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Question

Find the sum of all natural numbers from 100 to 300.
Which are divisible by 5.
Which are divisible by 6.
Which are divisible by 5 and 6.

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Solution

Numbers divisible by 5 between 100 and 300 are
105,110,295
a=100,d=5 last term l=295
tn=a+(n1)d295=100+(n1)×5
1955=n1
39=n1
n=39+1=40
Sum=n2(a+l)=402(105+295)
=20×400=8000

Numbers divisible by 6 are
102,108,114...294
a=102,l=294
and 294=102+(n+1)×6294102=(n1)×6
(n1)×6=192
n1=32
n=32+1=33
sum=332(102+294)=332×396=6534

Numbers divisible by both 5 and 6 are
120,150,180,210,240,270a=120,d=30
l=270
Also n=6
sum=62(120+270)=3×390=1170

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