(g) Consider the arithmetic series \( 4-2-8-\ldots-260 \). (1) Calculate the sum of the series. (2) Determine \( m \) if \( 4-2-8-\ldots \) (to \( m \) terms) \( =-1700 \). (3) Calculate the greatest value of \( k \) for which \( S_{k}>-800 \).

Let's analyze the arithmetic series \( 4, -2, -8, \ldots, -260 \) step by step. ### Given: - **First term (\( a_1 \))**: 4 - **Common difference (\( d \))**: \( -2 - 4 = -6 \) (since the second term is \(-2\)) The general term of the series is: \[ a_n = a_1 + (n-1)d = 4 + (n-1)(-6) = 10 - 6n \] ### Part (1): Calculate the Sum of the Series We need to find the sum of the series up to the term \(-260\). 1. **Find the number of terms (\( n \))**: \[ 10 - 6n = -260 \\ -6n = -270 \\ n = 45 \] 2. **Calculate the sum (\( S_{45} \))** using the formula for the sum of an arithmetic series: \[ S_n = \frac{n}{2}(a_1 + a_n) \\ S_{45} = \frac{45}{2}(4 + (-260)) = \frac{45}{2}(-256) = 45 \times (-128) = -5760 \] **Answer for Part (1):** The sum of the series is **\(-5760\)**. ### Part (2): Determine \( m \) if the Sum to \( m \) Terms is \(-1700\) We need to solve for \( m \) in: \[ S_m = \frac{m}{2}(a_1 + a_m) = -1700 \] Where: \[ a_m = 10 - 6m \] Substituting: \[ \frac{m}{2}(4 + (10 - 6m)) = -1700 \\ \frac{m}{2}(14 - 6m) = -1700 \\ m(14 - 6m) = -3400 \\ -6m^2 + 14m + 3400 = 0 \\ 6m^2 - 14m - 3400 = 0 \] Solving the quadratic equation: \[ m = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 6 \cdot (-3400)}}{2 \cdot 6} \\ m = \frac{14 \pm \sqrt{196 + 81600}}{12} \\ m = \frac{14 \pm \sqrt{81800}}{12} \\ m = \frac{14 \pm 286}{12} \] Only the positive root makes sense: \[ m = \frac{14 + 286}{12} = \frac{300}{12} = 25 \] **Answer for Part (2):** \( m = 25 \) ### Part (3): Find the Greatest \( k \) such that \( S_k > -800 \) We need to find the largest integer \( k \) where the sum \( S_k \) satisfies: \[ S_k = \frac{k}{2}(14 - 6k) > -800 \] Simplifying: \[ k(14 - 6k) > -1600 \\ -6k^2 + 14k + 1600 > 0 \\ 6k^2 - 14k - 1600 < 0 \] Solving the quadratic inequality: \[ k = \frac{14 \pm \sqrt{14^2 + 4 \cdot 6 \cdot 1600}}{2 \cdot 6} \\ k = \frac{14 \pm \sqrt{196 + 38400}}{12} \\ k = \frac{14 \pm \sqrt{38600}}{12} \approx \frac{14 \pm 196.46}{12} \] The positive root is approximately \( 17.54 \). Since \( k \) must be an integer: \[ k = 17 \] For \( k = 17 \): \[ S_{17} = \frac{17}{2}(14 - 6 \times 17) = \frac{17}{2}(14 - 102) = \frac{17}{2}(-88) = -748 > -800 \] For \( k = 18 \): \[ S_{18} = \frac{18}{2}(14 - 6 \times 18) = 9(-104) = -936 < -800 \] Thus, the greatest \( k \) is 17. **Answer for Part (3):** The greatest value of \( k \) is **17**. ### Summary of All Parts: 1. **Sum of the series:** \(-5760\) 2. **Value of \( m \):** \(25\) 3. **Greatest \( k \) with \( S_k > -800 \):** \(17\)