Chapter 7.8, Problem 35PPE

To determine

To write a recursive formula for a geometric sequence $\$48,\$36,\$27,\$20.25,\mathrm{.......}$ .

Answer to Problem 35PPE

The explicit rule is $a_n=48{\left(\frac{3}{4}\right)}^{n-1}$ and the recursive rule is $a_n=\left(\frac{3}{4}\right)a_{n-1},a_1=48,\text{where}n=2,3,4,5,\mathrm{.......}$

Explanation of Solution

Given information :

Givengeometric sequence $\$48,\$36,\$27,\$20.25,\mathrm{.......}$

In geometric sequence $\$48,\$36,\$27,\$20.25,\mathrm{.......}$ .

Where $a_1=48$ and common ratio $r=\frac{a_2}{a_1}=\frac{36}{48}=\frac{3}{4}$

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

  $\therefore a_n=48{\left(\frac{3}{4}\right)}^{n-1}$

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

  $\Rightarrow a_n=\left(\frac{3}{4}\right)a_{n-1},a_1=48,\text{where}n=2,3,4,5,\mathrm{.......}$

Hence, the explicit rule is $a_n=48{\left(\frac{3}{4}\right)}^{n-1}$ and the recursive rule is $a_n=\left(\frac{3}{4}\right)a_{n-1},a_1=48,\text{where}n=2,3,4,5,\mathrm{.......}$