find the 10 term of the given A.P 3,14,25,36,......
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3,14,25,36,47,58,69,80,91,102,113,124,135,146,157,168,.......
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39,.... will be 132 more than its 54th term?
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3, 15, 27, 39…….
a
n
=a+(n−1)d
a=3,d=15−3=12
n=54
a
54
=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.
a
n
=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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