节: 亚学和几何序列 | 3.7 测量和几何序列-interactive | The Way To Learn

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The Purpose of this Lesson
::本课程的目的

In this lesson, you will explore arithmetic and geometric sequences, distinguish them from other types of functions, and learn about their unique equations.
::在这个教训中,你将探索算术和几何序列,将它们与其他类型功能区分开来,并学习它们的独特方程。

Introduction : Writing Equations for Functions
::导言: 函数书写等量

Work it Out
::工作出来

The ability to translate a function will help you understand the structure of equations for sequences. For each of the following problems, a function is given. For each, you must translate the function as indicated and write the equation for the new function. A graph is not required.
::函数的翻译能力有助于您理解序列的方程式结构。对于以下每一个问题,都会给出一个函数。对于每一个问题,您必须按指示翻译函数,并为新函数写入方程式。不需要图形。

$\begin{aligned} \begin{array}{llcc} Function & Vertical Shift & Horizontal Shift \\ a. & f (x) = x^{2} & 3 & 5 \\ b. & g (x) = | x | & - 2 & 0 \\ c. & h (x) = 4 x^{2} & - 1 & - 1 \\ d. & f (x) = \sqrt{x} & 7 & - 4 \\ e. & y = g (x) & 10 & 15 \\ f. & h (x) = - \frac{1}{2} x + 5 & 7 & - 9 \end{array} \end{aligned}$
::函数Vertical ShiftHortical Shifta.f(x) =x235b.g(x) xx20c.h(x) =4x2 - 1 - 1d.f(x) =x7 - 4e.y=g(x) 1015f.h(x) 12x+57-9

Specifically, horizontal translations are most useful for understanding the equations of sequences. For each of the following problems, a function is given. For each, you must translate the function as indicated and write the equation for the new function. A graph is not required.
::具体地说,水平翻译对于理解序列的方程式最为有用。对于以下每一个问题,都指定了一个函数。对于每一个问题,必须按指示翻译函数,并写出新函数的方程式。不需要图表。

$\begin{aligned} \begin{array}{llcc} Function & Vertical Shift & Horizontal Shift \\ a. & f (x) = 2^{x} & 0 & 5 \\ b. & g (x) = 5 (3)^{x} & 0 & - 2 \\ c. & h (x) = 5 x - 4 & 0 & - 1 \\ d. & f (x) = \frac{2}{3} x & 0 & 4 \\ e. & y = g (x) & 0 & 25 \\ f. & h (x) = 12 {(\frac{1}{2})}^{x} & 0 & 9 \end{array} \end{aligned}$
::函数Vertical ShiftHortical Shifta.f(x)=2x05b.g(x)=5(3)x0-2c.h(x)=5x-40-1-dd.f(x)=23x04e.y=g(x)025f.h(x)=12(12)x09

Activity 1: Sequences
::活动1:顺序

Example 1-1
::例1-1

Steph and James play basketball on two different middle school teams. Steph scores 20 points in his pre-season game - that's game 0. In his first game of the regular season, he scores 24, in the next game 28, and so on. James, however, only scores 3 points in his pre-season game. In the first game of the regular season, he scores 6, in the next game 12, in the next game 24, and so on. Create functions that model the scoring for each player over the year. What do $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ represent? Given that they play 1 pre-season game and 8 regular season games per year, w hat is the domain for each function? What is the $\begin{aligned} y \end{aligned}$ - intercept for each function? Why? State the general form for each type of function and explain its parameters. Is this a completely realistic scenario? Why or why not?
::Steph 和 James 在两个不同的初中队打篮球。 Steph 和 James 在两场不同的初中队打篮球。 Steph 的赛前比赛中得分 20 分, 这就是游戏 0 。在他的第一个赛前比赛中, 他得分 24 分, 在下一个比赛中得分 28 等。然而, James 的赛前比赛中只得分 3 分。在第一个赛前比赛中, 他得分 6 分, 在下一个比赛中得分 12 , 在下一个比赛中得分 24 等。创建函数来模拟每年每个球员的评分。 X 和 Y 代表什么 ? 鉴于他们每年玩一个赛前比赛和八个赛季比赛, 每种功能的域是什么? 每个函数的域是什么? 为什么? 说明每种功能的一般形式并解释其参数。这是完全现实的场景吗为什么 ?

