Densider the arithmetic sequence: \( -\frac{7}{2} ;-3 ;-\frac{5}{2} \ldots \) 1.1 The sum of the first \( n \) terms of this sequence is 675 . Calculate the value of \( n \).

To solve the problem, we first need to find the common difference of the arithmetic sequence. The first term \( a_1 = -\frac{7}{2} \) and the second term \( a_2 = -3 \). The common difference \( d \) is calculated as follows: \[ d = a_2 - a_1 = -3 - \left(-\frac{7}{2}\right) = -3 + \frac{7}{2} = \frac{-6 + 7}{2} = \frac{1}{2} \] Now that we have \( d = \frac{1}{2} \), we can find the formula for the \( n \)-th term of the sequence, given by: \[ a_n = a_1 + (n-1)d = -\frac{7}{2} + (n-1)\frac{1}{2} \] This can be simplified to: \[ a_n = -\frac{7}{2} + \frac{n-1}{2} = \frac{-7 + n - 1}{2} = \frac{n - 8}{2} \] The sum of the first \( n \) terms of an arithmetic sequence (denoted as \( S_n \)) is given by: \[ S_n = \frac{n}{2} \cdot (a_1 + a_n) \] Substituting the expressions for \( a_1 \) and \( a_n \): \[ S_n = \frac{n}{2} \left(-\frac{7}{2} + \frac{n-8}{2}\right) = \frac{n}{2} \left(\frac{-7 + n - 8}{2}\right) = \frac{n}{2} \cdot \frac{n - 15}{2} = \frac{n(n - 15)}{4} \] We know that \( S_n = 675 \), so we set up the equation: \[ \frac{n(n - 15)}{4} = 675 \] Multiplying both sides by 4 to eliminate the fraction: \[ n(n - 15) = 2700 \] Rearranging gives us a standard quadratic equation: \[ n^2 - 15n - 2700 = 0 \] We can now solve this equation using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -15, c = -2700 \): \[ n = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 1 \cdot (-2700)}}{2 \cdot 1} \] \[ = \frac{15 \pm \sqrt{225 + 10800}}{2} = \frac{15 \pm \sqrt{11025}}{2} \] Calculating \( \sqrt{11025} = 105 \): \[ n = \frac{15 \pm 105}{2} \] This yields two possible solutions: 1. \( n = \frac{120}{2} = 60 \) 2. \( n = \frac{-90}{2} = -45 \) (not a valid solution since \( n \) must be positive) Thus, the value of \( n \) is: \[ \boxed{60} \]