▶️ Ejercicio 295 álgebra - Solucionario Baldor ✅✅

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CAPITULO XXXVII

Progresiones Geométricas
Suma de una progresión geométrica decreciente infinita

Ejercicio 295

Hallar la suma de las progresiones infinitas:

$\begin{matrix} \div \div 2 & : \frac{1}{2} : \frac{1}{8} ………. \\ S e a \\ a = 2, r = \frac{\frac{1}{2}}{2} = \frac{1}{4} \\ S & = \frac{a}{1 - r} \\ S & = \frac{2}{1 - \frac{1}{4}} \\ S & = \frac{2}{\frac{3}{4}} \\ S & = \frac{8}{3} \leftrightarrow 2 \frac{2}{3} \end{matrix}$
$\begin{matrix} \div \div \frac{1}{2} & : \frac{1}{6} : \frac{1}{18} ………. \\ S e a : \\ a = \frac{1}{2}, r = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3} \\ S & = \frac{a}{1 - r} \\ S & = \frac{\frac{1}{2}}{1 - \frac{1}{3}} \\ S & = \frac{\frac{1}{2}}{\frac{2}{3}} \\ S & = \frac{3}{4} \end{matrix}$
$\begin{matrix} \div \div - 5 & : - 2 : - \frac{4}{5} ………. \\ S e a : \\ a = - 5, r = \frac{2}{5} \\ S & = \frac{a}{1 - r} \\ S & = \frac{- 5}{1 - \frac{2}{5}} \\ S & = \frac{- 5}{\frac{3}{5}} \\ S & = - \frac{25}{3} \end{matrix}$
$\begin{matrix} \div \div - 4 & : - \frac{8}{3} : - \frac{16}{9} ………. \\ S e a : \\ a = - 4, r = \frac{- \frac{}{3}}{- 4} = \frac{2}{3} \\ S & = \frac{a}{1 - r} \\ S & = \frac{- 4}{1 - \frac{2}{3}} \\ S & = \frac{- 4}{\frac{1}{3}} \\ S & = - 12 \end{matrix}$
$\begin{matrix} \div \div \frac{3}{4} & : \frac{1}{4} : \frac{1}{12} ………. \\ S e a : \\ a = \frac{3}{4}, r = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \\ S & = \frac{a}{1 - r} \\ S & = \frac{\frac{3}{4}}{1 - \frac{1}{3}} \\ S & = \frac{\frac{3}{4}}{\frac{2}{3}} \\ S & = \frac{9}{8} \leftrightarrow 1 \frac{1}{8} \end{matrix}$
$\begin{matrix} \div \div \frac{1}{6} & : \frac{1}{7} : \frac{6}{49} ………. \\ S e a : \\ a = \frac{1}{6}, r = \frac{\frac{1}{7}}{\frac{1}{6}} = \frac{6}{7} \\ S & = \frac{a}{1 - r} \\ S & = \frac{\frac{1}{6}}{1 - \frac{6}{7}} \\ S & = \frac{\frac{1}{6}}{\frac{1}{7}} \\ S & = \frac{7}{6} \leftrightarrow 1 \frac{1}{6} \end{matrix}$
$\begin{matrix} \div \div 2 & : - \frac{2}{5} : \frac{2}{25} ………. \\ S e a : \\ a = 2, r = \frac{- \frac{2}{5}}{2} = - \frac{1}{5} \\ S & = \frac{a}{1 - r} \\ S & = \frac{2}{1 + \frac{1}{5}} \\ S & = \frac{2}{\frac{}{5}} \\ S & = \frac{5}{3} \leftrightarrow 1 \frac{2}{3} \end{matrix}$
$\begin{matrix} \div \div - 14 & : - 6 : - \frac{18}{7} ………. \\ S e a : \\ a = - 14, r = \frac{-}{-} = \frac{3}{7} \\ S & = \frac{a}{1 - r} \\ S & = \frac{- 14}{1 - \frac{3}{7}} \\ S & = \frac{-}{\frac{}{7}} \\ S & = - \frac{49}{2} \leftrightarrow - 24 \frac{1}{2} \end{matrix}$

Categories: Capítulo XXXVII