▶️ Ejercicio 291 álgebra - Solucionario Baldor ✅✅

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CAPITULO XXXVII

Progresiones Geométricas

Ejercicio 291

Hallar el 7° término de $\div \div 3 : 6 : 12….$
$\begin{matrix} S e a \\ a = 3, n = 7, r = 6 \div 3 = 2 \\ u & = a r^{n - 1} \\ u & = 3 (2^{7 - 1}) \\ u & = 3 (2^{6}) \\ u & = 3 (64) \\ u & = 192 \end{matrix}$
Hallar el 8° término de $\div \div \frac{1}{3} : 1 : 3….$
$\begin{matrix} S e a \\ a = \frac{1}{3}, n = 8, r = 3 \div 1 = 3 \\ u & = a r^{n - 1} \\ u & = \frac{1}{3} (3^{8 - 1}) \\ u & = \frac{1}{3} (3^{}) \\ u & = 3^{6} \\ u & = 729 \end{matrix}$
Hallar el 9° término de $\div \div 8 : 4 : 2….$
$\begin{matrix} S e a \\ a = 8, n = 9, r = 4 \div 8 = \frac{1}{2} \\ u & = a r^{n - 1} \\ u & = 8 {(\frac{1}{2})}^{9 - 1} \\ u & = 2^{3} (\frac{1}{2^{8}}) \\ u & = \frac{1}{2^{5}} \\ u & = \frac{1}{32} \end{matrix}$
Hallar el 6° término de $\div \div 1 : \frac{2}{5} : \frac{4}{25} ….$
$\begin{matrix} S e a \\ a = 1, n = 6, r = \frac{2}{5} \div 1 = \frac{2}{5} \\ u & = a r^{n - 1} \\ u & = 1 {(\frac{2}{5})}^{6 - 1} \\ u & = {(\frac{2}{5})}^{5} \\ u & = \frac{2^{5}}{5^{5}} \\ u & = \frac{32}{3125} \end{matrix}$
Hallar el 7° término de $\div \div 3 : 2 : \frac{4}{3} ….$
$\begin{matrix} S e a \\ a = 3, n = 7, r = 2 \div 3 = \frac{2}{3} \\ u & = a r^{n - 1} \\ u & = 3 {(\frac{2}{3})}^{7 - 1} \\ u & = 3 {(\frac{2}{3})}^{6} \\ u & = 3 (\frac{2^{6}}{3^{}}) \\ u & = \frac{2^{6}}{3^{5}} \\ u & = \frac{64}{243} \end{matrix}$
Hallar el 6° término de $\div \div \frac{1}{2} : \frac{1}{5} ….$
$\begin{matrix} S e a \\ a = \frac{1}{2}, n = 6, r = \frac{1}{5} \div \frac{1}{2} = \frac{2}{5} \\ u & = a r^{n - 1} \\ u & = \frac{1}{2} {(\frac{2}{5})}^{6 - 1} \\ u & = \frac{1}{2} {(\frac{2}{5})}^{5} \\ u & = \frac{1}{2} (\frac{2^{}}{5^{5}}) \\ u & = \frac{2^{4}}{5^{5}} \\ u & = \frac{16}{3125} \end{matrix}$
Hallar el 8° término de $\div \div 2 \frac{1}{4} : 3….$
$\begin{matrix} S e a \\ a = \frac{9}{4}, n = 8, r = 3 \div \frac{9}{4} = \frac{4}{3} \\ u & = a r^{n - 1} \\ u & = \frac{9}{4} {(\frac{4}{3})}^{8 - 1} \\ u & = \frac{3^{2}}{4} {(\frac{4}{3})}^{7} \\ u & = \frac{3^{2}}{4} (\frac{4^{}}{3^{}}) \\ u & = \frac{4^{6}}{3^{5}} \\ u & = \frac{4096}{243} \leftrightarrow 16 \frac{202}{243} \end{matrix}$
Hallar el 6° término de $\div \div - 3 : 6 : - 12….$
