Página — Matematica 3° Secundaria (pág. 28)

Recuerda

En una progresión geométrica,la razón es constante y se puede hallar dividiendo un término cualquiera entre su

antecesor.

r=\frac{\partial_{_2}}{\partial_{_1}}=\frac{\partial_{_3}}{\partial_{_2}}=\ldots=\frac{\partial_{_n}}{\partial_{_{n-1}}}

Ejemplo Determina la razón de las siguientes P. G.: a.-10;50;-250;1250; b 0,2;0,04;0,008;

C.2\sqrt{2};2\sqrt{6};6\sqrt{2};..

Resolución

a

r=\frac{50}{-10}=-5

b.

r=\frac{0.04}{0.2}=0.2

r=\frac{6\sqrt{2}}{2\sqrt{6}}=\sqrt{3}

Ten en cuenta

El término n-ésimo de una P. G.está dado por esta expresión: a_{n}=a_{1}\cdot r^{n-1} a: primer término r:razón n:posición

\begin{aligned}&\begin{aligned}\\ &\mathbf{E}\mathbf{j}\mathbf{e}\mathbf{m}\mathbf{p}\mathbf{l}\mathbf{o}\\&\mathbf{H}\mathbf{a}\mathbf{l}\mathbf{a}\mathbf{e}\mathbf{l}\mathbf{t}\mathbf{c}\mathbf{r}\mathbf{m}\mathbf{i}\mathbf{o}\mathbf{\rho}\mathbf{a}_{_{10}}\mathbf{d}\mathbf{e}\mathbf{l}\mathbf{a}\mathbf{s}\\&\mathbf{s}\mathbf{g}\mathbf{u}\mathbf{i}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{s}\mathbf{P}\mathbf{.}\mathbf{G}\mathbf{.}\mathbf{j}\\&\mathbf{a}\mathbf{.}\mathbf{\rho}\mathbf{-}\mathbf{l}\mathbf{0}\mathbf{;}\mathbf{5}\mathbf{0}\mathbf{;}\mathbf{\rho}\mathbf{-}\mathbf{2}5\mathbf{0}\mathbf{;}\mathbf{\rho}\mathbf{1}\mathbf{2}5\mathbf{0}\mathbf{;}\mathbf{\rho}\mathbf{s}\mathbf{\rho}\\&\mathbf{b}\mathbf{.}\mathbf{\rho}\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{;}\mathbf{\rho}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{4}\mathbf{;}\mathbf{\rho}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{0}\mathbf{8}\mathbf{;}\mathbf{\rho}\mathbf{s}\mathbf{\rho}\\&\mathbf{c}\mathbf{.}\mathbf{\rho}\mathbf{2}\sqrt{\mathbf{2}}\mathbf{;}\mathbf{\rho}\mathbf{2}\sqrt{\mathbf{6}}\mathbf{;}\mathbf{\rho}\mathbf{6}\sqrt{\mathbf{2}}\mathbf{;}\mathbf{\rho}\mathbf{s}\mathbf{\rho}\\&\mathbf{R}\mathbf{e}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{c}\mathbf{i}\mathbf{o}\mathbf{n}\\&\mathbf{a}\mathbf{.}\mathbf{\rho}\mathbf{a}_{_{10}}=(-10)(-5)^{\circ}\\&=19\ 531\ 250\\&\mathbf{b}\mathbf{.}\mathbf{\rho}\mathbf{a}_{_{10}}=(0\mathbf{,}\mathbf{2})(0\mathbf{,}\mathbf{2})^{\circ}\\&=\mathbf{1,}\mathbf{0}\mathbf{24}\times\mathbf{10^{-7}}\\&\mathbf{c}\mathbf{.}\mathbf{\rho}\mathbf{a}_{_{10}}=(2\sqrt{\mathbf{2}})(\sqrt{\mathbf{3}})^{\circ}\\&=\mathbf{162}\sqrt{\mathbf{6}}\\ &\end{aligned}\\ \end{aligned}

situación B: Comisión en Ia venta de motos

Rubén firma un contrato como vendedor de motos en un concesionario. En este documento,se contempla pagarle una comisión por la venta de la primera moto y

luego duplicarle la comisión anterior por cada moto adicional que venda. Si vende 9 motos y recibe S/12 775de comisión total, icuánto le pagaron de comisión por la cuarta moto que vendió?

