Which term of the AP 3,15,27,39,..... will be 132 more than its 54th term?

Given AP:

3,15,27,39,….

a<sub>n</sub>=a+(n-1)d

a=3

d=15-3=12

n=54

a<sub>54</sub>=a+(n-1)d

=3+(54-1)x12

=3+53×12

=639

Term which is 132 more than its 54th terms is -639+132=771

a<sub>n</sub>=771

771=a+(n-1)d

771=3+(n-1)12

771=3+(n-1)12

64=n-1

n=65

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