Course: 11.8 根据共同比率和第一任期,几何序列和查找第10学期

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参照共同比率和第一任期的几何序列和查找第n个学期
The following shows the distance (in centimeters) a pendulum travels with each successive swing. Write a general rule for the geometric sequence.
::以下显示时钟的间距(以厘米计) , 每次连续摆动时, 时钟的间距。写几何序列的一般规则。

80, 72, 64.8, 58.32, ...

Geometric Sequence
::几何序列

A geometric sequence is a sequence in which the ratio between any two consecutive terms, $\begin{aligned} \frac{a_{n}}{a_{n - 1}} \end{aligned}$ , is constant. This constant value is called the common ratio . Another way to think of this is that each term is multiplied by the same value, the common ratio, to get the next term.
::几何序列是一个序列, 任何两个连续任期( anan-1) 之间的比率都是恒定的。这个不变值被称为共同比率。另一种方式是, 每一术语乘以相同值, 即共同比率, 以获得下一个任期。

Let's consider the sequence $\begin{aligned} 2, 6, 18, 54, \dots \end{aligned}$
::让我们考虑一下顺序 2,6,18,54,...

Is this sequence geometric? If so, what is the common difference?
::这是几何序列吗?如果是,有什么共同的区别?

If we look at each pair of successive terms and evaluate the ratios, we get $\begin{aligned} \frac{6}{2} = \frac{18}{6} = \frac{54}{18} = 3 \end{aligned}$ which indicates that the sequence is geometric and that the common ratio is 3.
::如果我们每看一对相继条件并评估比率,我们就会得到62=186=5418=3,这表明顺序是几何,共同比率是3。

Now let’s see if we can develop a general rule ( $\begin{aligned} n^{t h} \end{aligned}$ term) for this sequence. Since we know that each term is multiplied by 3 to get the next term, let’s rewrite each term as a product and see if there is a pattern.
::现在让我们看看我们是否能够为这个序列制定一个通用规则( nth term) 。由于我们知道每个术语乘以3才能获得下一个术语,让我们将每个术语改写为产品,看看是否有模式。

$\begin{aligned} a_{1} & = 2 \\ a_{2} & = a_{1} (3) = 2 (3) = 2 (3)^{1} \\ a_{3} & = a_{2} (3) = 2 (3) (3) = 2 (3)^{2} \\ a_{4} & = a_{3} (3) = 2 (3) (3) (3) = 2 (3)^{3} \end{aligned}$

::a1=2a2=a1(3)=2(3)=2(3)(3)=2(3)1a3=a2(3)=2(3)=2(3)=2(3)(3)=2(3)(3)=2(3)=3(3)=3(3)=2(3)(3)=3(3)=3(3)=2(3)=3

This illustrates that the general rule is $\begin{aligned} a_{n} = a_{1} (r)^{n - 1} \end{aligned}$ , where $\begin{aligned} r \end{aligned}$ is the common ratio. This even works for the first term since $\begin{aligned} a_{1} = 2 (3)^{0} = 2 (1) = 2 \end{aligned}$ .
::这说明一般规则是 an=a1(r)n-1, 其中r为共同比率。这甚至适用于自 a1=2(3)0=2(1)=2以来的第一个学期。

Now, let's write a general rule for the geometric sequence $\begin{aligned} 64, 32, 16, 8, \dots \end{aligned}$
::现在,让我们为64,32,16,8的几何序列写出一条通则...

From the general rule above we can see that we need to know two things: the first term and the common ratio to write the general rule. The first term is 64 and we can find the common ratio by dividing a pair of successive terms, $\begin{aligned} \frac{32}{64} = \frac{1}{2} \end{aligned}$ . The $\begin{aligned} n^{t h} \end{aligned}$ term rule is thus $\begin{aligned} a_{n} = 64 {(\frac{1}{2})}^{n - 1} \end{aligned}$ .
::从上面的一般规则可以看出,我们需要知道两件事:第一个学期和编写一般规则的共同比率。第一个学期为64,我们可以通过分两个连续学期(3264=12)来找到共同比率。因此,n学期的规则是=64(12)n-1。

Finally, let's find the $\begin{aligned} n^{t h} \end{aligned}$ term rule for the sequence $\begin{aligned} 81, 54, 36, 24, \dots \end{aligned}$ and hence find the $\begin{aligned} 12^{t h} \end{aligned}$ term.
::最后,让我们找出第n个术语规则的顺序81,54,36,24... 并因此找到第12个术语。

