Course: 11.9 根据共同比率和任何任期或两个任期来寻找第九个任期

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根据共同比率和任何任期或两个任期
A bacteria sample doubles every hour. After four hours there are 64 bacteria in the sample. What is the $\begin{aligned} n^{t h} \end{aligned}$ term rule for the geometric represented by this situation?
::细菌样本每小时翻一番,经过四个小时后,样本中有64种细菌。

Finding the nth Term
::查找 nth 期数

We will be using the general rule for the $\begin{aligned} n^{t h} \end{aligned}$ term in a geometric sequence and the given term(s) to determine the first term and write a general rule to find any other term.
::我们将在几何序列和特定术语中使用第n个术语的一般规则来确定第一个术语,并写出一条一般规则,以找到任何其他术语。

Let's consider the geometric sequence in which the common ratio is $\begin{aligned} - \frac{4}{5} \end{aligned}$ and $\begin{aligned} a_{5} = 1280 \end{aligned}$ . We'll find the first term in the sequence and write the general rule for the sequence.
::让我们来考虑一下共同比率的几何序列- 45 和 a5= 1280。我们会在序列中找到第一个术语, 并写出序列的一般规则。

We will start by using the term we know, the common ratio and the general rule, $\begin{aligned} a_{n} = a_{1} r^{n - 1} \end{aligned}$ . By plugging in the values we know, we can then solve for the first term, $\begin{aligned} a_{1} \end{aligned}$ .
::首先,我们将使用我们所知道的术语,共同比率和一般规则, an=a1rn-1。通过插入我们所知道的值, 我们可以解决第一个术语,a1。

$\begin{aligned} a_{5} & = a_{1} {(- \frac{4}{5})}^{4} \\ 1280 & = a_{1} {(- \frac{4}{5})}^{4} \\ \frac{1280}{{(- \frac{4}{5})}^{4}} & = a_{1} \\ 3125 & = a_{1} \end{aligned}$

::a5=a1(- 45)-41280=a1(- 45)-41280(- 45)-41280(- 45)4=a13125=a1

Now, the $\begin{aligned} n^{t h} \end{aligned}$ term rule is $\begin{aligned} a_{n} = 3125 {(- \frac{4}{5})}^{n - 1} \end{aligned}$ .
::现在, nth 术语规则是 an=3125(- 45n- 1) 。

Now, let's find the $\begin{aligned} n^{t h} \end{aligned}$ term rule for a sequence in which $\begin{aligned} a_{1} = 16 \end{aligned}$ and $\begin{aligned} a_{7} = \frac{1}{4} \end{aligned}$ .
::现在,让我们找到 Nth 术语规则用于一个序列, 其中 a1=16 和 a7=14 。

Since $\begin{aligned} a_{7} = \frac{1}{4} \end{aligned}$ and we know the first term, we can write the equation $\begin{aligned} \frac{1}{4} = 16 r^{6} \end{aligned}$ and solve for the common ratio:
::从A7=14到我们知道第一个学期, 我们可以写14=16r6的方程式, 并解决共同比率:

$\begin{aligned} \frac{1}{4} & = 16 r^{6} \\ \frac{1}{64} & = r^{6} \\ \sqrt[6]{\frac{1}{64}} & = \sqrt[6]{r^{6}} \\ \frac{1}{2} & = r \end{aligned}$

::14=16r6164=r61646=r6612=r

The $\begin{aligned} n^{t h} \end{aligned}$ term rule is $\begin{aligned} a_{n} = 16 {(\frac{1}{2})}^{n - 1} \end{aligned}$ .
::nth 术语规则是 an=16 (12)n- 1 。

Finally, let's find the $\begin{aligned} n^{t h} \end{aligned}$ term rule for the geometric sequence in which $\begin{aligned} a_{5} = 8 \end{aligned}$ and $\begin{aligned} a_{10} = \frac{1}{4} \end{aligned}$ .
::最后,让我们来为A5=8和a10=14的几何序列找到nth术语规则。

Using the same method at the previous problem , we can solve for $\begin{aligned} r \end{aligned}$ and $\begin{aligned} a_{1} \end{aligned}$ . Then, write the general rule.
::在前一个问题上使用同样的方法,我们可以解决r和a1。然后,写出一般规则。

