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In the sequence 1, 3, 6, 10, 15, 21, ..., in the first 1000 terms of the sequence,
how many terms are multiples of 5?
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The n-th term of this sequence, , is the sum of the first n natural numbers
= 1,
= 3 = 1 + 2,
= 6 = 1 + 2 + 3,
= 10 = 1 + 2 + 3 + 4,
= 15 = 1 + 2 + 3 + 4 + 5,
and so on. For the sum of the first natural numbers, there is the formula
= .
So, the question is: how many terms of the first 1000 terms are multiples of 5.
In order for the term be a multiple of 5, EITHER n OR (n+1) must be multiple of 5
(here "either - or" is exclusive).
Of the first 1000 numbers n = 1, 2, 3, 4, 5, . . . , 1000, there are 1000/5 = 200 multiple of 5.
Of the first 1000 numbers n+1 = 2, 3, 4, 5, . . . , 1001, there are = 200 other numbers multiple of 5.
Thus, in all, among first 1000 numbers = , there are 200 + 200 = 400 numbers multiple of 5.
ANSWER. In the sequence 1, 3, 6, 10, 15, 21, ..., in the first 1000 terms of the sequence,
400 numbers are multiple of 5.
Solved.