Chapter 8.2, Problem 32E

To determine

To calculate: The rule for $nth$ term of the arithmetic sequence.

Answer to Problem 32E

The $nth$ term is $a_n=9n-5$ .

Explanation of Solution

Given information:

  $a_7=58,a_{11}=94$ .

Formula used:

Arithmetic difference − the difference of consecutive term is constant. The constant difference is called the common difference and is denoted by d.

Therefore,

  $d=a_{n+1}-a_n;n\ge 1$ .

The $nth$ tem of an arithmetic sequence with first term $a_1$ and common difference $d$ is given by : $a_n=a_1+(n-1)d$ .

Calculation:

Consider,

  $a_7=58,a_{11}=94$

Step 1. System of equations using formula $a_n=a_1+(n-1)d$ .

  $\begin{array}{l}=a_7=a_1+(7-1)d\\ =a_7=a_1+6d\\ =58=a_1+6d\text{ (i) }\end{array}$

  $\begin{array}{l}=a_{11}=a_1+(11-1)d\\ =a_{11}=a_1+10d\\ \text{= 94}=a_1+10d\text{ (ii)}\end{array}$

Step 2. Solving equation (i) and (ii) we get,

  $\begin{array}{l}=\begin{array}{cccc}58& =& a_1+& 6d\end{array}\\ \begin{array}{ccccc}=& 94& =& a_1+& 10d\end{array}\to \text{Subtract}\\ =\begin{array}{cccc}-36& =& 0& -\text{ 4}d\end{array}\end{array}$

  $\begin{array}{l}=-36=-4d\\ =\frac{-36}{-4}=\frac{-4d}{-4}\\ =9=d\end{array}$

Putting value d in one the equation we get,

  $\begin{array}{l}=58=a_1+6(9)\\ =58=a_1+54\\ =58-54=a_1\\ =4=a_1\end{array}$

Step 3. The general rule.

  $\begin{array}{l}{a}_n=4+(n-1)9\\ =4+9n-9\\ =9n-5\end{array}$

Hence, the $nth$ term is $a_n=9n-5$ .