▶️ Ejercicio 294 álgebra - Solucionario Baldor ✅✅

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CAPITULO XXXVII

Progresiones Geométricas

Ejercicio 294

Interpolar:

3 medios geométricos entre 5 y 3125.
$\begin{matrix} S e a : \\ a = 5, u = 3125, n = 5 \\ \div \div 5…………………3125 \\ r & = \sqrt[n - 1]{\frac{u}{a}} \\ r & = \sqrt[5 - 1]{\frac{}{5}} \\ r & = \sqrt[4]{625} \\ r & = \sqrt[4]{5^{4}} \\ r & = \pm 5 \\ 5 \times \pm 5 = \pm 25 \\ 25 \times \pm 5 = \pm 125 \\ 125 \times \pm 5 = \pm 625 \\ \div \div 5 : \pm 25 : \pm 125 : \pm 625 : 3125 \end{matrix}$
4 medios geométricos entre -7 y -224.
$\begin{matrix} S e a : \\ a = - 7, u = - 224, n = 6 \\ \div \div - 7………………… - 224 \\ r & = \sqrt[n - 1]{\frac{u}{a}} \\ r & = \sqrt[6 - 1]{\frac{-}{- 7}} \\ r & = \sqrt[5]{32} \\ r & = \sqrt[5]{2^{5}} \\ r & = 2 \\ - 7 \times 2 = - 14 \\ - 14 \times 2 = - 28 \\ - 28 \times 2 = - 56 \\ - 56 \times 2 = - 112 \\ \div \div - 7 : - 14 : - 28 : - 56 : - 112 : - 224 \end{matrix}$
5 medios geométricos entre 128 y 2.
$\begin{matrix} S e a : \\ a = 128, u = 2, n = 7 \\ \div \div 128…………………2 \\ r & = \sqrt[n - 1]{\frac{u}{a}} \\ r & = \sqrt[7 - 1]{\frac{2}{}} \\ r & = \sqrt[6]{\frac{1}{64}} \\ r & = \sqrt[6]{\frac{1}{2^{6}}} \\ r & = \pm \frac{1}{2} \\ 128 (\pm \frac{1}{2}) = \pm 64 \\ 64 (\pm \frac{1}{2}) = \pm 32 \\ 32 (\pm \frac{1}{2}) = \pm 16 \\ 16 (\pm \frac{1}{2}) = \pm 8 \\ 8 (\pm \frac{1}{2}) = \pm 4 \\ \div \div 128 : \pm 64 : \pm 32 : \pm 16 : \pm 8 : \pm 4 : 2 \end{matrix}$
4 medios geométricos entre $4 \frac{1}{2}$ y $\frac{16}{27}$ .
$\begin{matrix} S e a : \\ a = \frac{9}{2}, u = \frac{16}{27}, n = 6 \\ \div \div \frac{9}{2} ………………… \frac{16}{27} \\ r & = \sqrt[n - 1]{\frac{u}{a}} \\ r & = \sqrt[6 - 1]{\frac{\frac{16}{27}}{\frac{9}{2}}} \\ r & = \sqrt[5]{\frac{\frac{2^{4}}{3^{3}}}{\frac{3^{2}}{2}}} \\ r & = \sqrt[5]{\frac{2^{5}}{3^{5}}} \\ r & = \frac{2}{3} \\ \frac{9}{2} \times \frac{2}{3} = 3 \\ 3 \times \frac{2}{3} = 2 \\ 2 \times \frac{2}{3} = \frac{4}{3} \\ \frac{4}{3} \times \frac{2}{3} = \frac{8}{9} \\ \div \div \frac{9}{2} : 3 : 2 : \frac{4}{3} : \frac{8}{9} : \frac{16}{27} \\ \div \div 4 \frac{1}{2} : 3 : 2 : 1 \frac{1}{3} : \frac{8}{9} : \frac{16}{27} \end{matrix}$
6 medios geométricos entre 2 y $34 \frac{11}{64}$ .