Solution: Here is selected data for each organized in tables:
::解决方案: 这是表格中每个表格的选定数据 :

$\begin{aligned} \begin{array}{cc} x & S (x) \\ 0 & 20 \\ 1 & 24 \\ 2 & 28 \\ 3 & 32 \end{array} \end{aligned}$

::xS(x) 020124228332

$\begin{aligned} \begin{array}{cc} x & J (x) \\ 0 & 3 \\ 1 & 6 \\ 2 & 12 \\ 3 & 24 \end{array} \end{aligned}$

::xJ(xx)0316212324

Here are the graphs. Note that the points are not connected, to emphasize the practical domain of the scenario. There are 9 games and no fractions of games.
::以下是图表。请注意, 点没有连接, 以强调此方案的实际领域。有 9 场游戏, 没有游戏的分数。

The first function is linear with a of 4 and a $\begin{aligned} y \end{aligned}$ -intercept of 20:
::第一个函数是线性函数, 4 和 Y 界面为 20 :

$\begin{aligned} S (x) = 4 x + 20 \end{aligned}$

::S(x)=4x+20

The second function is exponential, with an initial value of 3 and a multiplier of 2:
::第二个函数是指数函数,初始值为3,乘数为2:

$\begin{aligned} J (x) = 3 (2)^{x} \end{aligned}$

::J(x)=3(2)x

For each function, $\begin{aligned} x \end{aligned}$ represents the game number, and $\begin{aligned} y \end{aligned}$ represents the amount scored in each game. The domain for each is $\begin{aligned} {0, 1, 2, 3, 4, 5, 6, 7, 8} . \end{aligned}$
::对于每个函数,x 代表游戏编号,而y 代表每个游戏的分数。每个函数的域为 {0,1、2,3,4,5,6,6,7,8}。

The $\begin{aligned} y \end{aligned}$ -intercept for the linear function is 20, and the $\begin{aligned} y \end{aligned}$ -intercept for the exponential function is 3. Substituting 0 for $\begin{aligned} x \end{aligned}$ in each returns the $\begin{aligned} y \end{aligned}$ -intercept for each.
::线性函数的 Y 界面为 20 , 指数函数的 Y 界面为 3 。替换 0. x 键每返回 y 界面。

The linear function is in slope-intercept form : $\begin{aligned} y = m x + b, \end{aligned}$ where $\begin{aligned} m \end{aligned}$ is the slope and $\begin{aligned} b \end{aligned}$ is the $\begin{aligned} y \end{aligned}$ -intercept.
::线性函数以斜坡界面形式显示: y=mx+b, 其中 m 是斜坡, b 是 y 界面。

The exponential function is in standard form : $\begin{aligned} y = a (b)^{x}, \end{aligned}$ where $\begin{aligned} a \end{aligned}$ is the $\begin{aligned} y \end{aligned}$ -intercept and $\begin{aligned} b \end{aligned}$ is the multiplier.
::指数函数为标准格式:y=a(b)x,其中a为 Y 接口,b为乘数。

Of course, the scenario isn't very realistic, because according to the model, James scores 768 points in the 8th game!
::当然,情况并不很现实, 因为根据模型,詹姆斯在第八场比赛中得768分!

