A Finite Sequence
In a finite sequence of real numbers, the sum of any seven successive terms is negative, and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.
The answer is 16.
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2 solutions
For anyone interested in the inner workings of how I found that sequence, note that we can define the running sums S k = ∑ n = 1 k x n (where S 0 = 0 ) and then the two requirements given in the problem become S k + 7 − S k < 0 S k + 1 1 − S k > 0 0 ≤ k ≤ 9 0 ≤ k ≤ 5 Solving these gives the single chain S 1 0 < S 3 < S 1 4 < S 7 < 0 < S 1 1 < S 4 < S 1 5 < S 8 < S 1 < S 1 2 < S 5 < S 1 6 < S 9 < S 2 < S 1 3 < S 6 so we can find one solution by assigning respective to − 4 < − 3 < − 2 < − 1 < 0 < 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8 < 9 < 1 0 < 1 1 < 1 2 and then find the terms as x k = S k − S k − 1 . This gives exactly the sequence given in the solution.
f there was a sequence with 1 7 terms, arrange the terms (with some repetitions) as follows:
x 1 x 2 x 3 ⋮ x 7 x 2 x 3 x 4 ⋮ x 8 x 3 x 4 x 5 ⋮ x 9 . . . . . . . . . . . . . . . x 1 1 x 1 2 x 1 3 ⋮ x 1 7 The sum of the entries in any row is positive, and so the sum of all the terms in the array is positive. But the sum of the entries in any column is negative, and so the sum of all the terms in the array is negative. This contradiction shows that the sequence cannot contain 1 7 terms. A sequence with 1 6 terms is possible, as in 1 2 1 2 − 3 1 1 2 1 2 1 2 − 3 1 1 2 1 2 − 3 1 1 2 1 2 1 2 − 3 1 1 2 1 2
Write the sequence as x 1 , x 2 , … , x N and define S ( m , k ) = n = k ∑ k + m − 1 x n , which is the sum of the m consecutive terms starting with index k . Then the assumptions in the problem can be written as S ( 1 1 , k ) > 0 S ( 7 , k ) < 0 1 ≤ k ≤ N − 1 0 1 ≤ k ≤ N − 6
By using these, we can find that S ( 4 , k ) = S ( 1 1 , k − 7 ) − S ( 7 , k − 7 ) > 0 S ( 3 , k ) = S ( 7 , k − 4 ) − S ( 4 , k − 4 ) < 0 x k = S ( 4 , k − 3 ) − S ( 3 , k − 3 ) > 0 8 ≤ k ≤ N − 3 1 2 ≤ k ≤ N − 2 1 5 ≤ k ≤ N but if N ≥ 1 7 , this gives 0 < x 1 5 + x 1 6 + x 1 7 = S ( 3 , 1 5 ) < 0 which is a contradiction, so N ≤ 1 6 .
To show that this is in fact the maximum, it suffices to note that the sequence 5 , 5 , − 1 3 , 5 , 5 , 5 , − 1 3 , 5 , 5 , − 1 3 , 5 , 5 , 5 , − 1 3 , 5 , 5 has length 1 6 and is such that the sum of any seven consecutive terms is − 1 and the sum of any eleven consecutive terms is 1 .
Therefore, the answer is 1 6 .