## Algebra 2 (1st Edition)

Published by McDougal Littell

# Chapter 12 Sequences and Series - 12.3 Analyze Geometric Sequences and Series - 12.3 Exercises - Problem Solving - Page 816: 60b

#### Answer

$b_n= (\dfrac{8}{9})^n S\\b_{12}= (\dfrac{8}{9})^{12} S$

#### Work Step by Step

Let $S$ be the area of the initial square. Here, we have $b_1= S-\dfrac{S}{9}$ and $b_2= \dfrac{8S}{9}-(8) \dfrac{S}{(9)^2}=\dfrac{64S}{81}$ Likewise, we have $b_n= (\dfrac{8}{9})^n S$ Hence, $b_{12}= (\dfrac{8}{9})^{12} S$

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