If $\frac{-3+12}{1+3+...+(2l-1)} = \frac{\text{some}}{(2l-1)\text{ some}}$, then $n$ is equal to
Step-by-Step Solution
Key Concept: Use the sum formula for arithmetic progressions and properties of individual terms to set up equations
Given $S_9 = 13$ and the sum of 7th and (n-1)th terms equals some value. Using the arithmetic progression formula, we set up equations: $S_9 = \frac{9}{2}[2a + 8d] = 13$, which gives $2a + 8d = \frac{26}{9}$. The problem involves finding $n$ where specific terms satisfy given conditions. Through algebraic manipulation and solving the resulting quadratic $15n^2 - 88n - 119 = 0$, we get $(n-7)(15n+17) = 0$, yielding $n = 7$ (since $n$ must be positive).
Correct Answer: 7