Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - 12.3 Analyze Geometric Sequences and Series - 12.3 Exercises - Quiz for Lessons 12.1-12.3 - Page 817: 14


$a_n=2 (\dfrac{2}{3})^{n-1}$ $a_{15}=\dfrac{32768}{4,782, 969}$ $S_{15}= 5.99$

Work Step by Step

The general formula for the nth term of a geometric series is given by $a_n= a_1r^{n-1}$ ...(1) The ratio of the successive terms is $\dfrac{2}{3}$. Equation (1) gives: $a_n= a_1 \times \dfrac{2}{3}^{n-1}=2 (\dfrac{2}{3})^{n-1}$ Plugging in $n =15$, we have $a_{15}=2 (\dfrac{2}{3})^{15-1}=2 (\dfrac{2}{3})^{14}=\dfrac{32768}{4,782, 969}$ We know that $S_{n}=\dfrac{a_1(1-r^n)}{1-r}$ Now, $S_{15}=\dfrac{2(1-\dfrac{2}{3}^{15})}{1-\dfrac{2}{3}} = 5.99$
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