3.4 A quadratic sequence, with general term \( T_{n} \), has the following properties - \( T_{11}=125 \) - \( T_{n}-T_{n-1}=3 n-4 \) Determine the first term of the sequence.

To solve for the first term of the quadratic sequence, we start with the information given: 1. We know that \( T_{n} - T_{n-1} = 3n - 4 \). 2. This means that the difference between consecutive terms is a linear expression. From the difference equation, we can express the terms in the sequence recursively: \[ T_n = T_{n-1} + (3n - 4) \] We can then derive the specific terms of the sequence step by step. Starting from \( T_1 \): - \( T_2 = T_1 + (3 \cdot 2 - 4) = T_1 + 2 \) - \( T_3 = T_2 + (3 \cdot 3 - 4) = T_1 + 2 + 5 = T_1 + 7 \) - \( T_4 = T_3 + (3 \cdot 4 - 4) = T_1 + 7 + 8 = T_1 + 15 \) Continuing this way, we can generate a pattern. Summing up the differences from \( T_1 \) to \( T_{11} \): Let’s denote the total number of terms \( n = 11 \), we know: \[ T_{11} = T_1 + \sum_{k=2}^{11} (3k - 4) \] Calculating the sum \( \sum_{k=2}^{11} (3k - 4) \): Break it into parts: \[ \sum_{k=2}^{11} (3k - 4) = \sum_{k=2}^{11} 3k - \sum_{k=2}^{11} 4 \] - The first part, \( \sum_{k=2}^{11} 3k = 3 \sum_{k=2}^{11} k\): \[ \sum_{k=2}^{11} k = \frac{(11)(12)}{2} - 1 = 66 - 1 = 65 \quad \Rightarrow \quad 3 \cdot 65 = 195 \] - The second part \( \sum_{k=2}^{11} 4\) has 10 terms (from \( k=2 \) to \( k=11 \)), thus: \[ \sum_{k=2}^{11} 4 = 4 \cdot 10 = 40 \] So, we can combine these: \[ \sum_{k=2}^{11} (3k - 4) = 195 - 40 = 155 \] Thus, \[ T_{11} = T_1 + 155 \] Given \( T_{11} = 125 \), we equate: \[ 125 = T_1 + 155 \] Solving for \( T_1 \): \[ T_1 = 125 - 155 = -30 \] So the first term of the sequence is \( \boxed{-30} \).