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# Finding the nth Term Given Two Terms for an Arithmetic Sequence

## Identify common difference, use it to state a general rule

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Practice Finding the nth Term Given Two Terms for an Arithmetic Sequence
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Finding the nth Term Given Two Terms

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 $16th$ payment, your remaining balance is$12,000. What is the $n^{th}$ term rule for the sequence represented by this situation?

### Guidance

In the last concept 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 $n^{th}$ term rule given any two terms in the sequence.

#### Example A

Find the common difference, first term and $n^{th}$ term rule for the arithmetic sequence in which $a_7=17$ and $a_{20}=82$ .

Solution: We will start by using the $n^{th}$ term rule for an arithmetic sequence to create two equations in two variables:

$a_7=17$ , so $a_1+(7-1)d=17$ or more simply: $a_1+6d=17$

$a_{20}=82$ , so $a_1+(20-1)d=82$ or more simply: $a_1+19d=82$

Solve the resulting system:

$& \qquad a_1+6d=17 \qquad \qquad \quad \cancel{a}_1+6d \ =17 \\&\underline{-1(a_1+19d=82)} \quad \Rightarrow \quad \underline{- \cancel{a}_1-19d=-82} \\&\qquad \qquad \qquad \qquad \qquad \qquad \quad \ -13d=-65 \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ d=5$ , replacing $d$ with 5 in one of the equations we get $a_1+6(5)&=7 \\a_1+30&=17 \\a_1&=-13$ .

Using these values we can find the $n^{th}$ term rule:

$a_n&=-13+(n-1)(5) \\a_n&=-13+5n-5 \\a_n&=5n-18$

#### Example B

Find the common difference, first term and $n^{th}$ term rule for the arithmetic sequence in which $a_{11}=-13$ and $a_{40}=-71$ .

Solution: Though this is exactly the same type of problem as Example A, we are going to use a different approach. We discovered in the last concept that the $n^{th}$ term rule is really just using the first term and adding $d$ to it $n-1$ times to find the $n^{th}$ term. We are going to use that idea to find the common difference. To get from the $11^{th}$ term to the $40^{th}$ term, the common difference is added $40-11$ or 29 times. The difference in the term values is $-71-(-13)$ 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.

$\frac{-71- \left(-13\right)}{40-11}= \frac{-71+13}{29}=\frac{-58}{29}=-2. \ \text{So} \ d=-2.$

Now we can use the common difference and one of the terms to find the first term as we did previously.

$a_1+(11-1)(-2)&=-13 \\a_1+(-20)&=-13 \\a_1&=7$

Writing the $n^{th}$ term rule we get: $a_n=7+(n-1)(-2)=7-2n+2=-2n+9$ .

### More Guidance

Before we look at the final example for this concept, we are going to connect the $n^{th}$ term rule for an arithmetic sequence to the equation of a line. Have you noticed that the simplified $n^{th}$ term rule, $a_n=pn+q$ , where $p$ and $q$ represent constants, looks a little like $y=mx+b$ , the slope-intercept form of the equation of a line? Let’s explore why this is the case using the arithmetic sequence $1, 4, 7, 10, \ldots$ If we create points by letting the $x$ – coordinate be the term number and the $y$ – coordinate be the term, we get the following points and can plot them in the coordinate plane as shown below,

The points are: $(1, 1), (2, 4), (3, 7), (4, 10)$

Notice, that all of these points lie on the same line. This happens because for each increase of one in the term number $(x)$ , the term value $(y)$ increases by 3. This common difference is actually the slope of the line.

We can find the equation of this line using the slope, 3, and the point $(1, 1)$ in the equation $y=mx+b$ as follows:

$1&=3(1)+b \\1&=2+b \quad , \text{so the equation of the line is} \ y=3x-1 \\-1&=b$

The $n^{th}$ term rule for the sequence is thus: $a_n=3n-1$ .

#### Example C

Find the common difference, first term and $n^{th}$ term rule for the arithmetic sequence in which $a_{10}=-50$ and $a_{32}=-182$ .

