Course: 11.6 以两个任期为第九个任期

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找到第n期有两个任期
You are paying off a student loan in monthly installments. After your fifth payment your remaining balance on the loan is $17,500. After your $\begin{align*}16th\end{align*}$ payment, your remaining balance is $12,000. What is the $\begin{align*}n^{th}\end{align*}$ term rule for the represented by this situation?
::您每月分期偿还学生贷款。在第五次支付之后, 您的剩余贷款余额为 17 500美元。在第16次支付之后, 您的剩余余额为 12 000美元。对于这种情况, 您的第n期规则是什么 ?

Finding the nth Term
::查找 nth 期数

Previously, we were given the common difference directly or two consecutive terms from which we could determine the common difference. In this concept we will find the common difference and write $\begin{align*}n^{th}\end{align*}$ term rule given any two terms in the sequence.
::以前,我们直接获得或连续获得共同的区别,从中我们可以决定共同的区别。在这个概念中,我们会找到共同的区别,并写下Nth 规则,在顺序中给出任何两个条件。

Let's find the common difference, first term and $\begin{align*}n^{th}\end{align*}$ term rule for the arithmetic sequence in which $\begin{align*}a_7=17\end{align*}$ and $\begin{align*}a_{20}=82\end{align*}$ .
::让我们找到一个常见的区别, 第一个术语和 nth 术语规则用于计算序列, 其中 a7=17 和 a20=82 。

We will start by using the $\begin{align*}n^{th}\end{align*}$ term rule for an arithmetic sequence to create two equations in two variables:
::我们将首先使用 nth 术语规则的算术序列规则, 在两个变量中创建两个方程式 :

$\begin{align*}a_7=17\end{align*}$ , so $\begin{align*}a_1+(7-1)d=17\end{align*}$ or more simply: $\begin{align*}a_1+6d=17\end{align*}$
::a7=17, 所以 a1+( 7- 1) d=17 或更多简单 : a1+6d=17

$\begin{align*}a_{20}=82\end{align*}$ , so $\begin{align*}a_1+(20-1)d=82\end{align*}$ or more simply: $\begin{align*}a_1+19d=82\end{align*}$
::a20=82, 所以 a1+( 20- 1) d=82 或更简单 : a1+19d=82

Solve the resulting system:
::解决产生的系统 :

$\begin{aligned} a_{1} + 6 d = 17 a_{1} + 6 d = 17 \\ \underline{- 1 (a_{1} + 19 d = 82)} \Rightarrow \underline{- a_{1} - 19 d = - 82} \\ - 13 d = - 65 \\ d = 5 \end{aligned}$
$\begin{align*}& \qquad a_1+6d=17 \qquad \qquad \quad \cancel{a}_1+6d \ =17 \\ &\underline{-1(a_1+19d=82)} \quad \Rightarrow \quad \underline{- \cancel{a}_1-19d=-82} \\ &\qquad \qquad \qquad \qquad \qquad \qquad \quad \ -13d=-65 \\ &\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ d=5\end{align*}$ Replacing $\begin{align*}d\end{align*}$ with 5 in one of the equations we get
$\begin{aligned} a_{1} + 6 (5) & = 7 \\ a_{1} + 30 & = 17 \\ a_{1} & = - 13 \end{aligned}$
$\begin{align*}a_1+6(5)&=7 \\ a_1+30&=17 \\ a_1&=-13\end{align*}$
::a1+6d=17a1+6d=17-1(a1+19d=82) @a1-19d=82_-13d65 d=5 将 d 替换为 5 的方程中, 我们得到 a1+6(5)=7a1+30=17a1}13

Using these values we can find the $\begin{align*}n^{th}\end{align*}$ term rule:
::使用这些价值,我们可以找到nth术语规则:

$\begin{aligned} a_{n} & = - 13 + (n - 1) (5) \\ a_{n} & = - 13 + 5 n - 5 \\ a_{n} & = 5 n - 18 \end{aligned}$
$\begin{align*}a_n&=-13+(n-1)(5) \\ a_n&=-13+5n-5 \\ a_n&=5n-18\end{align*}$
::an13+5n=5n-18

