Find the sum of all 3 digit natural numbers,which are divisible by 7
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1st 3 digit no. divisible by 13=104
last 3 digit no. divisible by 13=988
l=a+(n-1)d
988=104+(n-1)13
884/13=n-1
68+1=n
n=69
sn=n(a+l)/2
=69(1092)/2
=69*546
=37674
last 3 digit no. divisible by 13=988
l=a+(n-1)d
988=104+(n-1)13
884/13=n-1
68+1=n
n=69
sn=n(a+l)/2
=69(1092)/2
=69*546
=37674
yashlevel2:
hey the question is divisible by 7
Yashlevel12 is correct
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