$\begin{matrix} S e a : \\ a = 2, u = \frac{2187}{64}, n = 8 \\ \div \div 2………………… \frac{2187}{64} \\ r & = \sqrt[n - 1]{\frac{u}{a}} \\ r & = \sqrt[8 - 1]{\frac{\frac{2187}{64}}{2}} \\ r & = \sqrt[7]{\frac{2187}{128}} \\ r & = \sqrt[7]{\frac{3^{7}}{2^{7}}} \\ r & = \frac{3}{2} \\ 2 \times \frac{3}{2} = 3 \\ 3 \times \frac{3}{2} = \frac{9}{2} \\ \frac{9}{2} \times \frac{3}{2} = \frac{27}{4} \\ \frac{27}{4} \times \frac{3}{2} = \frac{81}{8} \\ \frac{81}{8} \times \frac{3}{2} = \frac{243}{16} \\ \frac{243}{16} \times \frac{3}{2} = \frac{729}{32} \\ \div \div 2 : 3 : \frac{9}{2} : \frac{27}{4} : \frac{81}{8} : \frac{243}{16} : \frac{729}{32} : \frac{2187}{64} \\ \div \div 2 : 3 : 4 \frac{1}{2} : 6 \frac{3}{4} : 10 \frac{1}{8} : 15 \frac{3}{16} : 22 \frac{25}{32} : 34 \frac{11}{64} \end{matrix}$
4 medios geométricos entre $\frac{4}{9}$ y $\frac{27}{256}$ .
$\begin{matrix} S e a : \\ a = \frac{4}{9}, u = \frac{27}{256}, n = 6 \\ \div \div \frac{4}{9} ………………… \frac{27}{256} \\ r & = \sqrt[n - 1]{\frac{u}{a}} \\ r & = \sqrt[6 - 1]{\frac{\frac{27}{256}}{\frac{4}{9}}} \\ r & = \sqrt[5]{\frac{\frac{3^{3}}{4^{4}}}{\frac{4}{3^{2}}}} \\ r & = \sqrt[5]{\frac{3^{5}}{4^{5}}} \\ r & = \frac{3}{4} \\ \frac{4}{9} \times \frac{3}{4} = \frac{1}{3} \\ \frac{1}{3} \times \frac{3}{4} = \frac{1}{4} \\ \frac{1}{4} \times \frac{3}{4} = \frac{3}{16} \\ \frac{3}{16} \times \frac{3}{4} = \frac{9}{64} \\ \div \div \frac{4}{9} : \frac{1}{3} : \frac{1}{4} : \frac{3}{16} : \frac{9}{64} : \frac{27}{256} \end{matrix}$
7 medios geométricos entre 8 y $\frac{1}{32}$ .
$\begin{matrix} S e a : \\ a = 8, u = \frac{1}{32}, n = 9 \\ \div \div 8………………… \frac{1}{32} \\ r & = \sqrt[n - 1]{\frac{u}{a}} \\ r & = \sqrt[9 - 1]{\frac{\frac{1}{32}}{8}} \\ r & = \sqrt[8]{\frac{\frac{1}{2^{5}}}{2^{3}}} \\ r & = \sqrt[8]{\frac{1}{2^{8}}} \\ r & = \pm \frac{1}{2} \\ 8 (\pm \frac{1}{2}) = \pm 4 \\ 4 (\pm \frac{1}{2}) = \pm 2 \\ 2 (\pm \frac{1}{2}) = \pm 1 \\ 1 (\pm \frac{1}{2}) = \pm \frac{1}{2} \\ \frac{1}{2} (\pm \frac{1}{2}) = \pm \frac{1}{4} \\ \frac{1}{4} (\pm \frac{1}{2}) = \pm \frac{1}{8} \\ \frac{1}{8} (\pm \frac{1}{2}) = \pm \frac{1}{16} \\ \div \div 8 : \pm 4 : \pm 2 : \pm 1 : \pm \frac{1}{2} : \pm \frac{1}{4} : \pm \frac{1}{8} : \pm \frac{1}{16} : \pm \frac{1}{32} \end{matrix}$

Categories: Capítulo XXXVII