find the sum of all 3digit natural number which are multiples of 11
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Answer:
a=110 an=990 d=11 x=2
No. of =2
an=a+(n−1)d
990=110+(n−1).11
880/11=x−1
80=x−1
x=81
Sn=[a+an]× n/2
=[110+990]× 81/2
=1100× 81/2
=550×81
=44550.
So ans is 44550.
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