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Question

Find the sum of all 3- digit natural numbers, which are multiples of 11.

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Solution

First three digit number divisible by 11 is 110
Last three digit numbers divisible by 11 is 990
a=110 an=990 d=11 x=2
No. of =2
Terms
an=a+(n1)d
990=110+(n1).11
88011=x1
80=x1
x=81
sum of all terms of A.P is given by
Sn=[a+an]×n2
=[110+990]×812
=1100×812
=550×81
=44550

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