▶️ Ejercicio 286 álgebra - Solucionario Baldor ✅✅

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CAPITULO XXXVII

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Ejercicio 286

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9° término de $\div$ 7.10.13….
$\begin{matrix} S e a & : \\ a = 7, n = 9, r = 10 - 7 = 3 \\ u & = a + (n - 1) r \\ u & = 7 + (9 - 1) 3 \\ u & = 7 + 8 (3) \\ u & = 7 + 24 \\ u & = 31 \end{matrix}$
12° término de $\div$ 5.10.15….
$\begin{matrix} S e a & : \\ a = 5, n = 12, r = 10 - 5 = 5 \\ u & = a + (n - 1) r \\ u & = 5 + (12 - 1) 5 \\ u & = 5 + 11 (5) \\ u & = 5 + 55 \\ u & = 60 \end{matrix}$
48° término de $\div$ 9.12.15….
$\begin{matrix} S e a & : \\ a = 9, n = 48, r = 12 - 9 = 3 \\ u & = a + (n - 1) r \\ u & = 9 + (48 - 1) 3 \\ u & = 9 + 47 (3) \\ u & = 9 + 141 \\ u & = 150 \end{matrix}$
63° término de $\div$ 3.10.17….
$\begin{matrix} S e a & : \\ a = 3, n = 63, r = 10 - 3 = 7 \\ u & = a + (n - 1) r \\ u & = 3 + (63 - 1) 7 \\ u & = 3 + 62 (7) \\ u & = 3 + 434 \\ u & = 437 \end{matrix}$
12° término de $\div$ 11.6.1….
$\begin{matrix} S e a & : \\ a = 11, n = 12, r = 6 - 11 = - 5 \\ u & = a + (n - 1) r \\ u & = 11 + (12 - 1) (- 5) \\ u & = 11 + 11 (- 5) \\ u & = 11 - 55 \\ u & = - 44 \end{matrix}$
28° término de $\div$ 19.12.5….
$\begin{matrix} S e a & : \\ a = 19, n = 28, r = 12 - 19 = - 7 \\ u & = a + (n - 1) r \\ u & = 19 + (28 - 1) (- 7) \\ u & = 19 + 27 (- 7) \\ u & = 19 - 189 \\ u & = - 170 \end{matrix}$
13° término de $\div$ 3.-1.-5….
$\begin{matrix} S e a & : \\ a = 3, n = 13, r = - 1 - 3 = - 4 \\ u & = a + (n - 1) r \\ u & = 3 + (13 - 1) (- 4) \\ u & = 3 + 12 (- 4) \\ u & = 3 - 48 \\ u & = - 45 \end{matrix}$
54° término de $\div$ 8.0.-8….
$\begin{matrix} S e a & : \\ a = 8, n = 54, r = 0 - 8 = - 8 \\ u & = a + (n - 1) r \\ u & = 8 + (54 - 1) (- 8) \\ u & = 8 + 53 (- 8) \\ u & = 8 - 424 \\ u & = - 416 \end{matrix}$
31° término de $\div$ -7.-3.1….
$\begin{matrix} S e a & : \\ a = - 7, n = 31, r = - 3 - (- 7) = 4 \\ u & = a + (n - 1) r \\ u & = - 7 + (31 - 1) (4) \\ u & = - 7 + 30 (4) \\ u & = - 7 + 120 \\ u & = 113 \end{matrix}$
17° término de $\div$ -8.2.12….