The first term here is 81 and the common ratio, $\begin{aligned} r \end{aligned}$ , is $\begin{aligned} \frac{54}{81} = \frac{2}{3} \end{aligned}$ . The $\begin{aligned} n^{t h} \end{aligned}$ term rule is $\begin{aligned} a_{n} = 81 {(\frac{2}{3})}^{n - 1} \end{aligned}$ . Now we can find the $\begin{aligned} 12^{t h} \end{aligned}$ term $\begin{aligned} a_{12} = 81 {(\frac{2}{3})}^{12 - 1} = 81 {(\frac{2}{3})}^{11} = \frac{2048}{2187} \end{aligned}$ . Use the graphing calculator for the last step and MATH > Frac your answer to get the fraction. We could also use the calculator and the general rule to generate terms $\begin{aligned} s e q (81 (2 / 3)^{\land} (x - 1), x, 12, 12) \end{aligned}$ . Reminder: the $\begin{aligned} s e q () \end{aligned}$ function can be found in the LIST ( $\begin{aligned} 2^{n d} \end{aligned}$ STAT ) Menu under OPS . Be careful to make sure that the entire exponent is enclosed in parenthesis.
::第一个术语是81, 共同比率为5481=23。 nth 术语规则是 an=81(23)n-1。现在我们可以找到第12个术语 a12=81(23)12-1=81(23)11=81(23)11=20482187。使用图形计算计算器来最后一步, 并使用 MATH > Frac 您的答案来获取分数。我们还可以使用计算器和一般规则来生成后续术语( 81( 2/3)( x- 1)x, 12, 12) 。提醒器: 后函数可以在OPS 下的菜单列表中找到。要小心确保整个列表都包含在括号中。

Examples
::实例

Example 1
::例1

Earlier, you were asked to write a general rule for the sequence 80, 72, 64.8, 58.32, ...
::早些时候,你被要求写出一条 80,72,648,5832,... 80,72,648,5832的一般规则

We need to know two things, the first term and the common ratio, to write the general rule. The first term is 80 and we can find the common ratio by dividing a pair of successive terms, $\begin{aligned} \frac{72}{80} = \frac{9}{10} \end{aligned}$ . The $\begin{aligned} n^{t h} \end{aligned}$ term rule is thus $\begin{aligned} a_{n} = 80 {(\frac{9}{10})}^{n - 1} \end{aligned}$ .
::我们需要知道两件事,第一个学期和共同比率,以写出一般规则。第一个学期为80,我们可以通过对两个学期的相继学期(7280=910)来找到共同比率。因此,第n学期规则是 an=80(910n)-1。

For Examples 2-4, identify which of the sequences are . If the sequence is geometric, find the common ratio.
::对于例2-4,请指明序列是哪个序列。如果序列是几何,请找到共同比率。

Example 2
::例2

$\begin{aligned} 5, 10, 15, 20, \dots \end{aligned}$

arithmetic
::算法

Example 3
::例3

$\begin{aligned} 1, 2, 4, 8, \dots \end{aligned}$

geometric, $\begin{aligned} r = 2 \end{aligned}$
::r=2, r=2 几何

Example 4
::例4

$\begin{aligned} 243, 49, 7, 1, \dots \end{aligned}$

geometric, $\begin{aligned} r = \frac{1}{7} \end{aligned}$
::r=17 几何,r=17

Example 5
::例5

Find the general rule and the $\begin{aligned} 20^{t h} \end{aligned}$ term for the sequence $\begin{aligned} 3, 6, 12, 24, \dots \end{aligned}$
::找到3,6,12,24的顺序的一般规则和第20个学期...

The first term is 3 and the common ratio is $\begin{aligned} r = \frac{6}{3} = 2 \end{aligned}$ so $\begin{aligned} a_{n} = 3 (2)^{n - 1} \end{aligned}$ .
::第一个学期是3学期,共同比率为r=63=2,a=3(2)n-1。

The $\begin{aligned} 20^{t h} \end{aligned}$ term is $\begin{aligned} a_{20} = 3 (2)^{19} = 1, 572, 864 \end{aligned}$ .
::第20个学期是a20=3(2)19=1 572 864。

Example 6
::例6

Find the $\begin{aligned} n^{t h} \end{aligned}$ term rule and list terms 5 thru 11 using your calculator for the sequence $\begin{aligned} - 1024, 768, - 432, - 324, \dots \end{aligned}$
::使用您的计算器查找 nth 术语规则并列出 5 xu 11 术语, 序列 : - 1024, 768, - 432, - 324 。...

