find the sum of all the natural numbers between hundred- 1000 which are multiples of 3
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first multiple=102
last multiple=999
n=(999-102)/3+1=299+1=300
S=(999+102)*(300/2)
=165150
last multiple=999
n=(999-102)/3+1=299+1=300
S=(999+102)*(300/2)
=165150
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