Algebra 2 (1st Edition)

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

Chapter 12 Sequences and Series - Mixed Review of Problem Solving - Lessons 12.1-12.3 - Page 818: 7d


$n \gt 7.044$

Work Step by Step

From the previous part it can be seen that $a_1=66; r=\dfrac{1}{2}$. and $a_n=66 (\dfrac{1}{2})^{n-1}$ We need the n-th term to be less than 1 gram. Thus, $a_n \lt 1 \implies 66 (\dfrac{1}{2})^{n-1} \lt 1$ or, $(n-1) \log (\dfrac{1}{2}) \lt \log (\dfrac{1}{66})$ We will divide the inequality with $\log (1/2)$, which is negative. Thus, we have $n-1 \gt \dfrac{\log (1/66)}{\log (1/2)}$ or, $n \gt 7.044$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.