The first term is -1024 and the common ratio is $\begin{aligned} r = \frac{768}{- 1024} = - \frac{3}{4} \end{aligned}$ so $\begin{aligned} a_{n} = - 1024 {(- \frac{3}{4})}^{n - 1} \end{aligned}$ .
::第一个学期为-1024,共同比率为 r=768-102434, 即 an1024(- 34n-1)。

Using the calculator sequence function to find the terms and MATH > Frac ,
::使用计算器序列函数查找术语和 MATH > Frac,

$\begin{aligned} s e q (- 1024 (- 3 / 4)^{\land} (x - 1), x, 5, 11) = {- 324 243 - \frac{729}{4} \frac{2187}{16} - \frac{6561}{256} \frac{19683}{256} - \frac{59049}{1024}} \end{aligned}$
:-1024(--3/4) (x-1),x,5,11) 324243-7294218716-656125619683256-590491024)]

Example 7
::例7

Find the value of a 10 year old car if the purchase price was $22,000 and it depreciates at a rate of 9% per year.
::如果购买价为22,000美元,并且每年以9%的折旧率贬值,则可以找到一辆10年之久的汽车的价值。

The first term (value of the car after 0 years) is $22,000. The common ratio is $\begin{aligned} 1 - .09 \end{aligned}$ or $\begin{aligned} 0.91 \end{aligned}$ . The value of the car after $\begin{aligned} n \end{aligned}$ years can be determined by $\begin{aligned} a_{n} = 22, 000 (0.91)^{n} \end{aligned}$ . For 10 years we get $\begin{aligned} a_{10} = 22, 000 (0.91)^{10} = 8567.154599 \approx $ 8567 \end{aligned}$ .
::第一个术语(汽车在0年后的价值)为22,000美元,共同比率为1-.09或0.91,零年之后的汽车价值可以由 =22,000(0.91)n确定,10 =22,000(0.91)10 =8567.154599=8567。

Review
::回顾

Identify which of the following sequences are arithmetic, geometric or neither.
::标明下列哪些序列是算术、几何或两者兼有。

$\begin{aligned} 2, 4, 6, 8, \dots \end{aligned}$

$\begin{aligned} \frac{1}{2}, \frac{3}{2}, \frac{9}{2}, \frac{27}{2}, \dots \end{aligned}$

$\begin{aligned} 1, 2, 4, 7, \dots \end{aligned}$

$\begin{aligned} 24, - 16, \frac{32}{3}, - \frac{64}{9}, \dots \end{aligned}$

$\begin{aligned} 10, 5, 0, - 5, \dots \end{aligned}$

$\begin{aligned} 3, 4, 7, 11, \dots \end{aligned}$

Given the first term and common ratio, write the $\begin{aligned} n^{t h} \end{aligned}$ term rule and use the calculator to generate the first five terms in each sequence.
::鉴于第一个任期和共同比率,请写下 nth 术语规则,并使用计算器在每个序列中生成前五个术语。

$\begin{aligned} a_{1} = 32 \end{aligned}$ and $\begin{aligned} r = \frac{3}{2} \end{aligned}$
::a1=32和r=32

$\begin{aligned} a_{1} = - 81 \end{aligned}$ and $\begin{aligned} r = - \frac{1}{3} \end{aligned}$
::a181和r13。

$\begin{aligned} a_{1} = 7 \end{aligned}$ and $\begin{aligned} r = 2 \end{aligned}$
::a1=7和r=2

$\begin{aligned} a_{1} = \frac{8}{125} \end{aligned}$ and $\begin{aligned} r = - \frac{5}{2} \end{aligned}$
::a1=8125和r52

Find the $\begin{aligned} n^{t h} \end{aligned}$ term rule for each of the following geometric sequences.
::为以下各几何序列中的每一序列查找 nth 术语规则。

$\begin{aligned} 162, 108, 72, \dots \end{aligned}$

$\begin{aligned} - 625, - 375, - 225, \dots \end{aligned}$

$\begin{aligned} \frac{9}{4}, - \frac{3}{2}, 1, \dots \end{aligned}$

$\begin{aligned} 3, 15, 75, \dots \end{aligned}$

$\begin{aligned} 5, 10, 20, \dots \end{aligned}$

$\begin{aligned} \frac{1}{2}, - 2, 8, \dots \end{aligned}$

Use a geometric sequence to solve the following word problems.
::使用几何序列解决以下字词问题。

Rebecca inherited some land worth $50,000 that has increased in value by an average of 5% per year for the last 5 years. If this rate of appreciation continues, about how much will the land be worth in another 10 years?
::Rebecca继承了一些价值50 000美元的土地,在过去5年中,其价值平均每年增加5%。如果这一升值率继续下去,那么这块土地在未来10年中的价值是多少?

A farmer buys a new tractor for $75,000. If the tractor depreciates in value by about 6% per year, how much will it be worth after 15 years?
::如果拖拉机每年贬值约6%,那么15年后价值是多少?

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

参照共同比率和第一任期的几何序列和查找第n个学期

Geometric Sequence ::几何序列

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Example 7 ::例7

Review ::回顾

Review (Answers) ::回顾(答复)

Geometric Sequence
::几何序列

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Example 6
::例6

Example 7
::例7

Review
::回顾

Review (Answers)
::回顾(答复)