(e) How many terms of each the following arithmetic series must be added to obtain the given sum? \( \begin{array}{ll}\text { (1) } 5+9+13+\ldots=860 & \text { (2) } 10+7+4+\ldots=-238 \\ \text { (3) } 3-2-7-\ldots=-24450 & \text { (4) }-1-3-5-\ldots=-2500\end{array} \)

We use the sum formula for an arithmetic series:   Sₙ = (n/2) × [2a + (n – 1)d], where a is the first term, d is the common difference, and n is the number of terms. Let’s solve each case: ──────────────────────── (1) Series: 5 + 9 + 13 + …, Sum = 860  • a = 5, d = 9 – 5 = 4  • Sum formula:   n/2 [2·5 + (n – 1)·4] = 860   n/2 (10 + 4n – 4) = 860   n/2 (4n + 6) = 860  Multiply both sides by 2:   n(4n + 6) = 1720  Expand:   4n² + 6n – 1720 = 0  Divide by 2 for simplicity:   2n² + 3n – 860 = 0  Solve using the quadratic formula:   n = [–3 ± √(3² – 4·2·(–860))] / (2·2)   n = [–3 ± √(9 + 6880)] / 4   n = [–3 ± √6889] / 4 Since 83² = 6889, we take the positive solution:   n = (–3 + 83) / 4 = 80/4 = 20 ──────────────────────── (2) Series: 10 + 7 + 4 + …, Sum = –238  • a = 10, d = 7 – 10 = –3  • Sum formula:   n/2 [2·10 + (n – 1)(–3)] = –238   n/2 (20 – 3n + 3) = –238   n/2 (23 – 3n) = –238  Multiply both sides by 2:   n(23 – 3n) = –476  Expand:   23n – 3n² = –476  Rearrange:   –3n² + 23n + 476 = 0 Multiply by –1:   3n² – 23n – 476 = 0 Solve using the quadratic formula:   n = [23 ± √(23² – 4·3·(–476))] / (2·3)   n = [23 ± √(529 + 5712)] / 6   n = [23 ± √6241] / 6 Note that 79² = 6241 so:   n = [23 ± 79] / 6 Taking the positive solution:   n = (23 + 79)/6 = 102/6 = 17 ──────────────────────── (3) Series: 3 – 2 – 7 – …, Sum = –24450  • a = 3, d = (–2) – 3 = –5  • Sum formula:   n/2 [2·3 + (n – 1)(–5)] = –24450   n/2 (6 – 5n + 5) = –24450   n/2 (11 – 5n) = –24450 Multiply both sides by 2:   n(11 – 5n) = –48900 Expand:   11n – 5n² = –48900 Rearrange:   –5n² + 11n + 48900 = 0 Multiply by –1:   5n² – 11n – 48900 = 0 Solve by the quadratic formula:   n = [11 ± √(11² – 4·5·(–48900))] / (2·5)   n = [11 ± √(121 + 978000)] / 10   n = [11 ± √978121] / 10 Since 989² = 978121, we take the positive solution:   n = (11 + 989)/10 = 1000/10 = 100 ──────────────────────── (4) Series: –1 – 3 – 5 – …, Sum = –2500  • a = –1, d = (–3) – (–1) = –2  • Sum formula:   n/2 [2(–1) + (n – 1)(–2)] = –2500   n/2 [–2 – 2n + 2] = –2500 Notice that –2 + 2 = 0, so:   n/2 [–2n] = –2500   –n² = –2500 Thus:   n² = 2500 Taking the positive square root:   n = 50 ──────────────────────── Final Answers:   (1) 20 terms   (2) 17 terms   (3) 100 terms   (4) 50 terms