Find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 3
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We have,
first term(a)=252
last term(an)=999
common differnce(d)=3
Now,
an=a+(n-1)d
999-252=(n-1)3
n=249+1
n=250
therefore,
sum=n/2{a+an]
=156375
first term(a)=252
last term(an)=999
common differnce(d)=3
Now,
an=a+(n-1)d
999-252=(n-1)3
n=249+1
n=250
therefore,
sum=n/2{a+an]
=156375
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