$\begin{matrix} S e a \\ a = - 3, n = 6, r = 6 \div - 3 = - 2 \\ u & = a r^{n - 1} \\ u & = - 3 {(- 2)}^{6 - 1} \\ u & = - 3 {(- 2)}^{5} \\ u & = - 3 (- 32) \\ u & = \frac{4^{6}}{3^{5}} \\ u & = 96 \end{matrix}$
Hallar el 9° término de $\div \div 3 : - 1 : \frac{1}{3} ….$
$\begin{matrix} S e a \\ a = 3, n = 9, r = - 1 \div 3 = - \frac{1}{3} \\ u & = a r^{n - 1} \\ u & = 3 {(- \frac{1}{3})}^{9 - 1} \\ u & = 3 {(- \frac{1}{3})}^{8} \\ u & = 3 (\frac{1}{3^{}}) \\ u & = \frac{1}{3^{7}} \\ u & = \frac{1}{2187} \end{matrix}$
Hallar el 5° término de $\div \div \frac{5}{6} : \frac{1}{2} ….$
$\begin{matrix} S e a \\ a = \frac{5}{6}, n = 5, r = \frac{1}{2} \div \frac{5}{6} = \frac{3}{5} \\ u & = a r^{n - 1} \\ u & = \frac{5}{6} {(\frac{3}{5})}^{5 - 1} \\ u & = \frac{5}{6} {(\frac{3}{5})}^{4} \\ u & = \frac{5}{2. 3} (\frac{3^{}}{5^{}}) \\ u & = \frac{3^{3}}{2. 5^{3}} \\ u & = \frac{27}{250} \end{matrix}$
Hallar el 8° término de $\div \div 16 : - 4 : 1….$
$\begin{matrix} S e a \\ a = 16, n = 8, r = - 4 \div 16 = - \frac{1}{4} \\ u & = a r^{n - 1} \\ u & = 16 {(- \frac{1}{4})}^{8 - 1} \\ u & = 4^{2} {(- \frac{1}{4})}^{7} \\ u & = 4^{2} (- \frac{1}{4^{}}) \\ u & = - \frac{1}{4^{5}} \\ u & = - \frac{1}{1024} \end{matrix}$
Hallar el 8° término de $\div \div \frac{3}{4} : - \frac{1}{2} : \frac{1}{3} ….$
$\begin{matrix} S e a \\ a = \frac{3}{4}, n = 8, r = - \frac{1}{2} \div \frac{3}{4} = - \frac{2}{3} \\ u & = a r^{n - 1} \\ u & = \frac{3}{4} {(- \frac{2}{3})}^{8 - 1} \\ u & = \frac{3}{2^{2}} {(\frac{2}{3})}^{7} \\ u & = \frac{3}{2^{2}} (- \frac{2^{}}{3^{}}) \\ u & = - \frac{2^{5}}{3^{6}} \\ u & = - \frac{32}{29} \end{matrix}$
Hallar el 5° término de $\div \div - \frac{3}{5} : \frac{3}{2} : - \frac{15}{4} ….$
$\begin{matrix} S e a \\ a = - \frac{3}{5}, n = 5, r = \frac{3}{2} \div - \frac{3}{5} = - \frac{5}{2} \\ u & = a r^{n - 1} \\ u & = - \frac{3}{5} {(- \frac{5}{2})}^{5 - 1} \\ u & = - \frac{3}{5} {(- \frac{5}{2})}^{4} \\ u & = - \frac{3}{5} (\frac{5^{}}{2^{4}}) \\ u & = - \frac{3. 5^{3}}{2^{4}} \\ u & = - \frac{375}{16} \leftrightarrow - 23 \frac{7}{16} \end{matrix}$
Hallar el 10° término de $\div \div - \frac{3}{4} : - \frac{1}{4} : - \frac{1}{12} ….$
$\begin{matrix} S e a \\ a = - \frac{3}{4}, n = 10, r = - \frac{1}{4} \div - \frac{3}{4} = \frac{1}{3} \\ u & = a r^{n - 1} \\ u & = - \frac{3}{4} {(\frac{1}{3})}^{10 - 1} \\ u & = - \frac{3}{4} {(\frac{1}{3})}^{9} \\ u & = - \frac{3}{4} (\frac{1}{3^{}}) \\ u & = - \frac{1}{4. 3^{8}} \\ u & = - \frac{1}{26244} \end{matrix}$