You buy new furniture at zero percent interest on a monthly installment plan. The total of your furniture is $4800. The following sequence shows the balance you still owe on the furniture at the beginning of each month. How would you write a general rule for the sequence?

4800, 4600, 4400, 4200,...

### nth Term

Recursive rules can help us generate multiple sequential terms in a sequence but are not helpful in determining a particular single term. Consider the sequence: \begin{align*} 3, 5, 7, \ldots, a_n \end{align*}. The recursive rule for this sequence is \begin{align*}a_n=a_{n-1}+2\end{align*}. What if we want to find the \begin{align*}100^{th}\end{align*} term? The recursive rule only allows us to find a term in the sequence if we know the previous term. An \begin{align*}n^{th}\end{align*} **term** or **general rule,** however, will allow us to find the \begin{align*}100^{th}\end{align*} term by replacing \begin{align*}n\end{align*} in the formula with 100.

Let's solve the following problems.

- Write the first three terms, the \begin{align*}15^{th}\end{align*} term and the \begin{align*}40^{th}\end{align*} term of the sequence with the general rule: \begin{align*}a_n=n^2-1\end{align*}.

We can find each of these terms by replacing \begin{align*}n\end{align*} with the appropriate term number:

\begin{align*}a_1 &=(1)^2-1=0 \\ a_2 &=(2)^2-1=3 \\ a_3 &=(3)^2-1=8 \\ a_{15} &=(15)^2-1=224 \\ a_{40} &=(40)^2-1=1599\end{align*}

These terms can also be found using a graphing calculator. First press **\begin{align*}2^{nd}\end{align*} STAT** (to get to the **List** menu). Arrow over to **OPS,** select option **5: seq(** and type in (expression, variable, begin, end). For this particular problem, the calculator yields the following:

\begin{align*}seq\left(x^2-1,x,1,3 \right) =\{0 \ 3 \ 8\}\end{align*} for the first three terms

\begin{align*}seq\left(x^2-1,x,15,15 \right)=\{224\}\end{align*} for the \begin{align*}15^{th}\end{align*} term

\begin{align*}seq\left(x^2-1,x,40,40 \right)=\{1599\}\end{align*} for the \begin{align*}40^{th}\end{align*} term

- Write a general rule for the sequence: \begin{align*} 5, 10, 15, 20,\ldots\end{align*}

The previous problem illustrates how a general rule maps a term number directly to the term value. Another way to say this is that the general rule expresses the \begin{align*}n^{th}\end{align*} term as a function of \begin{align*}n\end{align*}. Let’s put the terms in the above sequence in a table with their term numbers to help identify the rule.

Looking at the terms and term numbers together helps us to see that each term is the result of multiplying the term number by 5. The general rule is \begin{align*}a_n=5n\end{align*}

\begin{align*}n\end{align*} | 1 | 2 | 3 | 4 |
---|---|---|---|---|

\begin{align*}a\end{align*} | 5 | 10 | 15 | 20 |

- Find the \begin{align*}n^{th}\end{align*} term rule for the sequence: \begin{align*}0, 2, 6, 12,\ldots\end{align*}

Let’s make the table again to begin to analyze the relationship between the term number and the term value.

\begin{align*}n\end{align*} | 1 | 2 | 3 | 4 |
---|---|---|---|---|

\begin{align*}a_n\end{align*} | 0 | 2 | 6 | 12 |

\begin{align*}n(?)\end{align*} | \begin{align*}(1)(0)\end{align*} | \begin{align*}(2)(1)\end{align*} | \begin{align*}(3)(2)\end{align*} | \begin{align*}(4)(3)\end{align*} |

This time the pattern is not so obvious. To start, write each term as a product of the term number and a second factor. Then it can be observed that the second factor is always one less that the term number and the general rule can be written as \begin{align*}a_n=n(n-1)\end{align*}

### Examples

#### Example 1

Earlier, you were asked how would you write a general rule for the sequence 4800, 4600, 4400, 4200, ...

Let’s put the terms in the sequence in a table with their term numbers to help identify the rule.

\begin{align*}n\end{align*} | 1 | 2 | 3 | 4 |
---|---|---|---|---|

\begin{align*}a\end{align*} | 4800 | 4600 | 4400 | 4200 |

Looking at the terms and term numbers together helps us to see that each term is the result of subtracting 200 times one less than the term from the first term. The general rule is \begin{align*}a_n=a_{n-1} - 200(n - 1)\end{align*}.