$\begin{matrix} S e a & : \\ a = - 8, n = 17, r = 2 - (- 8) = 10 \\ u & = a + (n - 1) r \\ u & = - 8 + (17 - 1) (10) \\ u & = - 8 + 16 (10) \\ u & = - 8 + 160 \\ u & = 152 \end{matrix}$
12° término de $\div \frac{1}{2} . \frac{3}{4} .1…$
$\begin{matrix} S e a & : \\ a = \frac{1}{2}, n = 12, r = \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \\ u & = a + (n - 1) r \\ u & = \frac{1}{2} + (12 - 1) (\frac{1}{4}) \\ u & = \frac{1}{2} + 11 (\frac{1}{4}) \\ u & = \frac{1}{2} + \frac{11}{4} \\ u & = \frac{13}{4} \leftrightarrow 3 \frac{1}{4} \end{matrix}$
17° término de $\div \frac{2}{3} . \frac{5}{6} .1…$
$\begin{matrix} S e a & : \\ a = \frac{2}{3}, n = 17, r = \frac{5}{6} - \frac{2}{3} = \frac{1}{6} \\ u & = a + (n - 1) r \\ u & = \frac{2}{3} + (17 - 1) (\frac{1}{6}) \\ u & = \frac{2}{3} + 16 (\frac{1}{6}) \\ u & = \frac{2}{3} + \frac{8}{3} \\ u & = \frac{10}{3} \leftrightarrow 3 \frac{1}{3} \end{matrix}$
25° término de $\div \frac{3}{8} . \frac{11}{24} …$
$\begin{matrix} S e a & : \\ a = \frac{3}{8}, n = 25, r = \frac{11}{24} - \frac{3}{8} = \frac{1}{12} \\ u & = a + (n - 1) r \\ u & = \frac{3}{8} + (25 - 1) (\frac{1}{12}) \\ u & = \frac{3}{8} + 124 (\frac{1}{12}) \\ u & = \frac{3}{8} + 2 \\ u & = \frac{19}{8} \leftrightarrow 2 \frac{3}{8} \end{matrix}$
19° término de $\div \frac{1}{3} . \frac{7}{8} ….$
$\begin{matrix} S e a & : \\ a = \frac{1}{3}, n = 19, r = \frac{7}{8} - \frac{1}{3} = \frac{13}{24} \\ u & = a + (n - 1) r \\ u & = \frac{1}{3} + (19 - 1) (\frac{13}{24}) \\ u & = \frac{1}{3} + 18 (\frac{13}{24}) \\ u & = \frac{1}{3} + \frac{39}{4} \\ u & = \frac{121}{12} \leftrightarrow 10 \frac{1}{12} \end{matrix}$
27° término de $\div 3 \frac{1}{2} .5 \frac{1}{4} ….$
$\begin{matrix} S e a & : \\ a = 3 \frac{1}{2}, n = 27, r = 5 \frac{1}{4} - 3 \frac{1}{2} = \frac{7}{4} \\ u & = a + (n - 1) r \\ u & = 3 \frac{1}{2} + (27 - 1) (\frac{7}{4}) \\ u & = \frac{7}{2} + 26 (\frac{7}{4}) \\ u & = \frac{7}{2} + \frac{91}{2} \\ u & = \frac{}{2} \\ u & = 49 \end{matrix}$
36° término de $\div \frac{7}{9} . \frac{1}{3} ….$
$\begin{matrix} S e a & : \\ a = \frac{7}{9}, n = 36, r = \frac{1}{3} - \frac{7}{9} = - \frac{4}{9} \\ u & = a + (n - 1) r \\ u & = \frac{7}{9} + (36 - 1) (- \frac{4}{9}) \\ u & = \frac{7}{9} + 35 (- \frac{4}{9}) \\ u & = \frac{7}{9} - \frac{140}{9} \\ u & = - \frac{133}{9} \leftrightarrow - 14 \frac{7}{9} \end{matrix}$
15° término de $\div \frac{2}{7} . \frac{1}{8} ….$
$\begin{matrix} S e a & : \\ a = \frac{2}{7}, n = 15, r = \frac{1}{8} - \frac{2}{7} = - \frac{9}{56} \\ u & = a + (n - 1) r \\ u & = \frac{2}{7} + (15 - 1) (- \frac{9}{56}) \\ u & = \frac{2}{7} + 14 (- \frac{9}{56}) \\ u & = \frac{2}{7} - \frac{9}{4} \\ u & = - \frac{55}{28} \leftrightarrow - 1 \frac{27}{28} \end{matrix}$
21° término de $\div - \frac{3}{5} . - \frac{14}{15} ….$
