Calculate the sum of each of the following arithmetic series: \( \begin{array}{ll}\text { (1) } 7+9+11+\ldots+105 & \text { (2) } 50+43+36+\ldots-643 \\ \text { (3) } 2-3-8-\ldots-368 & \text { (4) }-15-12-9-\ldots+432 \\ \text { (5) } 2+\frac{13}{4}+\frac{9}{2}+\ldots+82 & \text { (6) } \frac{11}{2}+\frac{7}{2}+\frac{3}{2}+\ldots-\frac{65}{2} \\ \text { (7) }-2 p+2 p+6 p+\ldots+102 p & \text { (8) } x-2 x-5 x-\ldots-56 x\end{array} \)

To calculate the sums of the given arithmetic series, we can utilize the formula for the sum of an arithmetic series which is: \[ S_n = \frac{n}{2} \times (a + l) \] where \( S_n \) is the sum of the series, \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term. Let's find the sums step by step for each series provided: ### (1) \( 7+9+11+\ldots+105 \) - **First term (a)**: 7 - **Last term (l)**: 105 - **Common difference (d)**: 2 To find the number of terms (n): \[ n = \frac{l - a}{d} + 1 = \frac{105 - 7}{2} + 1 = 50 \] Now, substituting into the sum formula: \[ S_n = \frac{50}{2} \times (7 + 105) = 25 \times 112 = 2800 \] ### (2) \( 50+43+36+\ldots-643 \) - **First term (a)**: 50 - **Last term (l)**: -643 - **Common difference (d)**: -7 To find the number of terms (n): \[ n = \frac{l - a}{d} + 1 = \frac{-643 - 50}{-7} + 1 = 99 + 1 = 100 \] Now, substituting into the sum formula: \[ S_n = \frac{100}{2} \times (50 - 643) = 50 \times (-593) = -29650 \] ### (3) \( 2-3-8-\ldots-368 \) - **First term (a)**: 2 - **Last term (l)**: -368 - **Common difference (d)**: -5 To find the number of terms (n): \[ n = \frac{l - a}{d} + 1 = \frac{-368 - 2}{-5} + 1 = 74 + 1 = 75 \] Now, substituting into the sum formula: \[ S_n = \frac{75}{2} \times (2 - 368) = 37.5 \times (-366) = -13777.5 \] ### (4) \( -15-12-9-\ldots+432 \) - **First term (a)**: -15 - **Last term (l)**: 432 - **Common difference (d)**: 3 To find the number of terms (n): \[ n = \frac{l - a}{d} + 1 = \frac{432 + 15}{3} + 1 = 149 + 1 = 150 \] Now, substituting into the sum formula: \[ S_n = \frac{150}{2} \times (-15 + 432) = 75 \times 417 = 31275 \] ### (5) \( 2+\frac{13}{4}+\frac{9}{2}+\ldots+82 \) - **First term (a)**: 2 - **Last term (l)**: 82 - **Common difference (d)**: \(\frac{13}{4} - 2 = \frac{5}{4}\) To find the number of terms (n): Convert terms to fractions: - 2 = \(\frac{8}{4}\) - Last term \(l = 82\) in quarters is \(\frac{328}{4}\) Calculate n: \[ n = \frac{\frac{328}{4} - \frac{8}{4}}{\frac{5}{4}} + 1 = \frac{\frac{320}{4}}{\frac{5}{4}} + 1 = 16 + 1 = 17 \] Now, substituting: \[ S_n = \frac{17}{2} \times \left( 2 + 82 \right) = \frac{17}{2} \times 84 = 714 \] ### (6) \( \frac{11}{2}+\frac{7}{2}+\frac{3}{2}+\ldots-\frac{65}{2} \) - **First term (a)**: \(\frac{11}{2}\) - **Last term (l)**: \(-\frac{65}{2}\) - **Common difference