Chapter 6, Problem 9CT

To determine

To write an explicit and recursive rule for the given sequence.

Answer to Problem 9CT

The explicit rule is $a_n=14n-20$ and recursive rule is $a_1=-6,a_n=a_{n-1}+14.$

Explanation of Solution

Given information:

Given table that represents a sequence is

$n$	$1$	$2$	$3$	$4$
$a_n$	$-6$	$8$	$22$	$36$

The sequence obtained from above table is $-6,8,22,36$ is an arithmetic sequence with first term $a_1=-6$ and common difference $d=8-(-6)=8+6=14$

Thus, the explicit rule or $n^{th}\text{term}$ for an arithmetic sequence is $a_n=a_1+\left(n-1\right)d$

  $\begin{array}{l}\Rightarrow a_n=\left(-6\right)+\left(n-1\right)14\\ =-6+14n-14\\ \therefore a_n=14n-20\end{array}$

Thus, recursive rule for an arithmetic sequence is $a_n=a_{n-1}+d$

  $\begin{array}{l}\Rightarrow a_n=a_{n-1}+(14)\\ \therefore a_n=a_{n-1}+14.\end{array}$

Hence, the explicit rule is $a_n=14n-20$ and recursive rule is $a_1=-6,a_n=a_{n-1}+14.$