Interactive
::交互式互动

Steph and James both get injured over the summer! They won't be able to start playing until later in the season, but they will score as before . This means that Steph will still score 20 points in his first game and increase by 4 points every game thereafter and James will still score 3 points in his first game and double his points every game thereafter.
::Steph和James在夏天都受伤了! 他们要到赛季晚些时候才能开始比赛, 但是他们会和以前一样得分。这意味着 Steph在第一次比赛中仍得20分, 其后每场比赛增加4分, 而James在第一次比赛中仍得3分, 其后每场比赛都得2倍。

Use the interactive below to explore what Steph and James' scores will be depending on which game they start playing in.
::使用下面的互动式来探索斯蒂芬和詹姆斯的分数取决于他们开始玩什么游戏。

Example 1- 2
::例1-2

The next year , Steph and James both miss the pre-season game, but start playing in game 1. They score as before. Create functions for the scoring of each by modifying the original function as in the previous interactive. What is the domain for each?
::下一年, Steph 和 James 都错过了季节前的游戏, 但开始玩游戏 1 。它们和以前一样得分。通过修改前一次互动中的原有函数, 为每个函数的评分创建函数。每种函数的域是什么 ?

Solution: The original functions must be shifted 1 unit to the right. The domain for each is $\begin{aligned} {1, 2, 3, 4, 5, 6, 7, 8} . \end{aligned}$
::解决方案: 原始函数必须移到右边 1 个单位。每个单位的域名为 {1, 2, 3, 4, 5, 6, 7, 8} 。

$\begin{aligned} \begin{array}{ll} Original Function & New Function \\ S (x) = 4 x + 20 & S (x) = 4 (x - 1) + 20 \\ J (x) = 3 (2)^{x} & J (x) = 3 (2)^{x - 1} \end{array} \end{aligned}$

::原始函数新函数S( x)=4x+20S( x)=4( x-1)+20J( x)=3(2)xJ( x)=3(2)x- 1

Sequences
::序列

Sequences have the set of natural numbers, or a subset of the natural numbers, as their domain.
::序列将自然数字的一组或自然数字的一个子集作为其域。

They can be written as an ordered list of numbers.
::它们可以写成一份有命令的号码清单。

Each number in the list is called a term .
::名单上的每个号码都称为术语。

PLIX Interactive
::PLIX 交互式互动

Activity 2: Defining Sequences and Using Sequence Notation
::活动2:界定序列和使用序列编号

A sequence is a function like those in the very last example. Its domain is either the infinite set of natural numbers: $\begin{aligned} {1, 2, 3, . . ., \infty}, \end{aligned}$ or a finite consecutive subset of the natural numbers starting with 1. In the last example, the domain was $\begin{aligned} {1, 2, 3, 4, 5, 6, 7, 8} . \end{aligned}$ The sequence itself is the ordered list of $\begin{aligned} y \end{aligned}$ -values for each function. There were two sequences in the last example:
::序列是一个函数, 类似上次示例中的函数。它的域是无限的自然数字集 : {1, 2, 3,...,...,\\\\\\\\\\\\ n或自然数字中从 1 开始的有限连续子集。在最后一个示例中, 域是 {1, 2, 3, 4, 5, 6, 7, 8}。序列本身是每个函数的 y 值的顺序列表。在最后一个示例中有两个序列 :

$\begin{aligned} \begin{array}{ll} Steph: & {20, 24, 28, 32, 36, 40, 44, 48} \\ James: & {3, 6, 12, 24, 48, 96, 192, 384} \end{array} \end{aligned}$

::斯蒂芬: {20,24,28,32,36,40,44,48}詹姆斯: {3,6,12,24,48,96,192,384}

The $\begin{aligned} y \end{aligned}$ -values for a sequence are called its terms . The term number is usually denoted by the variable $\begin{aligned} n . \end{aligned}$ Instead of using for sequences, you will use subscripts to write the equation for a sequence:
::序列的 y 值称为其术语。术语数通常由变量 n 表示。您将使用下标来为序列写方程 :

$\begin{aligned} \begin{array}{l} s_{n} = 20 + 4 (n - 1) \\ j_{n} = 3 (2)^{n - 1} \end{array} \end{aligned}$

