QUESTION 6 Given the arithmetic series: \( -7-3+1+\cdots+173 \) 6.1 How many terms are there in the series? 6.2 Calculate the sum of the series. 6.3 Write the series in sigma notation.

1. **Determine the common difference and first term:** The series is \[ -7,\, -3,\, 1,\, \dots,\, 173 \] The first term is \[ a = -7 \] and the common difference is \[ d = (-3) - (-7) = 4. \] 2. **Find the number of terms (\( n \)):** The general term of an arithmetic series is given by \[ a_n = a + (n-1)d. \] We know the last term is \[ a_n = 173. \] Substitute the known values: \[ -7 + (n-1) \cdot 4 = 173. \] Solve for \( n \): \[ \begin{aligned} 4(n-1) &= 173 + 7 \\ 4(n-1) &= 180 \\ n - 1 &= \frac{180}{4} = 45 \\ n &= 45 + 1 = 46. \end{aligned} \] There are thus \(\boxed{46}\) terms in the series. 3. **Calculate the sum of the series:** The sum of the first \( n \) terms of an arithmetic series is given by \[ S_n = \frac{n}{2}(a + L), \] where \( L \) is the last term. Substitute the values: \[ S_{46} = \frac{46}{2}(-7 + 173) = 23 \times 166. \] Compute the product: \[ 23 \times 166 = 3818. \] Thus, the sum of the series is \(\boxed{3818}\). 4. **Express the series in sigma notation:** The general term of the series can be written as \[ a_k = -7 + 4k, \quad \text{for } k = 0, 1, 2, \dots, 45. \] Therefore, the series in sigma notation is: \[ \sum_{k=0}^{45} \left(-7 + 4k\right). \]