Two sequences in just One!

We conjecture that $17$ terms is impossible. (Question for thought: Why is this a sound conjecture?)

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$\textbf{Solution 1: }$ Let's assume that the sequence has at least 17 terms. Let's denote the $k^{\text{th}}$ term $x_k$ . Thus you can make the matrix: $\begin{pmatrix} x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & x_7 \\ x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & x_8 \\ x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ x_{11} & x_{12} & x_{13} & x_{14} & x_{15} & x_{16} & x_{17} \end{pmatrix}$

Now consider the sum of all the terms by rows. Since the sum of each consecutive seven numbers is positive, you know that the sum of all of the rows must be $\textbf{positive.}$

Now consider the sum of all the terms by columns. Since the sum of each consecutive eleven numbers is negative, you know that the sum of all of the rows must be $\textbf{negative.}$

However these two sums are equal and they cannot be both positive and negative at the same time, which implies that having at least $17$ terms is impossible.

We can easily find a 16-term sequence: $-5, - 5, 13, -5, -5, -5, 13, -5, -5, 13, -5, -5, -5, 13, -5, -5$ .

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Here is a more generalized solution.

$\textbf{Solution 2:}$ Denote $S_n$ as the partial sums of terms, that is: $S_1 = x_1$ $S_2 = x_1 + x_2$ $S_3 = x_1 + x_2 + x_3$ $\vdots$ $S_n = \sum_{k = 1}^{n}{x_k}$

This implies that $x_1 = S_1$ $x_2 = S_2 - S_1$ $\vdots$ $x_n = S_n - S_{n-1}$

Notice that since the sum of every seven terms is positive, we have that: $S_{n+7} > S_n$

Notice that since the sum of every eleven terms is negative, we have that: $S_{n+11} < S_n$

Thus, assume that there are at least $17$ terms, we can create this looping inequality and noting that $S_0 = 0$ : $0 < S_7 < S_{14} < S_3 < S_{10} < \boxed{S_{17}} < S_6 < S_{13} < S_2 < S_9 < S_{16} < S_5 < S_{12} < S_1 < S_8 < S_{15} < S_4 < S_{11} < 0$

Looking at the leftmost number and rightmost number, this is obviously a contradiction. Therefore, there can be at most $16$ terms.

In the long inequality, remove $S_{17}$ , and move the right side to the left of the left side, to get that: $S_6 < S_{13} < S_2 < S_9 < S_{16} < S_5 < S_{12} < S_1 < S_8 < S_{15} < S_4 < S_{11} < 0 < S_7 < S_{14} < S_3 < S_{10}$ $-12 < - 11 < -10 < -9 < -8 < -7 < -6 < -5 < -4 < -3 < -2 < -1 < 0 < 1 < 2 < 3 < 4$ It is easy to find a $16$ sequence, as we can just assign random values to the $S_n$ that fufill the inequality, as I had done below the inequality. This combination gets us $-5, - 5, 13, -5, -5, -5, 13, -5, -5, 13, -5, -5, -5, 13, -5, -5$ .