::sn=20+4(n-1)jn=3(2)n-1

Notice that the variables $\begin{aligned} s \end{aligned}$ and $\begin{aligned} j \end{aligned}$ are usually lower-case when writing sequences in sequence notation.
::注意变量 s 和 j 通常在以序列符号写入序列序列时为小写字母。

Also notice that while these expressions could easily be simplified, the convention is to write them as linear or shifted 1 to the right.
::还注意,虽然这些表达方式可以很容易地简化,但公约是将它们写成线性文字或移到右边。

Finally, notice that the sequence drawn from a linear function is written such that the initial term is first.
::最后,请注意,从线性函数中得出的顺序是写成的,因此最初的术语是第一个词。

Example 2-1
::例2-1

The following sequence is based on a linear function. Write the equation for the sequence using sequence notation.
::以下序列以线性函数为基础。使用序列符号写出序列的方程。

$\begin{aligned} {11, 16, 21, 26, 31, . . .} \end{aligned}$

Solution: The initial value for the sequence, that is, $\begin{aligned} a_{1}, \end{aligned}$ is 11.
::解答:序列的初始值,即 a1,为 11。

The values for the sequence increase by $\begin{aligned} 5 \end{aligned}$ , which means the slope is 5.
::顺序增加 5 的值,这意味着斜坡是 5 。

Using the structure developed above:
::使用上述结构:

$\begin{aligned} a_{n} = 11 + 5 (n - 1) \end{aligned}$

::an=11+5(n- 1)

Example 2-2
::例2-2

The following sequence is based on an exponential function. Write the equation for the sequence using sequence notation.
::以下序列以指数函数为基础。使用序列符号写出序列的方程式。

$\begin{aligned} {5, 15, 45, 135, 405, . . .} \end{aligned}$

Solution: The initial value for the sequence, that is, $\begin{aligned} a_{1}, \end{aligned}$ is 5.
::解答: 序列的初始值, 即 a1, 是 5 。

Each value in the sequence is 3 times the previous value, so the multiplier is 3.
::序列中的每个值是先前值的3倍,因此乘数是3。

Using the structure developed above:
::使用上述结构:

$\begin{aligned} a_{n} = 5 (3)^{n - 1} \end{aligned}$

::an=5(3)n-1

Arithmetic and Geometric Sequences
::亚学和几何序列

The equations for the sequences explored so far show that they are analogous to linear and exponential functions, shifted 1 unit to the right.
::迄今为止所探讨的序列方程式表明,它们与线性函数和指数函数相似,将1个单位移到右边。

The formula for an arithmetic sequence is $\begin{aligned} a_{n} = a_{1} + d (n - 1), \end{aligned}$ where $\begin{aligned} a_{1} \end{aligned}$ is the initial value, and $\begin{aligned} d \end{aligned}$ is the common difference , that is, the amount added to advance from one term to the next.
::算术序列的公式是 an=a1+d(n-1),其中 a1 是初始值, d 是共同差, 即为从一个任期向下一个任期前进而增加的数额。

The formula for a geometric sequence is $\begin{aligned} a_{n} = a_{1} (r)^{n - 1}, \end{aligned}$ where $\begin{aligned} a_{1} \end{aligned}$ is the initial value, and $\begin{aligned} r \end{aligned}$ is the common ratio , that is, the amount by which each term is multiplied to advance to the next term.
::几何序列的公式是 an=a1(r)n-1, 其中 a1 是初始值, r 是共同比率, 即每个术语的乘数乘以到下一任期。

PLIX Interactive
::PLIX 交互式互动

Activity 3 : Finding the Equations for Sequences
::活动3:寻找序列的等号

Example 3-1
::例3-1

A sequence can be described with an explicit formula like those you created above. An explicit formula returns the term for a given term number. A sequence can also be described with a recursive formula that gives the initial term and the relationship between consecutive terms. Write the explicit formula for the sequence described by the following recursive formula:
::序列可以用一个清晰的公式来描述一个序列, 和您在上面创建的公式一样。明确的公式返回给定的术语编号的术语。也可以用一个循环的公式来描述一个序列, 给出初始术语和连续术语之间的关系。写下以下递归公式描述的序列的清晰公式 :