Now, let's find the common difference, first term and $\begin{align*}n^{th}\end{align*}$ term rule for the arithmetic sequence in which $\begin{align*}a_{11}=-13\end{align*}$ and $\begin{align*}a_{40}=-71\end{align*}$ .
::现在,让我们找出共同的区别, 第一个任期和 nth 术语规则用于算术序列, 在其中 a11\\\\\13 和 a40\\\ 71 。

Though this is exactly the same type of problem as the previous problem above, we are going to use a different approach. We know that the $\begin{align*}n^{th}\end{align*}$ term rule is really just using the first term and adding $\begin{align*}d\end{align*}$ to it $\begin{align*}n-1\end{align*}$ times to find the $\begin{align*}n^{th}\end{align*}$ term. We are going to use that idea to find the common difference. To get from the $\begin{align*}11^{th}\end{align*}$ term to the $\begin{align*}40^{th}\end{align*}$ term, the common difference is added $\begin{align*}40-11\end{align*}$ or 29 times. The difference in the term values is $\begin{align*}-71-(-13)\end{align*}$ or -58. What must be added 29 times to create a difference of -58? We can subtract the terms and divide by the difference in term number to determine the common difference.
::虽然这是与前一个问题完全相同的问题类型,但我们将使用不同的方法。我们知道,n一词规则其实只是使用第一个术语,并且给它添加 d- 1 次来找到 n- 1 次来找到 nth 术语。我们将使用这个概念来找到共同的差别。从第11个术语到第40个术语,共同的差别会增加40- 11次或29次。术语的差别是- 71- (- 13) 或 - 58。产生 - 58 差必须增加29次什么? 我们可以用术语和数字的差别来减去这些条件和差别,以确定共同的差别。

$\begin{aligned} \frac{- 71 - (- 13)}{40 - 11} = \frac{- 71 + 13}{29} = \frac{- 58}{29} = - 2. So d = - 2. \end{aligned}$
$\begin{align*}\frac{-71- \left(-13\right)}{40-11}= \frac{-71+13}{29}=\frac{-58}{29}=-2. \ \text{So} \ d=-2.\end{align*}$
::- 71-(-)1340-1171+132958292. So d2。

Now we can use the common difference and one of the terms to find the first term as we did previously.
::现在,我们可以使用共同的区别和其中的一个术语来找到第一个任期,就像我们以前那样。

$\begin{aligned} a_{1} + (11 - 1) (- 2) & = - 13 \\ a_{1} + (- 20) & = - 13 \\ a_{1} & = 7 \end{aligned}$
$\begin{align*}a_1+(11-1)(-2)&=-13 \\ a_1+(-20)&=-13 \\ a_1&=7\end{align*}$
::a1+(11-1)(-2) 13a1+(-20) 13a1=7

Writing the $\begin{align*}n^{th}\end{align*}$ term rule we get: $\begin{align*}a_n=7+(n-1)(-2)=7-2n+2=-2n+9\end{align*}$ .
::写入 nth 术语规则 : an= 7+(n- 1)(-2)= 7- 2n+2 @ @ @ @ 2n+9 。

Before we look at the next problem for this concept, we are going to connect the $\begin{align*}n^{th}\end{align*}$ term rule for an arithmetic sequence to the equation of a line. Have you noticed that the simplified $\begin{align*}n^{th}\end{align*}$ term rule, $\begin{align*}a_n=pn+q\end{align*}$ , where $\begin{align*}p\end{align*}$ and $\begin{align*}q\end{align*}$ represent constants, looks a little like $\begin{align*}y=mx+b\end{align*}$ , the slope-intercept form of the equation of a line? Let’s explore why this is the case using the arithmetic sequence $\begin{align*}1, 4, 7, 10, \ldots\end{align*}$ If we create points by letting the $\begin{align*}x\end{align*}$ – coordinate be the term number and the $\begin{align*}y\end{align*}$ – coordinate be the term, we get the following points and can plot them in the coordinate plane as shown below,
::在我们研究这个概念的下一个问题之前, 我们将会将计算序列的 nth 术语规则与行的方程连接起来。你有没有注意到, 简化nth 术语规则, an=pn+q, 其中 p和 q 代表常数, 看上去有点像 y=mx+b , 线的方程的斜度- 交叉形式? 我们来探讨为什么使用算术序列 1, 4, 7, 10, ... 如果我们通过让 x - 协调为术语编号, y - 协调为术语, 来创建点, 我们得到以下的点, 我们可以在下面的坐标平面上绘制这些点,