A continuación, analizamos los procedimientos planteados.

Resolución

Comisión por la primera venta:x

Comisión por la segunda venta:2\cdot x=2x=2^{2-1}\cdot x

●Comisión por la tercera venta:2\cdot(2x)=2^{2}\cdot x=2^{3-1}\cdot x

Comisión por la cuarta venta:2\cdot(2^{2}\cdot x)=2^{3}\cdot x=\sqrt[3]{} X

Se obtiene esta P.G.:x;2x;2^{2}\cdot x;2^{3}\cdot x;

Entonces, el término general es el siguiente:

a_{n}=2^{n-1}\cdot x

Para calcular cuánto le pagaron de comisión por la cuarta moto, se necesita hallar el valor de x, es decir, la comisión que recibió por la venta de la primera moto.

Dada la progresión geométrica :\partial_{1};\partial_{2};\partial_{3};\partial_{4};\partial_{5};\ldots;\partial_{n}..., podemos afirmar que la suma de sus n primeros +=1=.1==.. S_{n}=\bar{\alpha}_{1}+\bar{\alpha}_{2}+\bar{\alpha}_{3}+\bar{\alpha}_{4}+\bar{\alpha}_{5}+\ldots+\bar{\alpha}_{n-1}+\bar{\alpha}_{n}=\frac{\bar{\alpha}_{1}(r^{n}-1)}{r-1}

S_{n}=\bar{\alpha}_{1}+\bar{\alpha}_{2}+\bar{\alpha}_{3}+\bar{\alpha}_{4}+\bar{\alpha}_{5}+\ldots+\bar{\alpha}_{n-1}+\bar{\alpha}_{n}=\frac{\bar{\alpha}_{1}(r^{n}-1)}{r-1}

Esta expresión se obtiene de la siguiente manera:multiplicamos ambos miembros de la primera igualdad anterior por r:

\begin{array}{l}r\cdot S_{n}=\underbrace{r\cdot\bar{\alpha}_{1}+r\cdot\bar{\alpha}_{2}+r\cdot\bar{\alpha}_{3}+r\cdot\bar{\alpha}_{4}+r\cdot\bar{\alpha}_{5}+\ldots+r\cdot\bar{\alpha}_{n-1}+r\cdot\bar{\alpha}_{n}}_{\downarrow}\\r\cdot S_{n}=\underbrace{\bar{\alpha}_{2}+\bar{\alpha}_{3}+\bar{\alpha}_{4}+\bar{\alpha}_{5}+\bar{\alpha}_{6}+\ldots+\bar{\alpha}_{n}+r\cdot\bar{\alpha}_{n}}_{\downarrow}\\(-)S_{n}=\underbrace{\bar{\alpha}_{1}+\bar{\alpha}_{2}+\bar{\alpha}_{3}+\bar{\alpha}_{4}+\bar{\alpha}_{5}+\bar{\alpha}_{6}+\ldots+\bar{\alpha}_{n<1}+\bar{\alpha}_{n}}_{\downarrow}\\(-1)\cdot S_{n}=r\cdot\bar{\alpha}_{n}-\bar{\alpha}_{1}\\\overbrace{r\cdot r^{n-1}\cdot\bar{\alpha}_{1}\cdot r^{n}-\bar{\alpha}_{1}}^{\text{ 一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一�

Luego, aplicamos la fórmula de la suma de términos:

S_{n}=\frac{\partial_{1}\left(r^{n}-1\right)}{r-1}\rightarrow S_{n}=\frac{\partial_{1}\left(2^{9}-1\right)}{2-1}





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