Question:

<div class="solution"> <p><strong>Step 1:</strong> To solve this problem, we first need to understand the properties of an arithmetic sequence and its sum. The sum of the first \(n\) terms of an arithmetic sequence can be given by the formula \(S_n = \frac{n}{2}[2a + (n-1)d]\), where \(a\) is the first term and \(d\) is the common difference.</p> <p><strong>Step 2:</strong> Given \(S_6 > S_7 > S_5\), let's express these sums using the formula. \(S_6 = \frac{6}{2}[2a + (6-1)d] = 3(2a + 5d)\), \(S_7 = \frac{7}{2}[2a + (7-1)d] = \frac{7}{2}(2a + 6d)\), and \(S_5 = \frac{5}{2}[2a + (5-1)d] = \frac{5}{2}(2a + 4d)\). The condition \(S_6 > S_7 > S_5\) implies that \(3(2a + 5d) > \frac{7}{2}(2a + 6d) > \frac{5}{2}(2a + 4d)\).</p> <p><strong>Step 3:</strong> Simplifying the inequalities, we get \(6a + 15d > 7a + 21d\) and \(7a + 21d > 5a + 10d\), which further simplify to \(a + 6d < 0\) and \(2a + 11d > 0\). These inequalities provide conditions on \(a\) and \(d\), but to find \(n\) for which \(S_n S_{n+1} < 0\), we need to consider when the sum changes sign.</p> <p><strong>Step 4:</strong> For \(S_n S_{n+1} < 0\), one sum must be positive and the next negative, or vice versa. This happens when the sequence crosses zero, meaning the sum changes from positive to negative or vice versa. The sum \(S_n = \frac{n}{2}[2a + (n-1)d]\) changes sign when \(2a + (n-1)d = 0\), giving \(n = 1 - \frac{2a}{d}\). However, since \(n\) must be an integer and we're looking for a specific \(n\) that satisfies the given conditions, we must analyze how \(S_n\) behaves as \(n\) increases, considering the constraints on \(a\) and \(d\).</p> <p><strong>Step 5:</strong> Given \(a + 6d < 0\) and \(2a + 11d > 0\), let's solve these inequalities to understand the range of \(a\) and \(d\). From \(a + 6d < 0\), we get \(a < -6d\), and from \(2a + 11d > 0\), we get \(2a > -11d\) or \(a > -\frac{11}{2}d\). Thus, \(-\frac{11}{2}d < a < -6d\). This range indicates that \(a\) is negative and \(d\) could be positive or negative, but given the constraints, \(d\) must be such that it allows for the sequence to have both positive and negative sums as \(n\) varies.</p> <p><strong>Step 6:</strong> To find the integral value of \(n\) satisfying \(S_n S_{n+1} < 0\), we recognize that since \(S_n\) is a quadratic function of \(n\), it can change sign at most twice. Given \(S_6 > S_7 > S_5\), and knowing that for large \(n\), \(S_n\) will be dominated by the \(n^2\) term which is positive if \(d > 0\) or negative if \(d < 0\), we need to consider how many times \(S_n\) can change sign. Since \(S_6\) and \(S_7\) have opposite behaviors compared to \(S_5\), and considering the parabolic nature of \(S_n\), there must be a change in sign around \(n = 6\) or \(7\), but to satisfy \(S_n S_{n+1} < 0\), we look for \(n\) where this transition happens, taking into account the constraints on \(a\) and \(d\).</p> <p><strong>Answer:</