The points are: $\begin{align*}(1, 1), (2, 4), (3, 7), (4, 10)\end{align*}$
::要点是1,1,(2,4),(3,7),(4,10)

Notice, that all of these points lie on the same line. This happens because for each increase of one in the term number $\begin{align*}(x)\end{align*}$ , the term value "> $\begin{align*}(y)\end{align*}$ increases by 3. This common difference is actually the slope of the line.
::请注意,所有这些点都位于同一行。这是因为在数字( x) 中每次增加一个, 术语值增加3 。这个共同差实际上是线的斜坡。

We can find the equation of this line using the slope, 3, and the point $\begin{align*}(1, 1)\end{align*}$ in the equation $\begin{align*}y=mx+b\end{align*}$ as follows:
::在y=mx+b的方程式中,用斜度3和点(1,1)来找到这条线的方程式如下:

$\begin{aligned} 1 & = 3 (1) + b \\ 1 & = 2 + b, so the equation of the line is y = 3 x - 1 \\ - 1 & = b \end{aligned}$
$\begin{align*}1&=3(1)+b \\ 1&=2+b \quad , \text{so the equation of the line is} \ y=3x-1 \\ -1&=b\end{align*}$
::1=3(1)+b1=2+b,因此线的方程式是y=3x-1-1=b

The $\begin{align*}n^{th}\end{align*}$ term rule for the sequence is thus: $\begin{align*}a_n=3n-1\end{align*}$ .
::因此,序列的 nth 术语规则是: an=3n-1。

Finally, let's find the common difference, first term and $\begin{align*}n^{th}\end{align*}$ term rule for the arithmetic sequence in which $\begin{align*}a_{10}=-50\end{align*}$ and $\begin{align*}a_{32}=-182\end{align*}$ .
::最后,让我们找到一个共同的区别, 第一个任期和 nth 术语规则用于a1050 和 a32182 的算术序列。

This time we will use the concept that the terms in an arithmetic sequence are actually points on a line to write an equation. In this case our points are $\begin{align*}(10, -50)\end{align*}$ and $\begin{align*}(32, -182)\end{align*}$ . We can find the slope and the equation as shown.
::这次我们将使用一个概念,即算术序列中的术语实际上是线上的点来写一个方程。在这种情况下,我们的点是(10,-50)和(32,-182),我们可以找到所显示的斜坡和方程。

$\begin{aligned} m = \frac{- 182 - (- 50)}{32 - 10} = \frac{- 132}{22} = - 6 \end{aligned}$
$\begin{align*}m=\frac{-182- \left(-50\right)}{32-10}=\frac{-132}{22}=-6\end{align*}$
::má182-(- 50)32-101322226

Use the point $\begin{align*}(10, -50)\end{align*}$ so find the $\begin{align*}y\end{align*}$ -intercept:
$\begin{aligned} - 50 & = - 6 (10) + b \\ - 50 & = - 60 + b \\ 10 & = b \end{aligned}$
$\begin{align*}-50&=-6(10)+b \\ -50&=-60+b \\ 10&=b\end{align*}$ So $\begin{align*}y=-6x+10\end{align*}$ and $\begin{align*}a_n=-6n+10\end{align*}$ .
::使用点 (10,-50) , 找到 Y 的截取点 : - 506(10)+b- 5060+b10=bso y6x+10 和 an6n+10 。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the $\begin{align*}n^{th}\end{align*}$ term rule for the sequence represented by the student loan situation.
::先前,你被要求找到学生贷款情况所代表顺序的第n期规则。

In this case our points are $\begin{align*}(5, 17,500)\end{align*}$ and $\begin{align*}(16, 12,000)\end{align*}$ . We can find the slope and the equation as shown.
::在这种情况下,我们的观点是(5 17 500美元)和(16 12 000美元),我们可以找到斜坡和方程。

$\begin{aligned} m = \frac{12, 000 - 17, 500}{16 - 5} = \frac{- 5500}{11} = - 500 \end{aligned}$
$\begin{align*}m=\frac{12,000-17,500}{16-5}=\frac{-5500}{11}=-500\end{align*}$
::=12,000 - 17 50016 - 5550011500

