Question

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


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Answer 1

Answer:

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3, 15, 27, 39…….

an=a+(n−1)d

a=3,d=15−3=12

n=54

a54=a+(n−1)d

=3+(54−1)×12

=3+53×12

=639

Term which is 132 more than its 54th term is – 639 +132 = 771.

an=771

771=a+(n−1)d

771=3+(n−1)12

771=3+(n−1)12

64=n−1

n=65

Answer 2

Answer:

3, 15, 27, 39…….

an=a(n−1)da=3,

d=15−3=12n=54a54=a+(n−1)d=3+(54−1)×12=3+53×12=639

Term which is 132 more than its 54th term is – 639 +132 = 771.771=a+(n−1)12

a+(n−1)d771=3+(n−1)12771=3+(n−1)1264=n−1n=65

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