which term of the AP : 3,15,27,39.... will be 132 more than its 54th term
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d=12
a= 3
n=54
Therefore,
an=a+(n-1)d
a 54= 3+53x 12
a 54=3+636
So, a 54=639.
Therefore, 132 more than 639= 639+132=771
Now,
an= 771
So, 771=3+(n-1)12
771=3+12n-12
768=12n- 12
780=12n
So, n=65.
Done:) Plz comment positively
a= 3
n=54
Therefore,
an=a+(n-1)d
a 54= 3+53x 12
a 54=3+636
So, a 54=639.
Therefore, 132 more than 639= 639+132=771
Now,
an= 771
So, 771=3+(n-1)12
771=3+12n-12
768=12n- 12
780=12n
So, n=65.
Done:) Plz comment positively
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