Question 979148: the least integer of a set of consecutive even integers is -30. if the sum of these integers is 66, how any integers are in the set?
Found 2 solutions by CubeyThePenguin, greenestamps:
Answer by CubeyThePenguin(3113)   (Show Source):
You can put this solution on YOUR website! (-30 -28 -... - 2) + 0 + (2 + ... + 30) + 32 + 34 = 66
The sum of the integers from -30 to -2 and the sum of the integers from 2 to 30 cancel each other out.
# of integers: (from -30 to -2) + (the number 0) + (from 2 to 30) + (32 and 34)
= 15 + 1 + 15 + 2
= 33
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The sequence is
-30, -28, -26, ..., 0, ..., 26, 28, 30, ......
The sum of those terms of the sequence is 0, so the sum of 66 we are looking for comes from the subsequent terms of the sequence. The next terms are 32 and 34, and that's as far as we need to go, because 32+34=66.
So the first term is -30 and the last term is 34; the number of terms in the sequence is
(last minus first) divided by (the common difference), plus 1:
ANSWER: there are 33 terms in the sequence.
For a more traditional algebraic approach....
Let n be the number of terms in the sequence.
The first term is -30; the common difference is 2; so the n-th term is -30 plus the common difference (n-1) times. That makes the last term -30+2(n-1) = 2n-32.
The sum of the terms of the sequence is (number of terms) times (average of all the terms). Since the sequence is arithmetic, this is the same as (number of terms) times (average of first and last terms).
The number of terms is either 33 or -2; and obviously the negative answer makes no sense.
ANSWER (as before, of course!): There are 33 terms in the sequence.
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