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# Finding the nth Term Given the Common Difference and a Term

## Sequences where difference between any two consecutive terms is constant.

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Practice Finding the nth Term Given the Common Difference and a Term
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Arithmetic Sequences and Finding the nth Term Given the Common Difference and a Term

Halley's Comet appears in the sky approximately every 76 years. The comet was first spotted in the year 1531. Find the $n^{th}$ term rule and the $10^{th}$ term for the sequence represented by this situation.

### Guidance

In this concept we will begin looking at a specific type of sequence called an arithmetic sequence . In an arithmetic sequence the difference between any two consecutive terms is constant. This constant difference is called the common difference . For example, question one in the Review Queue above is an arithmetic sequence. The difference between the first and second terms is $(5 - 3) = 2$ , the difference between the second and third terms is $(7 - 5) = 2$ and so on. We can generalize this in the equation below:

$a_n-a_{n-1}=d$ , where $a_{n-1}$ and $a_n$ represent two consecutive terms and $d$ represents the common difference.

Since the same value, the common difference, $d$ , is added to get each successive term in an arithmetic sequence we can determine the value of any term from the first term and how many time we need to add $d$ to get to the desired term as illustrated below:

Given the sequence: $22, 19, 16, 13, \ldots$ in which $a_1=22$ and $d=-3$

$a_1&=22 \ or \ 22+(1-1)(-3)=22+0=22 \\a_2&=19 \ or \ 22+(2-1)(-3)=22+(-3)=19 \\a_3&=16 \ or \ 22+(3-1)(-3)=22+(-6)=16 \\a_4&=13 \ or \ 22+(4-1)(-3)=22+(-9)=13 \\&\qquad \qquad \vdots \\a_n&=22+(n-1)(-3) \\a_n&=22-3n+3 \\a_n&=-3n+25$

Now we can generalize this into a rule for the $n^{th}$ term of any arithmetic sequence:

$a_{n}=a_1+(n-1)d$

#### Example A

Find the common difference and $n^{th}$ term rule for the arithmetic sequence: $2, 5, 8, 11 \ldots$

Solution: To find the common difference we subtract consecutive terms.

$5-2&=3 \\8-5&=3 \ ,\text{thus the common difference is} \ 3. \\11-8&=3$

Now we can put our first term and common difference into the $n^{th}$ term rule discovered above and simplify the expression.

$a_n&=2+(n-1)(3) \\&=2+3n-3 \quad ,\text{so} \ a_n=3n-1. \\&=3n-1$

#### Example B

Find the $n^{th}$ term rule and thus the $100^{th}$ term for the arithmetic sequence in which $a_1=-9$ and $d=2$ .

Solution: We have what we need to plug into the rule:

$a_n&=-9+(n-1)(2) \\&=-9+2n-2 \quad , \text{thus the} \ n^{th} \ \text{term rule is} \ a_n=2n-11. \\&=2n-11$

Now to find the $100^{th}$ term we can use our rule and replace $n$ with 100: $a_{100}=2(100)-11=200-11=189$ .

#### Example C

Find the $n^{th}$ term rule and thus the $100^{th}$ term for the arithmetic sequence in which $a_3=8$ and $d=7$ .

Solution: This one is a little less straightforward as we will have to first determine the first term from the term we are given. To do this, we will replace $a_n$ with $a_3=8$ and use 3 for $n$ in the formula to determine the unknown first term as shown:

$a_1+(3-1)(7)&=8 \\a_1+2(7)&=8 \\a_1+14&=8 \\a_1&=-6$

Now that we have the first term and the common difference we can follow the same process used in the previous example to complete the problem.

$a_n&=-6+(n-1)(7) \\&=-6+7n-7 \quad , \text{thus} \ a_n=7n-13. \\&=7n-13$

Now we can find the $100^{th}$ term: $a_{100}=7(100)-13=687$ .

Intro Problem Revisit From the information given, we can conclude that $a_1=1531$ and $d=76$ .

We now have what we need to plug into the rule:

$a_n&=1531+(n-1)(76) \\&= 1531+76n-76 \quad , \text{thus the} \ n^{th} \ \text{term rule is} \ a_n=76n+1455$

Now to find the $10^{th}$ term we can use our rule and replace $n$ with 10: $a_{10}=76(10) + 1455=760 + 1455 =2215$ .

### Guided Practice

1. Find the common difference and the $n^{th}$ term rule for the sequence: $5, -3, -11,\ldots$

2. Write the $n^{th}$ term rule and find the $45^{th}$ term for the arithmetic sequence with $a_{10}=1$ and $d=-6$ .

3. Find the $62^{nd}$ term for the arithmetic sequence with $a_1=-7$ and $d=\frac{3}{2}$ .

1. The common difference is $-3-5=-8$ . Now $a_n=5+(n-1)(-8)=5-8n+8=-8n+13$ .

2. To find the first term:

$a_1+(10-1)(-6)&=1 \\a_1-54&=1 \\a_1&=55$

Find the $n^{th}$ term rule: $a_n=55+(n-1)(-6)=55-6n+6=-6n+61$ .

Finally, the $45^{th}$ term: $a_{45}=-6(45)+61=-209$ .

3. This time we will not simplify the $n^{th}$ term rule, we will just use the formula to find the $62^{rd}$ term: $a_{62}=-7+(62-1) \left(\frac{3}{2}\right)=-7+61 \left(\frac{3}{2}\right)=- \frac{14}{2}+ \frac{183}{2}= \frac{169}{2}$ .

### Vocabulary

Arithmetic Sequence
A sequence in which the difference between any two consecutive terms is constant.
Common Difference
The value of the constant difference between any two consecutive terms in an arithmetic sequence.

### Practice

Identify which of the following sequences is arithmetic. If the sequence is arithmetic find the $n^{th}$ term rule.

1. $2, 3, 4, 5, \ldots$
2. $6, 2, -1, -3, \ldots$
3. $5, 0, -5, -10, \ldots$
4. $1, 2, 4, 8, \ldots$
5. $0, 3, 6, 9, \ldots$
6. $13, 12, 11, 10, \ldots$
7. $4, -3, 2, -1, \ldots$
8. $a, a+2, a+4, a+6, \ldots$

Write the $n^{th}$ term rule for each arithmetic sequence with the given term and common difference.

1. $a_1=15$ and $d=-8$
2. $a_1=-10$ and $d= \frac{1}{2}$
3. $a_3=24$ and $d=-2$
4. $a_5=-3$ and $d=3$
5. $a_{10}=-15$ and $d=-11$
6. $a_7=32$ and $d=7$
7. $a_{n-2}=3n+2$ , find $a_n$