Equation 1: $\begin{aligned} a_{5} = 8 \end{aligned}$ , so $\begin{aligned} 8 = a_{1} r^{4} \end{aligned}$ , solving for $\begin{aligned} a_{1} \end{aligned}$ we get $\begin{aligned} a_{1} = \frac{8}{r^{4}} \end{aligned}$ .
::等式 1 : a5= 8, 所以 8 = a1r4, 解决 a1, 我们得到 a1 = 8r4 。

Equation 2: $\begin{aligned} a_{10} = \frac{1}{4} \end{aligned}$ , so $\begin{aligned} \frac{1}{4} = a_{1} r^{9} \end{aligned}$ , solving for $\begin{aligned} a_{1} \end{aligned}$ we get $\begin{aligned} a_{1} = \frac{\frac{1}{4}}{r^{9}} \end{aligned}$ .
::方程式2: a10=14, 所以14=a1r9, 解决a1, 我们得到a1=14r9。

$\begin{aligned} \frac{8}{r^{4}} & = \frac{\frac{1}{4}}{r^{9}} \\ 8 r^{9} & = \frac{1}{4} r^{4} \\ \frac{8 r^{9}}{8 r^{4}} & = \frac{\frac{1}{4} r^{4}}{8 r^{4}} \\ r^{5} & = \frac{1}{32} \\ \sqrt[5]{r^{5}} & = \sqrt[5]{\frac{1}{32}} \\ r & = \frac{1}{2} \end{aligned}$

::8r4=14r98r9=14r48r98r4=14r48r4r5=132r55=1325r=12

Thus, $\begin{aligned} a_{1} = \frac{8}{{(\frac{1}{2})}^{4}} = \frac{8}{\frac{1}{16}} = \frac{8}{1} \cdot \frac{16}{1} = 128 \end{aligned}$ .
::因此,a1=8(12)4=8116=81161=128。

The $\begin{aligned} n^{t h} \end{aligned}$ term rule is $\begin{aligned} a_{n} = (\frac{3}{8}) (2)^{n - 1} \end{aligned}$ .
::Nth术语规则是 an=(38)(2)n-1。

$\begin{aligned} ^{*} \end{aligned}$ Note: In solving the equation above for $\begin{aligned} r \end{aligned}$ we divided both sides by $\begin{aligned} r^{4} \end{aligned}$ . In general it is not advisable to divide both sides of an equation by the variable because we may lose a possible solution, $\begin{aligned} r = 0 \end{aligned}$ . However, in this case, $\begin{aligned} r \neq 0 \end{aligned}$ since it is the common ratio in a geometric sequence.
::* 注:在解决r的上述等式时,我们将双方除以r4。一般而言,不宜将等式的两边除以变量,因为我们可能会失去一个可能的解决方案,r=0。然而,在本案中,r=0,因为它是几何序列中的共同比率。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the $\begin{aligned} n^{t h} \end{aligned}$ term rule of the geometric sequence represented by a bacteria sample that doubles every hour.
::早些时候,有人要求你找到以细菌样本为代表的几何序列的 nth 术语规则每小时翻一番。

We are given that $\begin{aligned} a_{4} = 64 \end{aligned}$ and because the sample doubles every hour we know that the common ratio is 2. We can therefore plug the known values into the equation $\begin{aligned} a_{n} = a_{1} r^{n - 1} \end{aligned}$ to get $\begin{aligned} a_{1} \end{aligned}$ .
::我们得到A4=64, 因为样本每小时翻一番, 我们知道共同比率是2, 因此我们可以将已知值插入 an=1rn-1 方程式中, 以获得 a1 。

$\begin{aligned} a_{4} = a_{1} r^{n - 1} \end{aligned}$ .
::a4 = a1n- 1。

$\begin{aligned} 64 & = a_{1} (2)^{3} \\ 64 & = 8 a_{1} \\ 8 & = a_{1} \end{aligned}$

::64=a1(2)364=8a18=a1

Therefore, there are 8 bacteria in the sample to begin with and the $\begin{aligned} n^{t h} \end{aligned}$ term rule is $\begin{aligned} a_{n} = 8 \cdot 2^{n - 1} \end{aligned}$ .
::因此,样本中从8种细菌开始,N值规则为=82n-1。

