Question 1164999: In an arithmetic series, the 3rd term is equal to 114 and the last term is equal to -27. The sum of all terms in the series is equal to 2325. What are the last three terms in the series?
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
Clearly, the common difference is negative.
The sum of the 3rd and last terms is 114-27 = 87; since the common difference is negative, we know the sum of the first and last terms is a bit more than 87.
The sum of the terms of an arithmetic sequence is the number of terms, multiplied by the average of the first and last. For this problem, it will be easier to think this way:
Twice the sum of the terms is the number of terms, multiplied by the sum of the first and last terms.
Twice the given sum is 4650; its prime factorization is 50*93 = 2*5*5*93.
Given that prime factorization of 4650, along with the fact that the sum of the first and last terms is a bit more than 87, and along with the fact that the number of terms is an integer, means the sum of the first and last terms is 93 and the number of terms is 50.
That means the first term is 93+27 = 120.
Then, since the third term (first term plus twice the common difference) is 114, the common difference is -3.
And finally, since the last term is -27, the preceding two terms are -24 and -21.
ANSWER: The last three terms of the series are -21, -24, and -27.
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