(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 \( S_{k}>-800 \).

To solve the problem regarding the arithmetic series \( S_n = 4, -2, -8, \ldots, -260 \), we begin by determining the common difference and the number of terms. **Step 1: Identify the common difference (d).** The first term \( a = 4 \), and the second term \( -2 \). To find the common difference: \[ d = -2 - 4 = -6. \] **Step 2: Find the number of terms, \( n \).** The \( n \)-th term of an arithmetic series can be found using the formula: \[ a_n = a + (n-1)d. \] We know that the last term \( a_n = -260 \). Setting this in the formula gives: \[ -260 = 4 + (n-1)(-6). \] Rearranging it: \[ -260 - 4 = (n-1)(-6) \implies -264 = (n-1)(-6). \] Solving for \( n \): \[ n - 1 = \frac{264}{6} = 44 \implies n = 45. \] **Step 3: Calculate the sum of the series.** The sum of an arithmetic series can be calculated by: \[ S_n = \frac{n}{2} \cdot (a + a_n). \] Plugging in the values we have: \[ S_{45} = \frac{45}{2} \cdot (4 + (-260)) = \frac{45}{2} \cdot (-256) = 45 \cdot (-128) = -5760. \] Thus, the sum of the series is: \[ \text{Sum} = -5760. \] **Step 4: Find \( m \) such that \( S_m = -1700 \).** Using the sum formula again for \( m \) terms: \[ S_m = \frac{m}{2} \cdot (4 + a_m) = \frac{m}{2} \cdot (4 + (4 + (m-1)(-6))). \] This simplifies to: \[ S_m = \frac{m}{2} \cdot (4 + 4 - 6(m-1)) = \frac{m}{2} \cdot (8 - 6m + 6) = \frac{m}{2} \cdot (14 - 6m). \] Setting this equal to -1700: \[ \frac{m(14 - 6m)}{2} = -1700 \implies m(14 - 6m) = -3400. \] This expands to: \[ -6m^2 + 14m + 3400 = 0. \] Multiplying through by -1 gives: \[ 6m^2 - 14m - 3400 = 0. \] Using the quadratic formula: \[ m = \frac{-(-14) \pm \sqrt{(-14)^2 - 4 \cdot 6 \cdot (-3400)}}{2 \cdot 6} = \frac{14 \pm \sqrt{196 + 81600}}{12}. \] Calculating the discriminant: \[ \sqrt{81896} \approx 286.5. \] Thus, \[ m = \frac{14 \pm 286.5}{12} \implies m_1 \approx 25.5 \quad m_2 \approx -22.66. \] We take the positive integer for \( m = 26 \). **Step 5: Find \( k \) such that \( S_k > -800 \).** We use the sum formula again: \[ S_k = \frac{k}{2} \cdot (4 + (4 + (k-1)(-6))) = \frac{k}{2} \cdot (8 - 6(k-1)). \] Setting the inequality: \[ \frac{k}{2} \cdot (14 - 6k) > -800 \implies k(14 - 6k) > -1600. \] This can be written as: \[ -6k^2 + 14k + 1600 > 0. \] Multiplying by -1 gives: \[ 6k^2 - 14k - 1600 < 0. \] Finding the roots: \[ k = \frac{-(-14) \pm \sqrt{(-14)^2 - 4 \cdot 6 \cdot (-1600