Chapter 6, Problem 10CT

To determine

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

Answer to Problem 10CT

The explicit rule is $a_n=100{\left(0.25\right)}^{n-1}$ and the recursive rule is $a_1=400,a_n=\left(0.25\right)a_{n-1}.$

Explanation of Solution

Given information :

Given table that represents a sequence is

$n$	$1$	$2$	$3$	$4$
$a_n$	$400$	$100$	$25$	$6.25$

The sequence obtained from above table is $400,100,25,6.25$ is a geometric sequence with first term $a_1=100$ and common ratio $r=\frac{100}{400}=\frac{1}{4}=0.25$

Thus, the explicit rule or $n^{th}\text{term}$ for a geometric sequence is $a_n=a_1{r}^{n-1}$

  $\therefore a_n=100{\left(0.25\right)}^{n-1}$

Thus, recursive rule for an arithmetic sequence is $a_n=r{a}_{n-1}$

  $\Rightarrow a_n=\left(0.25\right)a_{n-1}$

Hence, the explicit rule is $a_n=100{\left(0.25\right)}^{n-1}$ and the recursive rule is $a_1=400,a_n=\left(0.25\right)a_{n-1}.$