Use the point $\begin{align*}(5, 17,500)\end{align*}$ to find the $\begin{align*}y\end{align*}$ -intercept:
$\begin{aligned} 17, 500 & = - 500 (5) + b \\ 17, 500 & = - 2500 + b \\ 20, 000 & = b \end{aligned}$
$\begin{align*}17,500&=-500(5)+b \\ 17,500&=-2500+b \\ 20,000&=b\end{align*}$ So $\begin{align*}y=-500x+20,000\end{align*}$ and $\begin{align*}a_n=-500n+20,000\end{align*}$ .
::使用点(5,17,500)来查找 Y 接口: 17,500500(5)+b17,5002500+b20,000+20,000=bso y500x+20,000和 an500n+20,000。

Example 2
::例2

Find the $\begin{align*}n^{th}\end{align*}$ term rule for the arithmetic sequence with $\begin{align*}a_6=-13\end{align*}$ and $\begin{align*}a_{15}=-40\end{align*}$ .
::以 a6\\\13 和 a15\\\40 查找算术序列的 nth 术语规则。

From $\begin{align*}a_6=-13\end{align*}$ we get the equation $\begin{align*}a_1+(6-1)d=a_1+5d=-13\end{align*}$ .
::a613 我们从 a1+(6- 1)d=a1+5d}13得到方程式 a1+(6- 1)d=a1+5d}13。

From $\begin{align*}a_{15}=-40\end{align*}$ we get the equation $\begin{align*}a_1+(15-1)d=a_1+14d=-40\end{align*}$ .
::a15+40 我们从 a1+1+(15-1)d=a1+14d+40得到方程式。

Use the two equations to solve for $\begin{align*}a_1\end{align*}$ and $\begin{align*}d\end{align*}$ :
::使用两个方程式解析 a1 和 d:

$\begin{aligned} a_{1} + 5 d = - 13 a_{1} + 5 (- 3) = - 13 \\ \underline{- a_{1} - 14 d = 40} Use d to find a_{1} \Rightarrow a_{1} - 15 = - 13. \\ - 9 d = 27 a_{1} = 2 \\ d = - 3 \end{aligned}$
$\begin{align*}& \quad \ a_1+5d=-13 \qquad \qquad \qquad \qquad \quad a_1+5(-3)=-13 \\ &\underline{-a_1-14d=40 \;\;} \ \text{Use} \ d \ \text{to find} \ a_1 \Rightarrow \qquad a_1-15=-13. \\ & \qquad \ -9d=27 \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ a_1=2 \\ & \qquad \qquad d=-3\end{align*}$
::a1+5d13a1+5(-3)13-a1-a1-14d=40_使用 d来查找 a1a1-a1-1513.-9d=27 a1=2d}3

Find the $\begin{align*}n^{th}\end{align*}$ term rule: $\begin{align*}a_n=2+(n-1)(-3)=2-3n+3=-3n+5\end{align*}$ .
::查找 nth 术语规则 : an=2+(n- 1)(-3)=2- 3n+33n+5。

Example 3
::例3

F ind the $\begin{align*}n^{th}\end{align*}$ term rule for the arithmetic sequence with $\begin{align*}a_6=13\end{align*}$ and $\begin{align*}a_{22}=77\end{align*}$ .
::用 a6=13 和 a22=77 来查找算术序列的 nth 术语规则。

The common difference is $\begin{align*}\frac{77-13}{22-6}=\frac{64}{16}=4\end{align*}$ . The first term can be found using $\begin{align*}a_6=13\end{align*}$ :
$\begin{aligned} a_{1} + (6 - 1) (4) & = 13 \\ a_{1} + 20 & = 13 \\ a_{1} & = - 7 \end{aligned}$
$\begin{align*}a_1+(6-1)(4)&=13 \\ a_1+20&=13 \\ a_1&=-7\end{align*}$ Thus $\begin{align*}a_n=-7+(n-1)(4)=-7+4n-4=4n-11\end{align*}$ .
::共同的区别是77-1322-6=6416=4。第一个术语使用a6=13:a1+(6-1)(4)=13a1+20=13a1=13a1=7。因此,a+7+(n-1)(4)7+4(7+4n-4n=4n-11)。