Solution: 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 $(10, -50)$ and $(32, -182)$ . We can find the slope and the equation as shown.

$m=\frac{-182- \left(-50\right)}{32-10}=\frac{-132}{22}=-6$

Use the point $(10, -50)$ so find the $y$ -intercept: $-50&=-6(10)+b \\-50&=-60+b \\10&=b$ , so $y=-6x+10$ and $a_n=-6n+10$ .

Intro Problem Revisit We can use any of the methods learned in this lesson, but we'll employ the strategy used in Example C for illustrative purposes.

In this case our points are $(5, 17,500)$ and $(16, 12,000)$ . We can find the slope and the equation as shown.

$m=\frac{12,000-17,500}{16-5}=\frac{-5500}{11}=-500$

Use the point $(5, 17,500)$ to find the $y$ -intercept: $17,500&=-500(5)+b \\17,500&=-2500+b \\20,000&=b$ , so $y=-500x+20,000$ and $a_n=-500n+20,000$ .

### Guided Practice

1. Use the method in Example A to find the $n^{th}$ term rule for the arithmetic sequence with $a_6=-13$ and $a_{15}=-40$ .

2. Use the method in Example B to find the $n^{th}$ term rule for the arithmetic sequence with $a_6=13$ and $a_{22}=77$ .

3. Use the method in Example C to find the $n^{th}$ term rule for the arithmetic sequence with $a_7=-75$ and $a_{25}=-273$ .

1. From $a_6=-13$ we get the equation $a_1+(6-1)d=a_1+5d=-13$ .

From $a_{15}=-40$ we get the equation $a_1+(15-1)d=a_1+14d=-40$ .

Use the two equations to solve for $a_1$ and $d$ :

$& \quad \ a_1+5d=-13 \qquad \qquad \qquad \qquad \quad a_1+5(-3)=-13 \\&\underline{-a_1-14d=40 \;\;} \ \text{Use} \ d \ \text{to find} \ a_1 \Rightarrow \qquad a_1-15=-13. \\& \qquad \ -9d=27 \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ a_1=2 \\& \qquad \qquad d=-3$

Find the $n^{th}$ term rule: $a_n=2+(n-1)(-3)=2-3n+3=-3n+5$ .

2. The common difference is $\frac{77-13}{22-6}=\frac{64}{16}=4$ . The first term can be found using $a_6=13$ : $a_1+(6-1)(4)&=13 \\a_1+20&=13 \\a_1&=-7$ . Thus $a_n=-7+(n-1)(4)=-7+4n-4=4n-11$ .

3. From $a_7=-75$ we get the point $(7, -75)$ . From $a_{25}=-273$ we get the point $(25, -273)$ . The slope between these points is $\frac{-273- \left(-75\right)}{25-7}=\frac{-198}{18}=-11$ . The $y$ -intercept can be found next using the point $(7, -75)$ :

$-75&=-11(7)+b \\-75&=-77+b \\2&=b$

The final equation is $y=-11x+2$ and the $n^{th}$ term rule is $a_n=-11n+2$ .

### Practice

Use the two given terms to find an $n^{th}$ term rule for the sequence.

1. $a_7=-17$ and $a_{25}=-71$
2. $a_{11}=23$ and $a_{42}=85$
3. $a_3=-6$ and $a_{12}=-3$
4. $a_8=24$ and $a_2=9$
5. $a_6=-27$ and $a_{10}=-47$
6. $a_4=37$ and $a_{12}=85$
7. $a_{13}=-20$ and $a_{30}=-54$
8. $a_3=23$ and $a_9=65$
9. $a_{30}=-31$ and $a_{45}=-46$
10. $a_5=25$ and $a_{11}=73$
11. $a_{10}=-2$ and $a_{25}=-14$
12. $a_{16}=14$ and $a_{28}=23$
13. $a_2=4n+3$ and $a_{19}=4n+65$
14. $a_{10}=-3n+7$ and $a_{23}=-3n-16$
15. Which method do you prefer? Why?