$\begin{aligned} a_{1} = 7; a_{n} = a_{n - 1} + 3 \end{aligned}$

::a1=7;an=an - 1+3

Solution: The initial term is 7. From there, $\begin{aligned} a_{n} = a_{n - 1} + 3 \end{aligned}$ means:
::解决办法:最初的学期是7年,从此,a=an-1+3系指:

$\begin{aligned} \begin{array}{l} a_{2} = a_{1} + 3 \\ a_{3} = a_{2} + 3 \\ a_{4} = a_{3} + 3 \\ etc. \end{array} \end{aligned}$

::a2=a1+3a3=a2+3a4=a3+3etc。

This shows that each term is 3 more than the previous term, that is, the common difference, $\begin{aligned} d, \end{aligned}$ is 3.
::这表明,每个任期比前一个任期多3个,即共同差数(d)为3个。

The explicit formula is:
$\begin{aligned} a_{n} = 7 + 3 (n - 1) \end{aligned}$

::明确的公式是:an=7+3(n-1)

Interactive
::交互式互动

Use the interactive to explore explicit and recursive formulas for arithmetic and geometric sequences.
::使用互动来探索计算和几何序列的清晰和循环公式。

Work it Out
::工作出来

For each scenario, write the first few terms of the sequence, and then find the equation for the sequence.

Kevin scores 27 points in his first game, and he scores 2 more points per game every game thereafter.
::Kevin在第一次比赛中得27分每场比赛中得2分之后每场比赛都得2分

A vending machine sells 4 candy bars the first day it is installed and sells 5 more candy bars per day every day thereafter.
::一个自动售货机在安装第一天就卖掉4个糖果棒,此后每天又卖掉5个糖果棒。

The population of walruses on an island is 150 the first year they are counted. Their numbers double every year thereafter.
::岛屿海象人口在计算第一年为150人,其后每年翻一番。

A mountain of dirt exposed by fire is now eroding in the rain. In the first year, its height is 320 meters, but that is halved every year .
::被大火照射的一山土正在雨水中腐蚀。第一年,其高度是320米,但每年减少一半。

::对于每种情况,请写下顺序的最初几个条件, 然后找到序列的方程。 Kevin在第一次比赛中得27分, 然后每场比赛得2分。一个自动售货机在安装第一天卖4个糖果棒, 之后每天卖5个糖果棒。岛上的海象人口在计算的第一个年数是150个, 之后每年翻一番。火灾暴露的泥土现在在雨水中侵蚀。在第一年,它的高度是320米, 但每年减少一半。

The following sequences are either arithmetic or geometric. Use one or both of the following tactics to determine which type of sequence you have.
::以下的顺序是算术或几何。使用以下一种或两种策略来确定您拥有哪种序列。

Subtract each term from the previous term (to find a common difference).
::从上一个任期中减去每个任期(以找出共同的差别)。

OR
::或

Divide each term by the previous term (to find a common ratio).
::每一任期除以前一任期(以找到共同比率)。

Be sure and use the tactics with multiple pairs of consecutive terms. Explain which method you used and why it enabled you to determine which kind of sequence you were dealing with. Then write the equation for the sequence using sequence notation.
::请确定并使用策略, 使用多个连续的词组。解释您使用哪种方法, 以及为什么它使您能够确定您正在处理的序列。然后用序列符号写入序列的方程。

$\begin{aligned} \begin{array}{ll} a. & {2, 5, 8, 11, . . .} \\ b. & {2, 6, 18, 54, 162, . . .} \\ c. & {100, 90, 80, 70, . . .} \\ d. & {21, 22, 5, 24, 25, 5, . . .} \\ e. & {10, 15, 22.5, 33.75, . . .} \end{array} \end{aligned}$
::================================================ ==========================================================================================================================================================================================================================================================================================================================================================================================================================================

