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\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}

Ask by Donnelly Ayala. in South Africa
Jan 23,2025

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Answer

The sums of the arithmetic series are as follows: 1. \( 7 + 9 + 11 + \ldots + 105 = 2,\!800 \) 2. \( 2 - 3 - 8 - \ldots - 368 = -13,\!725 \) 3. \( 50 + 43 + 36 + \ldots -643 = -29,\!650 \) 4. \( 2 + \frac{13}{4} + \frac{9}{2} + \ldots + 82 = 2,\!730 \) 5. \( -2p + 2p + 6p + \ldots + 102p = 1,\!350p \) 6. \( -15 - 12 - 9 - \ldots +432 = 31,\!275 \)

Solution

Sure, let's calculate the sum of each of the given arithmetic series step by step. ### 1. \( 7 + 9 + 11 + \ldots + 105 \) **First Term (\(a\))**: 7 **Common Difference (\(d\))**: 9 - 7 = 2 **Last Term (\(l\))**: 105 **Number of Terms (\(n\))**: \[ a_n = a + (n-1)d \\ 105 = 7 + (n-1) \cdot 2 \\ 105 - 7 = 2(n-1) \\ 98 = 2(n-1) \\ n - 1 = 49 \\ n = 50 \] **Sum (\(S\))**: \[ S = \frac{n}{2} (a + l) = \frac{50}{2} (7 + 105) = 25 \times 112 = 2800 \] --- ### 2. \( 2 - 3 - 8 - \ldots - 368 \) **First Term (\(a\))**: 2 **Common Difference (\(d\))**: \(-3 - 2 = -5\) **Last Term (\(l\))**: -368 **Number of Terms (\(n\))**: \[ a_n = a + (n-1)d \\ -368 = 2 + (n-1)(-5) \\ -368 - 2 = -5(n-1) \\ -370 = -5n + 5 \\ -375 = -5n \\ n = 75 \] **Sum (\(S\))**: \[ S = \frac{n}{2} (a + l) = \frac{75}{2} (2 - 368) = 37.5 \times (-366) = -13725 \] --- ### 3. \( 50 + 43 + 36 + \ldots -643 \) **First Term (\(a\))**: 50 **Common Difference (\(d\))**: 43 - 50 = -7 **Last Term (\(l\))**: -643 **Number of Terms (\(n\))**: \[ a_n = a + (n-1)d \\ -643 = 50 + (n-1)(-7) \\ -643 - 50 = -7(n-1) \\ -693 = -7n + 7 \\ -700 = -7n \\ n = 100 \] **Sum (\(S\))**: \[ S = \frac{n}{2} (a + l) = \frac{100}{2} (50 - 643) = 50 \times (-593) = -29650 \] --- ### 4. \( 2 + \frac{13}{4} + \frac{9}{2} + \ldots + 82 \) **First Term (\(a\))**: 2 **Common Difference (\(d\))**: \(\frac{13}{4} - 2 = \frac{5}{4}\) **Last Term (\(l\))**: 82 **Number of Terms (\(n\))**: \[ a_n = a + (n-1)d \\ 82 = 2 + (n-1) \cdot \frac{5}{4} \\ 80 = (n-1) \cdot \frac{5}{4} \\ 80 \times \frac{4}{5} = n - 1 \\ 64 = n - 1 \\ n = 65 \] **Sum (\(S\))**: \[ S = \frac{n}{2} (a + l) = \frac{65}{2} (2 + 82) = 32.5 \times 84 = 2730 \] --- ### 5. \( -2p + 2p + 6p + \ldots + 102p \) **First Term (\(a\))**: \(-2p\) **Common Difference (\(d\))**: \(2p - (-2p) = 4p\) **Last Term (\(l\))**: 102p **Number of Terms (\(n\))**: \[ a_n = a + (n-1)d \\ 102p = -2p + (n-1) \cdot 4p \\ 102p + 2p = 4p(n-1) \\ 104p = 4pn - 4p \\ 108p = 4pn \\ n = \frac{108p}{4p} = 27 \] **Sum (\(S\))**: \[ S = \frac{n}{2} (a + l) = \frac{27}{2} (-2p + 102p) = \frac{27}{2} \times 100p = 27 \times 50p = 1350p \] --- ### 6. \( -15 - 12 - 9 - \ldots +432 \) **First Term (\(a\))**: -15 **Common Difference (\(d\))**: \(-12 - (-15) = 3\) **Last Term (\(l\))**: 432 **Number of Terms (\(n\))**: \[ a_n = a + (n-1)d \\ 432 = -15 + (n-1) \cdot 3 \\ 432 + 15 = 3(n-1) \\ 447 = 3n - 3 \\ 450 = 3n \\ n = 150 \] **Sum (\(S\))**: \[ S = \frac{n}{2} (a + l) = \frac{150}{2} (-15 + 432) = 75 \times 417 = 31275 \] --- ### **Summary of Sums** 1. \( 7 + 9 + 11 + \ldots + 105 = \mathbf{2,\!800} \) 2. \( 2 - 3 - 8 - \ldots - 368 = \mathbf{-13,\!725} \) 3. \( 50 + 43 + 36 + \ldots -643 = \mathbf{-29,\!650} \) 4. \( 2 + \frac{13}{4} + \frac{9}{2} + \ldots + 82 = \mathbf{2,\!730} \) 5. \( -2p + 2p + 6p + \ldots + 102p = \mathbf{1,\!350p} \) 6. \( -15 - 12 - 9 - \ldots +432 = \mathbf{31,\!275} \)

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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

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