Question 1201651: An arithmetic sequence of 15 terms has a sum of 3060. The common difference and each member of the arithmetic sequence are all positive even integers. If a is the first term of any such arithmetic sequence, find the sum of all distinct a in which at least one of the first four terms of the arithmetic sequence is the square of an integer.
Answer by greenestamps(13200)   (Show Source):
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The sum of the 15 terms is 3060, so the average is 3060/15 = 204.
In an arithmetic sequence of 15 terms, the 8th term (the one in the middle) is the average of all the terms, so in every sequence that satisfies the conditions of the problem, the 8th term is 204.
To find the sequences that satisfy all the conditions of the problem, look at each even perfect square less than 204: 196, 144, 100, 64, 36, 16, and 4.
For each of those perfect squares, the difference between 204 and the perfect square must be equal to n times the common difference d of the sequence.
The conditions of the problem place the following requirements on d and n:
(a) the common difference d must be even;
(b) since the perfect square must be one of the first four terms of the sequence, n must be at least 4 and at most 7; and
(c) since all the terms of the sequence must be positive, 204-n*d must be greater than 0
(1) 196
204-196 = 8 = 4 times 2
common difference of 2, 4 times; 196 is term number 8-4 = 4; 1st term is 204-7(2) = 190:
190, 192, 194, 196, 198, 200, 202, 204, ...
(2) 144
204-144 = 60 = 5 times 12
common difference of 12, 5 times; 144 is term number 8-5 = 3; 1st term is 204-7(12) = 120:
120, 132, 144, 156, 168, 180, 192, 204, ...
204-144 = 60 = 6 times 10
common difference 10, 6 times; 144 is term number 8-6 = 2; 1st term is 204-7(10) = 134:
134, 144, 154, 164, 174, 184, 194, 204, ...
(3) 100
204-100 = 104 = 4 times 26
common difference 26, 4 times; 100 is term number 8-4 = 4; 1st term is 204-7(26) = 22:
22, 48, 74, 100, 126, 152, 178, 204, ...
(4) 64
204-64 = 140 = 5 times 28
common difference 28, 5 times; 64 is term number 8-5 = 3; 1st term is 204-7(28) = 8:
8, 36, 64, 92, 120, 148, 176, 204, ...
(5) 36
204-36 = 168 = 6*28
common difference 28, 6 times; 36 is term number 8-6 = 2; 1st term is 204-7(28) = 8
8, 36, 64, 92, 120, 148, 176, 204, ...
(NOTE: we already have this sequence; in this sequence 2 of the terms are perfect squares...)
204-36 = 168 = 7 times 24
common difference 24, 7 times; 36 is term number 8-7 = 1; so 1st term is 36:
36, 60, 84, 108, 132, 156, 180, 204, ...
(others) For the perfect squares 16 and 4, there are no combinations of d and n that produce other arithmetic sequences with 8th term 204.
Summary...
The first terms of the sequences in which at least one of the terms is a perfect square are 190, 120, 134, 22, 8, and 36
ANSWER: 190 + 120 + 134 + 22 + 8 + 36 = 510
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