Chapter 7, Problem 11REAE

Solution

To state: The rule for the $n\text{th}$ term of the arithmetic sequence: $a_4=27,a_{11}=69$

The resultant rule is $a_n=3+6n$ .

Given information:

The given fourth and eleventh terms are $a_4=27,a_{11}=69$ .

Explanation:

Consider the given sequence,

  $a_4=27,a_{11}=69$

The 4th term of the sequence is: $a_4=27$

It can be written as:

  $\begin{array}{l}{a}_4=a_1+\left(4-1\right)d\\ 27=a_1+3d\cdots \left(1\right)\end{array}$

The 4th term of the sequence is: $a_{11}=69$

It can be written as:

  $\begin{array}{l}{a}_{11}=a_1+\left(11-1\right)d\\ 69=a_1+10d\cdots \left(2\right)\end{array}$

Subtract equation (1) from equation (2) to find the value of d ,

  $\begin{array}{l}\underset{¯}{\begin{array}{l}69=a_1+10d\\ 27=a_1+3d\end{array}}\\ 42=7d\end{array}$

  $\begin{array}{l}d=\frac{42}{7}\\ d=6\end{array}$

Now find the value of first term,

  $\begin{array}{c}69=a_1+10\left(6\right)\\ 69=a_1+60\\ 9=a_1\end{array}$

Write the general rule: $a_n=a_1+\left(n-1\right)d$

Substitute $a_1=9$ and $d=6$ in the general rule,

  $\begin{array}{c}{a}_n=9+\left(n-1\right)6\\ =9+6n-6\\ =3+6n\end{array}$

Therefore, the rule for the nth term is $a_n=3+6n$ .