// Numbas version: finer_feedback_settings {"name": "Encontrar la f\u00f3rmula para el t\u00e9rmino $n^{\\text {th}}$ de una secuencia lineal", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"parts": [{"type": "gapfill", "showFeedbackIcon": true, "scripts": {}, "stepsPenalty": 0, "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "customMarkingAlgorithm": "", "steps": [{"showCorrectAnswer": true, "prompt": "

La fórmula para el término $n^{\\text{th}}$ de una secuencia aritmética es: \\[a_n = a_1 + (n-1)d, \\]

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donde

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Para esta secuencia aritmética, ¿qué es $ a_1 $?

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¿Qué es $d$?

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$\\var{m[1]*2}, \\var{m[1]*3}, \\var{m[1]*4}, \\ldots$

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Término de lugar  $n^\\text{th}$ = [[0]]

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La fórmula para el término $ n^{\\text {th}}$ de una secuencia aritmética es $a_n = a_1 + (n-1) d $
dónde
$ a_n $ es el término $ n ^ \\ text {th} $;
$ a_1 $ es el primer término en la secuencia;
$ d $ es la diferencia común entre términos consecutivos.

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Para esta secuencia aritmética, ¿qué es $ a_1 $?

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¿Qué es $d$?

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$\\var{m[2]*8+2}, \\var{m[2]*7+2}, \\var{m[2]*6+2}...$

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Término $n^\\text{th}$ = [[0]]

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Dados los primeros tres términos de una secuencia, da una fórmula para el término $ n^\\text {th} $.

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En la primera secuencia, $d$ es positivo. En la segunda secuencia, $d$ es negativo.

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Una secuencia lineal es una serie de números que aumenta o disminuye en una cantidad constante en cada paso.

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Encuentre fórmulas para el término $n^ {\\text{th}} $ para cada una de las siguientes secuencias lineales, donde se dan los valores para $n = 1 \\text {,} 2 \\text {,}3 $:

", "variables": {"n": {"group": "Ungrouped variables", "description": "", "name": "n", "templateType": "anything", "definition": "repeat(random(1..4),7)"}, "ci": {"group": "Ungrouped variables", "description": "", "name": "ci", "templateType": "anything", "definition": "repeat(random(6..20),10)"}, "m": {"group": "Ungrouped variables", "description": "", "name": "m", "templateType": "anything", "definition": "repeat(random(2..10),5)"}, "c": {"group": "Ungrouped variables", "description": "", "name": "c", "templateType": "anything", "definition": "repeat(random(3..13 except[10]),8)"}, "b": {"group": "Ungrouped variables", "description": "", "name": "b", "templateType": "anything", "definition": "repeat(random(2..4), 5)"}, "ni": {"group": "Ungrouped variables", "description": "", "name": "ni", "templateType": "anything", "definition": "repeat(random(19..40),10)"}}, "advice": "

Ambas secuencias son lineales o aritméticas. Para encontrar fórmulas para estas secuencias necesitamos identificar sus primeros términos y diferencias comunes..

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a)

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La fórmula para el término $n^\\text {th} $ de una secuencia aritmética es: $a_n = a_1 + (n-1) d \\text {.}$

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$ a_1 $ es el primer término y $d$ la diferencia común entre términos consecutivos, que debemos identificar.

\n

Podemos encontrarlos dibujando una tabla de $a_n$ contra $ n $, y las diferencias entre términos consecutivos.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$n$123
$a_n$$\\pmb{\\var{m[1]*2}}$$\\var{m[1]*3}$$\\var{m[1]*4}$
Primeras diferencias$\\pmb{\\var{m[1]}}$$\\pmb{\\var{m[1]}}$
\n

El primer término y la diferencia común se han resaltado en negrita; Podemos usar estos para escribir la fórmula de la secuencia.

\n

\\begin{align}
a_n &= a_1+(n-1)d \\\\
&= \\var{m[1]*2}+(n-1)\\times\\var{m[1]} \\\\
&= \\var{m[1]}n + \\var{m[1]}\\text{.}
\\end{align}

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b)

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De manera similar a la parte a), podemos identificar $ a_1 $ y $ d $ para esta secuencia dibujando una tabla de $ a_n $ contra $ n $.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$a_n$$\\pmb{\\var{m[2]*8+2}}$$\\simplify{{m[2]}*7+2}$$\\simplify{{m[2]}*6+2}$
Primeras diferencias$\\pmb{\\var{-m[2]}}$$\\pmb{\\var{-m[2]}}$
\n

El primer término y la diferencia común se han resaltado en negrita; Podemos usar estos para formar la fórmula para la secuencia.

\n

\\begin{align}
a_n &=a_1+(n-1)d \\\\
&=\\var{m[2]*8+2}+(n-1)\\times\\var{-m[2]} \\\\
&=-\\var{m[2]}n + \\var{m[2]*9+2}\\text{.}
\\end{align}

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