SOLUTION: The question is: what is the sum of 4+11+18+....+4001 I have been taught to add the first and last number, so i get 4005. Then I usually divide by half of the last number, which w

Question 566162: The question is: what is the sum of 4+11+18+....+4001
I have been taught to add the first and last number, so i get 4005. Then I usually divide by half of the last number, which would be 2000.5
However, I know this only works if the numbers go up by one, not seven. This is where I get lost.
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Let S = the sum you want. Then, we have

S = 4 + 11+18+...+4001
S = 4001+3994 + ... + 4

If we add them up, we get 2S = 4005 + 4005 + ... + 4005. Now we have to count the number of "4005s" there are, which is equal to the size of the set {4, 11, 18, ..., 4001}. This is equal to the size of the sets {7, 14, 21, ..., 4004} and {1, 2, 3, ..., 572} (we are just adding or multiplying numbers in the sets by a fixed number; this does not change the size of the set).

Therefore, there are 572 "4005s." Hence, 2S = 572*4005, S = 572*4005/2 = 1145430.
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