Common Difference and Common Ratio
::共同差异和共同比率

The difference between consecutive values in an arithmetic sequence is called the common difference . It is represented by $\begin{aligned} d \end{aligned}$ in the formula for the sequence.
::算术序列中连续值之间的差称为共同差。在序列公式中以 d 表示。

The ratio of consecutive values in a geometric sequence is called the common ratio . It's represented by $\begin{aligned} r \end{aligned}$ in the formula for the sequence.
::几何序列中连续值的比率被称为共同比率。在序列的公式中以 r 表示。

Identify each formula as recursive or explicit, and as arithmetic or geometric. If a formula is recursive, find the explicit formula, and if it is explicit, find the recursive formula.
::指定每个公式为递归式或直线式,以及算术或几何公式。如果一个公式为递归式,请找到明确的公式,如果是明确的,则找到递归式。

$\begin{aligned} \begin{array}{ll} a. & a_{1} = 12; a_{n} = a_{n - 1} + 9 \\ b. & a_{n} = 15 + 5 (n - 1) \\ c. & a_{1} = 3; a_{n} = 8 a_{n - 1} \\ d. & a_{n} = 14 (5)^{n - 1} \\ e. & a_{n} = - 11 - 3 (n - 1) \\ f. & a_{1} = 1; a_{n} = a_{n - 1} + 1 \\ g. & a_{1} = 1; a_{n} = \frac{2}{3} a_{n - 1} \end{array} \end{aligned}$
::a.a1=12;an=an-1+9b.an=an-1+1+9b.an=an-1=15+5(n-11)c.a1=3;an=8an-1d.an=14(5)n-1e.an_11-1(n-1)f.a1=1;an=an-1+1g.a1=1;an=23an-1

Activity 4 : Finding Formulas Given Two Consecutive Terms
::活动4:基于两个连续术语的寻找公式

Example 4-1
::例4-1

Given an arithmetic sequence such that $\begin{aligned} a_{6} = 39 \end{aligned}$ and $\begin{aligned} a_{7} = 41, \end{aligned}$ find the common difference, $\begin{aligned} d . \end{aligned}$ Find the explicit formula for the sequence.
::a6=39和a7=41的算术序列发现共同差数,d. 找出序列的明确公式。

Solution: The common difference is 2.
::解决办法:共同的差别是2。

The formula so far is:
::迄今为止的公式是:

$\begin{aligned} a_{n} = a_{1} + 2 (n - 1) \end{aligned}$

::a=a1+2(n-1)

Since $\begin{aligned} a_{6} = 39 : \end{aligned}$
::从A6=39起:

$\begin{aligned} \begin{array}{l} 39 = a_{1} + 2 (6 - 1) \\ a_{1} = 29 \end{array} \end{aligned}$

::39=a1+2(6-1)a1=29

Therefore , the formula for the sequence is:
::因此,序列的公式是:

$\begin{aligned} a_{n} = 29 + 2 (n - 1) \end{aligned}$

::a=29+2(n-1)

Example 4-2
::例4-2

Given a geometric sequence such that $\begin{aligned} a_{4} = 324 \end{aligned}$ and $\begin{aligned} a_{5} = 486, \end{aligned}$ find the common ratio, $\begin{aligned} r . \end{aligned}$ Find the explicit formula for the sequence.
::以a4=324和a5=486等几何序列找到共同比率,r. 找到该序列的明确公式。

Solution: The common ratio is $\begin{aligned} \frac{486}{324} = 1.5. \end{aligned}$
::解决方案:共同比率为486324=1.5。

The formula so far is:
::迄今为止的公式是:

$\begin{aligned} a_{n} = a_{1} (1.5)^{n - 1} \end{aligned}$

::a=a1(1.5)n-1

Since $\begin{aligned} a_{4} = 324 : \end{aligned}$
::从A4=324起:

