Asked by Aille
Between 300 and 500 inclusive, find the sums of all the integers that are multiples of:
a. 3
a. 3
Answers
Answered by
oobleck
you want the sum of the terms of an arithmetic progression, where
a = 300
d = 3
so, what is the last term?
300 + 3n <= 500
3n <= 200
n <= 66
and the last term is thus 498
so the sum of the 67 terms is 67/2 (300 + 498) = ____
a = 300
d = 3
so, what is the last term?
300 + 3n <= 500
3n <= 200
n <= 66
and the last term is thus 498
so the sum of the 67 terms is 67/2 (300 + 498) = ____
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