$\begin{matrix} S e a & : \\ a = - \frac{3}{5}, n = 21, r = - \frac{14}{15} + \frac{3}{5} = - \frac{1}{3} \\ u & = a + (n - 1) r \\ u & = - \frac{3}{5} + (21 - 1) (- \frac{1}{3}) \\ u & = - \frac{3}{5} + 20 (- \frac{1}{3}) \\ u & = - \frac{3}{5} - \frac{20}{3} \\ u & = - \frac{109}{15} \leftrightarrow - 7 \frac{4}{15} \end{matrix}$
13° término de $\div - \frac{1}{4} . - 2 \frac{1}{4} ….$
$\begin{matrix} S e a & : \\ a = - \frac{1}{4}, n = 13, r = - 2 \frac{1}{4} + \frac{1}{4} = - 2 \\ u & = a + (n - 1) r \\ u & = - \frac{1}{4} + (13 - 1) (- 2) \\ u & = - \frac{1}{4} + 12 (- 2) \\ u & = - \frac{1}{4} - 24 \\ u & = - \frac{97}{4} \leftrightarrow - 24 \frac{1}{4} \end{matrix}$
19° término de $\div - \frac{5}{6} . - \frac{1}{3} ….$
$\begin{matrix} S e a & : \\ a = - \frac{5}{6}, n = 19, r = - \frac{1}{3} + \frac{5}{6} = \frac{1}{2} \\ u & = a + (n - 1) r \\ u & = - \frac{5}{6} + (19 - 1) (\frac{1}{2}) \\ u & = - \frac{5}{6} + 18 (\frac{1}{2}) \\ u & = - \frac{5}{6} + 9 \\ u & = \frac{49}{6} \leftrightarrow 8 \frac{1}{6} \end{matrix}$
33° término de $\div 3 \frac{2}{3} .2 \frac{11}{12} ….$
$\begin{matrix} S e a & : \\ a = 3 \frac{2}{3}, n = 33, r = 2 \frac{11}{12} - 3 \frac{2}{3} = - \frac{3}{4} \\ u & = a + (n - 1) r \\ u & = \frac{11}{3} + (33 - 1) (- \frac{3}{4}) \\ u & = \frac{11}{3} + 32 (- \frac{3}{4}) \\ u & = \frac{11}{3} - 24 \\ u & = - \frac{61}{3} \leftrightarrow - 20 \frac{1}{3} \end{matrix}$
41° término de $\div 2 \frac{4}{5} .2 \frac{7}{10} ….$
$\begin{matrix} S e a & : \\ a = 2 \frac{4}{5}, n = 41, r = 2 \frac{7}{10} - 2 \frac{4}{5} = - \frac{1}{10} \\ u & = a + (n - 1) r \\ u & = 2 \frac{4}{5} + (41 - 1) (- \frac{1}{10}) \\ u & = \frac{14}{5} + 40 (- \frac{1}{10}) \\ u & = \frac{14}{5} - 4 \\ u & = - \frac{6}{5} \leftrightarrow - 1 \frac{1}{5} \end{matrix}$
26° término de $\div - \frac{3}{5} . \frac{3}{10} ….$
$\begin{matrix} S e a & : \\ a = - \frac{3}{5}, n = 26, r = \frac{3}{10} + \frac{3}{5} = \frac{9}{10} \\ u & = a + (n - 1) r \\ u & = - \frac{3}{5} + (26 - 1) (\frac{9}{10}) \\ u & = - \frac{3}{5} + 25 (\frac{9}{10}) \\ u & = - \frac{3}{5} + \frac{45}{2} \\ u & = \frac{219}{10} \leftrightarrow 21 \frac{9}{10} \end{matrix}$
19° término de $\div - 4. - \frac{2}{3} ….$
$\begin{matrix} S e a & : \\ a = - 4, n = 19, r = - \frac{2}{3} + 4 = \frac{10}{3} \\ u & = a + (n - 1) r \\ u & = - 4 + (19 - 1) (\frac{10}{3}) \\ u & = - 4 + 18 (\frac{10}{3}) \\ u & = - 4 + 60 \\ u & = 56 \end{matrix}$
39° término de $\div 3. - 1 \frac{1}{4} ….$
$\begin{matrix} S e a & : \\ a = 3, n = 39, r = - 1 \frac{1}{4} - 3 = - \frac{17}{4} \\ u & = a + (n - 1) r \\ u & = 3 + (39 - 1) (- \frac{17}{4}) \\ u & = 3 + 38 (- \frac{17}{4}) \\ u & = 3 - \frac{323}{2} \\ u & = - \frac{317}{3} \leftrightarrow - 158 \frac{1}{2} \end{matrix}$

Categories: Capítulo XXXVII