Example 2
::例2

Find the first term and the $\begin{aligned} n^{t h} \end{aligned}$ term rule for the geometric sequence given that $\begin{aligned} r = - \frac{1}{2} \end{aligned}$ and $\begin{aligned} a_{6} = 3 \end{aligned}$ .
::根据 r12 和 a6=3 来查找第一个术语和 nth 术语对几何序列的规则。

Use the known quantities in the general form for the $\begin{aligned} n^{t h} \end{aligned}$ term rule to find $\begin{aligned} a_{1} \end{aligned}$ .
::在nth 规则中以一般形式使用已知数量来查找 a1。

$\begin{aligned} 3 & = a_{1} {(- \frac{1}{2})}^{5} \\ (- \frac{32}{1}) \cdot 3 & = a_{1} (- \frac{1}{32}) \cdot (- \frac{32}{1}) \\ a_{1} & = - 96 \end{aligned}$

::3=a1(-12)5(-321)3=a1(-132)_(-321)_(-321)1a196

Thus, $\begin{aligned} a_{n} = - 96 {(- \frac{1}{2})}^{n - 1} \end{aligned}$
::因此, an96(- 12- 12n- 1)

Example 3
::例3

Find the common ratio and the $\begin{aligned} n^{t h} \end{aligned}$ term rule for the geometric sequence given that $\begin{aligned} a_{1} = - \frac{16}{625} \end{aligned}$ and $\begin{aligned} a_{6} = - \frac{5}{2} \end{aligned}$ .
::根据 a116625 和 a652, 查找该几何序列的通用比率和 nth 术语规则。

Again, substitute in the known quantities to solve for $\begin{aligned} r \end{aligned}$ .
::同样,以已知数量取代r。

$\begin{aligned} - \frac{5}{2} & = (- \frac{16}{625}) r^{5} \\ - \frac{5}{2} (- \frac{625}{16}) & = r^{5} \\ \frac{3125}{32} & = r^{5} \\ \sqrt[5]{\frac{3125}{32}} & = \sqrt[5]{r^{5}} \\ r & = \frac{5}{2} \end{aligned}$

::-52=(-16625)r5-52(-62516)=r5312532=r5312525=r55r=52

So, $\begin{aligned} a_{n} = - \frac{16}{625} {(\frac{5}{2})}^{n - 1} \end{aligned}$
::那么, an16625( 52)n-1

Example 4
::例4

Find the $\begin{aligned} n^{t h} \end{aligned}$ term rule for the geometric sequence in which $\begin{aligned} a_{5} = 6 \end{aligned}$ and $\begin{aligned} a_{13} = 1536 \end{aligned}$ .
::查找 a5=6 和 a13=1536 几何序列的 nth 术语规则。

This time we have two unknowns, the first term and the common ratio. We will need to solve a system of equations using both given terms.
::这一次我们有两个未知因素,第一个任期和共同比率。我们需要用两个给定条件解决一个方程式系统。

Equation 1: $\begin{aligned} a_{5} = 6 \end{aligned}$ , so $\begin{aligned} 6 = a_{1} r^{4} \end{aligned}$ , solving for $\begin{aligned} a_{1} \end{aligned}$ we get $\begin{aligned} a_{1} = \frac{6}{r^{4}} \end{aligned}$ .
::等式 1 : a5= 6, 所以 6 = a1r4, 解决 a1 , 我们得到 a1= 6r4 。

Equation 2: $\begin{aligned} a_{13} = 1536 \end{aligned}$ , so $\begin{aligned} 1536 = a_{1} r^{12} \end{aligned}$ , solving for $\begin{aligned} a_{1} \end{aligned}$ we get $\begin{aligned} a_{1} = \frac{1536}{r^{12}} \end{aligned}$ .
::方程式2: a13=1536, 所以1536=a1r12, 解决a1, 我们得到a1=1536r12。

Now that both equations are solved for $\begin{aligned} a_{1} \end{aligned}$ we can set them equal to each other and solve for $\begin{aligned} r \end{aligned}$ .
::现在两个方程式都为A1解决了, 我们可以把它们等同起来, 并解决r。

$\begin{aligned} \frac{6}{r^{4}} & = \frac{1536}{r^{12}} \\ 6 r^{12} & = 1536 r^{4} \\ \frac{6 r^{12}}{6 r^{4}} & = \frac{1536 r^{4}}{6 r^{4}} \\ r^{8} & = 256 \\ \sqrt[8]{r^{8}} & = \sqrt[8]{256} \\ r & = 2 \end{aligned}$

