Fimd sum of natural no less than 100divisible by 5
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Step-by-step explanation:
5,10,15.......95 this sequence is an A.P
as,d=5
also,
a=5
tn=95
tn=a+(n-1)d
95=5+(n-1)5
95=5+5n-5
95=5n
n=19
thus,
Sn=n/2 [2a+(n-1)d]
=19/2 [2×5+(19-1)5]
=19/2 [10+(18×5)]
=19/2 [10+90]
=19/2×100
=19×50
=950
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