Chapter 7.8, Problem 33PPE

To determine

To write a recursive formula for a geometric sequence $192,128,85\frac{1}{3},56\frac{8}{9},\mathrm{.......}$ .

Answer to Problem 33PPE

A recursive rule for given sequence is $a_1=192,a_n=\left(\frac{2}{3}\right)a_{n-1},\text{where}n=2,3,4,\mathrm{.....}$ .

Explanation of Solution

Given information :

Givengeometric sequence $192,128,85\frac{1}{3},56\frac{8}{9},\mathrm{.......}$ ..

In geometric sequence $192,128,85\frac{1}{3},56\frac{8}{9},\mathrm{.......}$ ..

First term $a_1=192$ and common ratio $r=\frac{a_2}{a_1}=\frac{128}{192}=\frac{2}{3}$ .

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

  $\Rightarrow a_n=\left(\frac{2}{3}\right)a_{n-1}$

Hence, a recursive rule for given sequence is $a_1=192,a_n=\left(\frac{2}{3}\right)a_{n-1},\text{where}n=2,3,4,\mathrm{.....}$ .