::6r4=1536r126r12=1536r46r126r4=1536r46r4r8=256r88=2568r=2

Now use $\begin{aligned} r \end{aligned}$ to find $\begin{aligned} a_{1} \end{aligned}$ : $\begin{aligned} a_{1} = \frac{6}{(2^{4})} = \frac{6}{16} = \frac{3}{8} \end{aligned}$ .
::现在使用 r 查找 a1: a1=6( 24)=616=38 。

The $\begin{aligned} n^{t h} \end{aligned}$ term rule is $\begin{aligned} a_{n} = (\frac{3}{8}) (2)^{n - 1} \end{aligned}$ .
::Nth术语规则是 an=(38)(2)n-1。

Review
::回顾

Use the given information to find the $\begin{aligned} n^{t h} \end{aligned}$ term rule for each geometric sequence.
::使用给定信息为每个几何序列查找 nth 术语规则。

$\begin{aligned} r = \frac{2}{3} \end{aligned}$ and $\begin{aligned} a_{8} = \frac{256}{81} \end{aligned}$
::r=23和a8=25681

$\begin{aligned} r = - \frac{3}{4} \end{aligned}$ and $\begin{aligned} a_{5} = \frac{405}{8} \end{aligned}$
::r34和a5=4058

$\begin{aligned} r = \frac{6}{5} \end{aligned}$ and $\begin{aligned} a_{4} = 3 \end{aligned}$
::r=65和a4=3

$\begin{aligned} r = - \frac{1}{2} \end{aligned}$ and $\begin{aligned} a_{7} = 5 \end{aligned}$
::r 12和a7=5

$\begin{aligned} r = \frac{6}{7} \end{aligned}$ and $\begin{aligned} a_{0} = 1 \end{aligned}$
::r=67和a0=1

$\begin{aligned} a_{1} = \frac{11}{8} \end{aligned}$ and $\begin{aligned} a_{7} = 88 \end{aligned}$
::a1=118和A7=88

$\begin{aligned} a_{1} = 24 \end{aligned}$ and $\begin{aligned} a_{4} = 81 \end{aligned}$
::a1=24和a4=81

$\begin{aligned} a_{1} = 48 \end{aligned}$ and $\begin{aligned} a_{4} = \frac{3}{4} \end{aligned}$
::a1=48和a4=34

$\begin{aligned} a_{1} = \frac{343}{216} \end{aligned}$ and $\begin{aligned} a_{5} = \frac{6}{7} \end{aligned}$
::a1=343216和a5=67

$\begin{aligned} a_{6} = 486 \end{aligned}$ and $\begin{aligned} a_{10} = 39366 \end{aligned}$
::a6=486和a10=39366

$\begin{aligned} a_{5} = 648 \end{aligned}$ and $\begin{aligned} a_{10} = \frac{19683}{4} \end{aligned}$
::a5=648和a10=196834

$\begin{aligned} a_{3} = \frac{2}{3} \end{aligned}$ and $\begin{aligned} a_{5} = \frac{3}{2} \end{aligned}$
::a3=23和a5=32

$\begin{aligned} a_{5} = \frac{4}{3} \end{aligned}$ and $\begin{aligned} a_{10} = - \frac{128}{3} \end{aligned}$
::a5=43和a10=1283

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

Ricardo’s parents want to have $100,000 saved up to pay for college by the time Ricardo graduates from high school (16 years from now). If the investment vehicle they choose to invest in claims to yield 7% growth per year, how much should they invest today? Give your answer to the nearest one thousand dollars.
::里卡多的父母想要在里卡多高中毕业生(16年后的16年)之前有10万美元储蓄来支付大学学费。如果他们选择投资投资工具来投资每年增长7 % , 那么今天他们应该投资多少? 回答最接近1000美元的答案。

If a piece of machinery depreciates (loses value) at a rate of 6% per year, what was its initial value if it is 10 years old and worth $50,000? Give your answer to the nearest one thousand dollars.
::如果一台机器每年以6%的速率折旧(损耗价值),如果它有10年的历史,价值50 000美元,其最初价值是多少?回答最接近的1000美元。

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

根据共同比率和任何任期或两个任期

Finding the nth Term ::查找 nth 期数

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Finding the nth Term
::查找 nth 期数

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)