Question:

Step 1: Understand the general formula for sum of an arithmetic progression. For an arithmetic sequence $\{a_n\}$ with first term $a$ and common difference $d$, the sum of first $n$ terms is: $$S_n = \frac{n}{2}(2a + (n-1)d)$$ We are given that $S_6 > S_7 > S_5$. We will use this condition to determine properties of the sequence. Step 2: Analyze the condition $S_7 > S_5$. The difference $S_7 - S_5$ represents the sum of the 6th and 7th terms: $$S_7 - S_5 = a_6 + a_7 > 0$$ Expressing in terms of $a$ and $d$: $$a_6 + a_7 = (a + 5d) + (a + 6d) = 2a + 11d > 0$$ This simplifies to: $$a + 5.5d > 0$$ More directly, since $a_6 = a + 5d$ and $a_7 = a + 6d$, we have $a_6 + a_7 > 0$, which means: $$a_6 > 0$$ Step 3: Analyze the condition $S_6 > S_7$. The difference $S_6 - S_7$ equals the negative of the 7th term: $$S_6 - S_7 = -a_7 > 0$$ Therefore: $$a_7 < 0$$ This means $a + 6d < 0$. Step 4: Determine the sign pattern of the arithmetic sequence. From Steps 2 and 3, we have established: - $a_6 > 0$ (positive) - $a_7 < 0$ (negative) Since the sequence transitions from positive to negative between the 6th and 7th terms, the common difference must be negative: $d < 0$. The sequence is decreasing. Step 5: Determine when $S_n$ changes sign. Since $a_6 > 0$ and $a_7 < 0$, the sum $S_n$ is maximized at $n = 6$. The sum $S_n$ remains positive as long as the positive terms dominate. We need to find when $S_n$ becomes negative. Using the property that in an AP, equidistant terms from the ends sum to the same value: $$a_1 + a_{12} = a_6 + a_7$$ Since $a_6 + a_7 > 0$, we have $a_1 + a_{12} > 0$. Step 6: Calculate $S_{12}$ and $S_{13}$. For $S_{12}$, pairing terms symmetrically: $$S_{12} = \frac{12}{2}(a_1 + a_{12}) = 6(a_1 + a_{12}) = 6(a_6 + a_7) > 0$$ Since $a_6 + a_7 > 0$, we have $S_{12} > 0$. For $S_{13}$: $$S_{13} = S_{12} + a_{13}$$ Since $a_{13} = a + 12d$ and $d < 0$, the term $a_{13}$ is negative and sufficiently large in magnitude such that $S_{13} < 0$. Step 7: Identify the value of $n$ satisfying $S_n S_{n+1} < 0$. For the product $S_n S_{n+1}$ to be negative, the two consecutive sums must have opposite signs. From our analysis: - $S_{12} > 0$ - $S_{13} < 0$ Therefore, $S_{12} \cdot S_{13} < 0$, which means $n = 12$. **Final Answer: The integral value of $n$ which satisfies $S_n S_{n+1} < 0$ is $\boxed{12}$, which corresponds to Option 3.**