\begin{tabular}{l} culate the sum of each of the following arithmetic series: \\ \( 7+9+11+\ldots+105 \) \\ \( 2-3-8-\ldots-368 \) \\ (2) \( 50+43+36+\ldots-643 \) \\ \( 2+\frac{13}{4}+\frac{9}{2}+\ldots+82 \) \\ \( -2 p+2 p+6 p+\ldots+102 p \) \\ (4) \( -15-12-9-\ldots+432 \) \\ \\ \hline\end{tabular}
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Extra Insights
To sum each of the given arithmetic series, we will apply the formula for the sum of an arithmetic series: \[ S_n = \frac{n}{2} (a + l) \] where \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term. 1. For \( 7 + 9 + 11 + \ldots + 105 \): - First term \( a = 7 \), last term \( l = 105 \), and the common difference \( d = 2 \). - Number of terms \( n = \frac{(l - a)}{d} + 1 = \frac{(105 - 7)}{2} + 1 = 50 \). - Sum \( S_{50} = \frac{50}{2} (7 + 105) = 25 \times 112 = 2800 \). 2. For \( 2 - 3 - 8 - \ldots - 368 \): - First term \( a = 2 \), last term \( l = -368 \), and common difference \( d = -5 \). - \( n = \frac{(l - a)}{d} + 1 = \frac{(-368 - 2)}{-5} + 1 = 74 \). - Sum \( S_{74} = \frac{74}{2} (2 - 368) = 37 \times (-366) = -13642 \). 3. For \( 50 + 43 + 36 + \ldots - 643 \): - First term \( a = 50 \), last term \( l = -643 \), common difference \( d = -7 \). - \( n = \frac{(l - a)}{d} + 1 = \frac{(-643 - 50)}{-7} + 1 = 100 \). - Sum \( S_{100} = \frac{100}{2} (50 - 643) = 50 \times (-593) = -29650 \). 4. For \( 2 + \frac{13}{4} + \frac{9}{2} + \ldots + 82 \): - First term \( a = 2\), last term \( l = 82 \), converting \( \frac{9}{2} = 4.5 \) to decimal gives a common difference \( d = \frac{29}{4} \). - \( n = \frac{(l - a)}{d} + 1 = \frac{(82 - 2)}{\frac{29}{4}} + 1 = 14 \). - Sum \( S_{14} = \frac{14}{2} (2 + 82) = 7 \times 84 = 588 \). 5. For \( -2p + 2p + 6p + \ldots + 102p \): - First term \( a = -2p \), last term \( l = 102p \), and common difference \( d = 4p \). - \( n = \frac{(l - a)}{d} + 1 = \frac{(102p + 2p)}{4p} + 1 = 26 \). - Sum \( S_{26} = \frac{26}{2} (-2 + 102)p = 13 \times 100p = 1300p \). 6. For \( -15 - 12 - 9 - \ldots + 432 \): - First term \( a = -15 \), last term \( l = 432 \), common difference \( d = 3 \). - \( n = \frac{(l - a)}{d} + 1 = \frac{(432 + 15)}{3} + 1 = 149 \). - Sum \( S_{149} = \frac{149}{2} (-15 + 432) = 74.5 \times 417 = 31046.5 \). Thus, the final sums of the series are: 1. 2800 2. -13642 3. -29650 4. 588 5. \( 1300p \) 6. 31046.5