12⇒ (n−1)=64∴n=65Therefore, 65th term was 132 more than 54th term.