find the sum of all three digit natural numbers which leave a remainder 3 when divided by 5
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Answer:
Step-by-step explanation:
a=103 ; d=5; an or l=998
an=a+(n-1)d
998=103+(n-1)5
895=5n-5
900=5n
180=n
Sn=n/2(a+l)
S180=180/2(103+998)
S180=90(1101)
S180=99090
SUM OF ALL 3 DIGIT NATURAL NUMBER WHICH LEAVE A REMAINDER 3 WHEN DIVIDED BY 5 IS 99090
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