# 7.6.1: 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.

# Arithmetic Sequence

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

**. We can generalize the equation for an arithmetic sequence below:**

**common difference**\(\ a_{n}-a_{n-1}=d\), 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 \(\ d=-3\)

\(\ \begin{array}{l}

a_{1}=22 \text { or } 22+(1-1)(-3)=22+0=22 \\

a_{2}=19 \text { or } 22+(2-1)(-3)=22+(-3)=19 \\

a_{3}=16 \text { or } 22+(3-1)(-3)=22+(-6)=16 \\

a_{4}=13 \text { or } 22+(4-1)(-3)=22+(-9)=13 \\

\quad \vdots \\

a_{n}=22+(n-1)(-3) \\

a_{n}=22-3 n+3 \\

a_{n}=-3 n+25

\end{array}\)

Now we can generalize this into a rule for the \(\ n^{t h}\) term of any arithmetic sequence: \(\ a_{n}=a_{1}+(n-1) d\)

Let's find the common difference and \(\ n^{t h}\) term rule for the arithmetic sequence: \(\ 2,5,8,11 \ldots\)

To find the common difference we subtract consecutive terms.

\(\ \begin{aligned}

&5-2=3\\

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

&11-8=3

\end{aligned}\)

Now we can put our first term and common difference into the \(\ n^{t h}\) term rule discovered above and simplify the expression.

\(\ \begin{aligned}

a_{n} &=2+(n-1)(3) \\

&=2+3 n-3 \quad, \text { so } a_{n}=3 n-1 . \\

&=3 n-1

\end{aligned}\)

Now, let's find the \(\ n^{t h}\) term rule and thus the \(\ 100^{t h}\) term for the arithmetic sequence in which \(\ a_{1}=-9\) and \(\ d=2\).

We have what we need to plug into the rule:

\(\ \begin{aligned}

a_{n}&=-9+(n-1)(2)\\

&=-9+2 n-2 \quad, \text { thus the } n^{t h} \text { term rule is } a_{n}=2 n-11\\

&=2 n-11

\end{aligned}\)

Now to find the \(\ 100^{t h}\) term we can use our rule and replace \(\ n\) with 100:

\(\ a_{100}=2(100)-11=200-11=189\)

Finally, let's find the \(\ n^{t h}\) term rule and thus the \(\ 100^{t h}\) term for the arithmetic sequence in which \(\ a_{3}=8\) and \(\ d=7\).

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:

\(\ \begin{aligned}

a_{1}+(3-1)(7) &=8 \\

a_{1}+2(7) &=8 \\

a_{1}+14 &=8 \\

a_{1} &=-6

\end{aligned}\)

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.

\(\ \begin{aligned}

a_{n} &=-6+(n-1)(7) \\

&=-6+7 n-7 \quad, \text { thus } a_{n}=7 n-13 . \\

&=7 n-13

\end{aligned}\)

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

# Examples

Example 1

Earlier, you were asked to find the \(\ n^{t h}\) term rule and the \(\ 10^{t h}\) term for the sequence represented by Halley's Comet, which appears in the sky once approximately every 76 years and first appeared in 1531.

**Solution**

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:

\(\ \begin{aligned}

&a_{n}=1531+(n-1)(76)\\

&=1531+76 n-76 \quad\text { , thus the } n^{t h} \text { term rule is } a_{n}=76 n+1455

\end{aligned}\)

Now to find the \(\ 10^{t h}\) term we can use our rule and replace \(\ n\) with 10:

\(\ a_{10}=76(10)+1455=760+1455=2215\)

Example 2

Find the common difference and the \(\ n^{t h}\) term rule for the sequence: \(\ 5,−3,−11,…\)

**Solution**

The common difference is \(\ −3−5=−8\). Now

\(\ a_{n}=5+(n-1)(-8)=5-8 n+8=-8 n+13\)

Example 3

Write the \(\ n^{t h}\) term rule and find the \(\ 45^{t h}\) term for the arithmetic sequence with \(\ a_{10}=1\) and \(\ d=-6\).

**Solution**

To find the first term:

\(\ \begin{aligned}

a_{1}+(10-1)(-6) &=1 \\

a_{1}-54 &=1 \\

a_{1} &=55

\end{aligned}\)

Find the \(\ n^{t h}\) term rule: \(\ a_{n}=55+(n-1)(-6)=55-6 n+6=-6 n+61\).

Finally, the \(\ 45^{t h}\) term: \(\ a_{45}=-6(45)+61=-209\).

Example 4

Find the \(\ 62^{n d}\) term for the arithmetic sequence with \(\ a_{1}=-7\) and \(\ d=\frac{3}{2}\).

**Solution**

This time we will not simplify the \(\ n^{t h}\) term rule, we will just use the formula to find the \(\ 62^{r d}\) term:

\(\ a_{62}=-7+(62-1)\left(\frac{3}{2}\right)=-7+61\left(\frac{3}{2}\right)=-\frac{14}{2}+\frac{183}{2}=\frac{169}{2}\).

# Review

Identify which of the following sequences is arithmetic. If the sequence is arithmetic find the \(\ n^{t h}\) term rule.

- 2, 3, 4, 5, …
- 6, 2, −1, −3, …
- 5, 0, −5, −10, …
- 1, 2, 4, 8, …
- 0, 3, 6, 9, …
- 13, 12, 11, 10, …
- 4, −3, 2, −1, …
- a, a+2, a+4, a+6, …

Write the \(\ n^{t h}\) term rule for each arithmetic sequence with the given term and common difference.

- \(\ a_{1}=15\) and \(\ d=-8\)
- \(\ a_{1}=-10\) and \(\ d=\frac{1}{2}\)
- \(\ a_{3}=24\) and \(\ d=-2\)
- \(\ a_{5}=-3\) and \(\ d=3\)
- \(\ a_{10}=-15\) and \(\ d=-11\)
- \(\ a_{7}=32\) and \(\ d=7\)
- \(\ a_{n-2}=3 n+2\), find \(\ a_{n}\)

# Answers for Review Problems

To see the Review answers, open this PDF file and look for section 11.5.