Arithmetic Sequence III

Problem 222: Given $a = 4, 9, 14, 19, 24, \dots,$ .

a. Find the closed form: $a_n$

b. Use the closed form to find $a_{11}$ .

Solution:

a. The $n^{th}$ term of an arithmetic sequence is given by

(1) $\begin{equation*} a_n = a_0 + d(n-1)\end{equation*}$

where $a_n$ is called the closed form, $a_0$ is the first term in the sequence and $d$ is the common difference. To find the common difference, we just take the difference between the terms.

(2) $\begin{align*} 9 - 4 & =5 \\14 - 9 & = 5 \\ 19 - 14 & = 5 \\ 24-19 & = 5 \end{align*}$

This means that $d = 5$ . We are looking for $a_{n}$ . That is,

(3) $\begin{align*}a_n & = a_0 + d(n-1) \\ & = 4 + 5(n-1) \\ & = 4 + 5n - 5 \\ & = 5n - 1.\end{align*}$

Thus

(4) $\begin{equation*} a_n = 5n -1. \end{equation*}$

b. Using the closed form $a_n$ , we can now find $a_{11}$ .

(5) $\begin{align*}a_n & = 5n - 1\\ a_{11} & = 5(11)-1 \\ & = 55 -1 \\ & = 54. \end{align*}$

Therefore, $a_{11}=54$ .

Leave a Reply Cancel reply