$\begin{aligned} \begin{array}{ll} 324 = a_{1} (1.5)^{4 - 1} \\ 324 = a_{1} {(\frac{3}{2})}^{3} \\ 324 = a_{1} \frac{3^{3}}{2^{3}} \\ 324 = a_{1} \frac{27}{8} \\ a_{1} = \frac{8}{27} \cdot \frac{324}{1} \\ a_{1} = 96 \end{array} \end{aligned}$

::324=a1(1.5)4-1324=a1(32)3324=a13323324=a1278a1=827□3241a1=96

Therefore, the formula for the sequence is:
::因此,序列的公式是:

$\begin{aligned} a_{n} = 96 (1.5)^{n - 1} \end{aligned}$

::an=96(1.5)n-1

Work it Out
::工作出来

Given an arithmetic sequence such that $\begin{aligned} a_{9} = 100 \end{aligned}$ and $\begin{aligned} a_{10} = 114, \end{aligned}$ find the explicit formula for the sequence.
::考虑到一个算术序列,即a9=100和a10=114,找到该序列的明确公式。

Given a geometric sequence such that $\begin{aligned} a_{5} = 18 \end{aligned}$ and $\begin{aligned} a_{6} = 6, \end{aligned}$ find the explicit formula for the sequence.
::根据几何序列,a5=18和a6=6可以找到该序列的明确公式。

Activity 5 : Finding Formulas Given Any Two Terms
::活动5:基于任何两个条件的 " 寻找公式 "

Example 5-1
::例5-1

Given the arithmetic sequence with the following terms, find the explicit formula and the recursive formula for the sequence: $\begin{aligned} a_{4} = 14; a_{7} = 9. \end{aligned}$
::考虑到计算顺序,用以下术语计算,找到明确的公式和序列的递归公式:a4=14;a7=9。

Solution:
::解决方案 :

$\begin{aligned} \begin{array}{ll} Step & Explanation \\ d = \frac{9 - 14}{7 - 4} & Finding the common difference, i.e., the slope. \\ d = - \frac{5}{3} & Simplifying. \\ a_{n} = a_{1} + d (n - 1) & General formula for arithmetic sequence. \\ 14 = a_{1} - \frac{5}{3} (4 - 1) & Substituting a term number and term to find a_{1} . \\ 14 = a_{1} - 5 & Simplifying. \\ a_{1} = 19 & Solving for a_{1} . \\ a_{n} = 19 - \frac{5}{3} (n - 1) & Writing the explicit formula. \end{array} \end{aligned}$

::Steplanationd=9-147-4 确定共同差异,即斜坡.d53Simplication.an=a1+d(n-1) 算术序列的通用公式。 14=a1-53(4-1) 设置一个术语编号和术语以查找 a1.14=a1-5Simplication.a1=19Solving for a1.an=19=53(n-1) 明确公式。

The recursive formula is: $\begin{aligned} a_{n} = a_{n - 1} - \frac{5}{3} . \end{aligned}$
::循环公式是: a=an-1-53。

Example 5-2
::例5-2

Given the positive geometric sequence with the following terms, find the explicit formula and the recursive formula for the sequence: $\begin{aligned} a_{5} = 607.5; a_{7} = 1366.875. \end{aligned}$
::鉴于正几何序列,并有以下术语,请为该序列找到明确的公式和递合公式:a5=607.5;a7=1366.875。