#### Example 2

Given the general rule: \begin{align*}a_n=3n-13\end{align*}, write the first five terms, \begin{align*}25^{th}\end{align*} term and the \begin{align*}200^{th}\end{align*} term of the sequence.

Plug in the term numbers as shown:

\begin{align*}a_1 &=3(1)-13=-10 \\ a_2 &=3(2)-13=-7 \\ a_3 &=3(3)-13=-4 \\ a_4 &=3(4)-13=-1 \\ a_5 &=3(5)-13=2 \\ a_{25} &=3(25)-13=62 \\ a_{200} &=3(200)-13=587 \end{align*}

#### Example 3

Write the general rule for the sequence: \begin{align*}4, 5, 6, 7,\ldots\end{align*}

Put the values in a table with the term numbers and see if there is a way to write the term as a function of the term number.

\begin{align*}n\end{align*} | 1 | 2 | 3 | 4 |
---|---|---|---|---|

\begin{align*}a_n\end{align*} | 4 | 5 | 6 | 7 |

\begin{align*}n\pm(?)\end{align*} | \begin{align*}(1)+3\end{align*} | \begin{align*}(2)+3\end{align*} | \begin{align*}(3)+3\end{align*} | \begin{align*}(4)+3\end{align*} |

Each term appears to be the result of adding three to the term number. Thus, the general rule is \begin{align*}a_n=n+3\end{align*}

#### Example 4

Write the general rule and find the \begin{align*}35^{th}\end{align*} term of the sequence: \begin{align*}-1, 0, 3, 8, 15,\ldots\end{align*}

Put the values in a table with the term numbers and see if there is a way to write the term as a function of the term number.

\begin{align*}n\end{align*} | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|

\begin{align*}a_n\end{align*} | -1 | 10 | 3 | 8 | 15 |

\begin{align*}n(?)\end{align*} | \begin{align*}(1)(-1)\end{align*} | \begin{align*}(2)(0)\end{align*} | \begin{align*}(3)(1)\end{align*} | \begin{align*}(4)(2)\end{align*} | \begin{align*}(5)(3)\end{align*} |

Each term appears to be the result of multiplying the term number by two less than the term number. Thus, the general rule is \begin{align*}a_n=n(n-2)\end{align*}.

### Review

Use the \begin{align*}n^{th}\end{align*} term rule to generate the indicated terms in each sequence.

- \begin{align*}2n+7\end{align*}, terms \begin{align*}1-5\end{align*} and the \begin{align*}10^{th}\end{align*} term.
- \begin{align*}-5n-1\end{align*}, terms \begin{align*}1-3\end{align*} and the \begin{align*}50^{th}\end{align*} term.
- \begin{align*}2^n-1\end{align*}, terms \begin{align*}1-3\end{align*} and the \begin{align*}10^{th}\end{align*} term.
- \begin{align*}\left(\frac{1}{2}\right)^n\end{align*}, terms \begin{align*}1-3\end{align*} and the \begin{align*}8^{th}\end{align*} term.
- \begin{align*}\frac{n(n+1)}{2}\end{align*}, terms \begin{align*}1-4\end{align*} and the \begin{align*}20^{th}\end{align*} term.

Use your calculator to generate the first 5 terms in each sequence. Use **MATH > FRAC**, on your calculator to convert decimals to fractions.

- \begin{align*}4n-3\end{align*}
- \begin{align*} -\frac{1}{2}n+5\end{align*}
- \begin{align*}\left(\frac{2}{3}\right)^n+1\end{align*}
- \begin{align*}2n(n-1)\end{align*}
- \begin{align*}\frac{n(n+1)(2n+1)}{6}\end{align*}

Write the \begin{align*}n^{th}\end{align*} term rule for the following sequences.

- \begin{align*}3,5,7,9,\ldots\end{align*}
- \begin{align*}1,7,25,79,\ldots\end{align*}
- \begin{align*}6,14,24,36,\ldots\end{align*}
- \begin{align*}6,5,4,3,\ldots\end{align*}
- \begin{align*}2,5,9,14,\ldots\end{align*}

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 11.3.