Example 4
::例4

F ind the $\begin{align*}n^{th}\end{align*}$ term rule for the arithmetic sequence with $\begin{align*}a_7=-75\end{align*}$ and $\begin{align*}a_{25}=-273\end{align*}$ .
::查找算术序列的 nth 术语规则, 使用 a7\\\ 75 和 a25\\ 273 。

From $\begin{align*}a_7=-75\end{align*}$ we get the point $\begin{align*}(7, -75)\end{align*}$ . From $\begin{align*}a_{25}=-273\end{align*}$ we get the point $\begin{align*}(25, -273)\end{align*}$ . The slope between these points is $\begin{align*}\frac{-273- \left(-75\right)}{25-7}=\frac{-198}{18}=-11\end{align*}$
::从a7+75我们得到点(7,-75),从a25+273我们得到点(25,-273),两点之间的斜度是-273-(-75)25-7_19818*11。

The $\begin{align*}y\end{align*}$ -intercept can be found next using the point $\begin{align*}(7, -75)\end{align*}$ :
::y- interview 下一条使用点( 7,- 75) 找到 Y 界面 :

$\begin{aligned} - 75 & = - 11 (7) + b \\ - 75 & = - 77 + b \\ 2 & = b \end{aligned}$
$\begin{align*}-75&=-11(7)+b \\ -75&=-77+b \\ 2&=b\end{align*}$
::- 7511(7)+b-7577+b2=b

The final equation is $\begin{align*}y=-11x+2\end{align*}$ and the $\begin{align*}n^{th}\end{align*}$ term rule is $\begin{align*}a_n=-11n+2\end{align*}$ .
::最后的公式是 y11x+2, nth 术语规则是 an11n+2。

Review
::回顾

Use the two given terms to find an $\begin{align*}n^{th}\end{align*}$ term rule for the sequence.
::使用两个给定的术语来为序列找到 nth 术语规则。

$\begin{align*}a_7=-17\end{align*}$ and $\begin{align*}a_{25}=-71\end{align*}$
::a717和a2571

$\begin{align*}a_{11}=23\end{align*}$ and $\begin{align*}a_{42}=85\end{align*}$
::a11=23和a42=85

$\begin{align*}a_3=-6\end{align*}$ and $\begin{align*}a_{12}=-3\end{align*}$
::a3+6和a12+3

$\begin{align*}a_8=24\end{align*}$ and $\begin{align*}a_2=9\end{align*}$
::a8=24和a2=9

$\begin{align*}a_6=-27\end{align*}$ and $\begin{align*}a_{10}=-47\end{align*}$
::a627和a1047

$\begin{align*}a_4=37\end{align*}$ and $\begin{align*}a_{12}=85\end{align*}$
::a4=37和a12=85

$\begin{align*}a_{13}=-20\end{align*}$ and $\begin{align*}a_{30}=-54\end{align*}$
::a1320和a3054

$\begin{align*}a_3=23\end{align*}$ and $\begin{align*}a_9=65\end{align*}$
::a3=23和a9=65

$\begin{align*}a_{30}=-31\end{align*}$ and $\begin{align*}a_{45}=-46\end{align*}$
::a3031和a4546

$\begin{align*}a_5=25\end{align*}$ and $\begin{align*}a_{11}=73\end{align*}$
::a5=25和a11=73

$\begin{align*}a_{10}=-2\end{align*}$ and $\begin{align*}a_{25}=-14\end{align*}$
::a102和a2514

$\begin{align*}a_{16}=14\end{align*}$ and $\begin{align*}a_{28}=23\end{align*}$
::a16=14和a28=23

$\begin{align*}a_2=4n+3\end{align*}$ and $\begin{align*}a_{19}=4n+65\end{align*}$
::a2=4n+3和a19=4n+65

$\begin{align*}a_{10}=-3n+7\end{align*}$ and $\begin{align*}a_{23}=-3n-16\end{align*}$
::a10+3n+7和a23+3n-16

Which method do you prefer? Why?
::你喜欢哪种方法?

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

找到第n期 有两个任期

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)

找到第n期有两个任期

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)