find the sum of all natural numbers less than 100 and divisible by 6.
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323
6,12,18,...96 are all natural less than 100 and divisible by 6
first term =6
common difference= d = a2 -a1=12-6 =6
nth term = 96
a+(n-1)d=96
6+(n-1)6=96
(n-1)6=96-6
(n-1)6=90
n-1=90/6
n-1=15
n=15+1
n=16
sum = n/2[a+ nth term]
=16/2[6+96]
=8×102
=816
first term =6
common difference= d = a2 -a1=12-6 =6
nth term = 96
a+(n-1)d=96
6+(n-1)6=96
(n-1)6=96-6
(n-1)6=90
n-1=90/6
n-1=15
n=15+1
n=16
sum = n/2[a+ nth term]
=16/2[6+96]
=8×102
=816
Answered by
76
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