Find the sum of all 3digit natural number which are divisible by 8
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3 digit natural numbers, which are divisible by 8 are:-
104,112,120,128……………………………992.
a=104, d=8 , l or tn=992.
l or tn=a+(n-1).d
992=104+(n-1).8
992–104=(n-1).8
888/8=(n-1)
111+1=n
112=n
S=n/2.(a+l)
S=112/2.(104+992)
S=56×1096
S=61376 , Answer.
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104,112,120,128……………………………992.
a=104, d=8 , l or tn=992.
l or tn=a+(n-1).d
992=104+(n-1).8
992–104=(n-1).8
888/8=(n-1)
111+1=n
112=n
S=n/2.(a+l)
S=112/2.(104+992)
S=56×1096
S=61376 , Answer.
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104...........992
an=a+(n-1)d
992=104+(n-1)8
888/8=n-1
111=n-1
n= 112
sn = n/2(2a+(n-1)d)
56(2*104+111*8)
56(208+888)
56*1096
61376 is the answer
an=a+(n-1)d
992=104+(n-1)8
888/8=n-1
111=n-1
n= 112
sn = n/2(2a+(n-1)d)
56(2*104+111*8)
56(208+888)
56*1096
61376 is the answer
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