Solution: In a geometric sequence with common ratio $\begin{aligned} r, \end{aligned}$ the current term in the sequence is $\begin{aligned} r \end{aligned}$ times the previous term. That means that $\begin{aligned} a_{7} \end{aligned}$ is $\begin{aligned} r \end{aligned}$ times $\begin{aligned} a_{6}, \end{aligned}$ $\begin{aligned} a_{6} \end{aligned}$ is $\begin{aligned} r \end{aligned}$ times $\begin{aligned} a_{5}, \end{aligned}$ and so on. That means you can find the relationship between $\begin{aligned} a_{7} \end{aligned}$ and $\begin{aligned} a_{5} \end{aligned}$ as shown below, then use that relationship to find $\begin{aligned} r : \end{aligned}$
::解决方案 : 在具有 r 共比的几何序列中, 序列中的当前术语是乘以上一个术语的 r。这意味着 r xy a6, a6 r xy a5 等。这意味着您可以找到以下显示的 7 和 5 之间的关系, 然后使用此关系查找 r :

$\begin{aligned} \begin{array}{ll} Step & Explanation \\ a_{n} = a_{n - 1} (r) & Recursive formula for geometric sequence. \\ a_{6} = a_{5} (r) & Using the recursive formula. \\ a_{7} = a_{6} (r) & Using the recursive formula. \\ a_{7} = a_{5} (r) (r) & Substitution. \\ a_{7} = a_{5} (r)^{2} & Simplification. \\ 1366.875 = 607.5 r^{2} & Substitution. \\ r = 1.5 & Solving, only positive values for multiplier in positive geometric sequence. \end{array} \end{aligned}$

::Steplanationan=an-1(r) 几何序列的递解公式。 a6=a5(r) 表示递解公式。 a7=a6(r) 表示递解公式。 a7=a5(r)(r) Substitution.a7=a5(r) =a5(r)2 简化13666.875=607.5r2Substitution.r=1.5Solving, 仅表示正几何序列乘数的正正值。

Now that there is a common ratio, you can substitute it and $\begin{aligned} a_{5} = 607.5 \end{aligned}$ to solve for $\begin{aligned} a_{1} \end{aligned}$ and write the explicit formula for the sequence:
::现在有一个共同比率, 您可以替换它和 A5=607. 5 来解析 a1, 并写出序列的清晰公式 :

$\begin{aligned} \begin{array}{ll} Step & Explanation \\ a_{n} = a_{1} (r)^{n - 1} & Explicit formula for geometric sequence. \\ 607.5 = a_{1} (1.5)^{5 - 1} & Substituting. \\ a_{1} = 120 & Solving. \\ a_{n} = 120 (1.5)^{n - 1} & The explicit formula for this sequence. \end{array} \end{aligned}$

::Steplanationan=a1(r)n-1 几何序列的明显公式。 607.5=a1(1.5)5-1 替代公式。a1=120Solving.an=120(1.5)n-1 此序列的明显公式。

Work it Out
::工作出来

Given an arithmetic sequence such that $\begin{aligned} a_{4} = 12 \end{aligned}$ and $\begin{aligned} a_{6} = 17, \end{aligned}$ find the explicit formula for the sequence.
::给定一个算术序列,使 a4=12 和 a6=17 找到该序列的明确公式。

Given a geometric sequence such that $\begin{aligned} a_{5} = 405 \end{aligned}$ and $\begin{aligned} a_{7} = 567, \end{aligned}$ find the explicit formula for the sequence.
::根据几何序列,A5=405和A7=567为该序列找到明确的公式。

Summary
::摘要

The explicit formula for an arithmetic sequence is $\begin{aligned} a_{n} = a_{1} + d (n - 1), \end{aligned}$ where $\begin{aligned} a_{1} \end{aligned}$ is the initial value and $\begin{aligned} d \end{aligned}$ is the common difference.
::算术序列的明确公式是 an=a1+d(n-1),其中 a1 是初始值, d 是共同差数。

The explicit formula for a geometric sequence is $\begin{aligned} a_{n} = a_{1} (r)^{n - 1}, \end{aligned}$ where $\begin{aligned} a_{1} \end{aligned}$ is the initial value and $\begin{aligned} r \end{aligned}$ is the common ratio.
::几何序列的明确公式是 an=a1(r)n-1, 其中 a1 是初始值, r 是共同比率。

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