Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Section 11.1 - Sequences - Exercise Set - Page 639: 45

Answer

There will be around 50 sparrows in 2004. The sparrow will be extinct by 2010.

Work Step by Step

RECALL: (1) The $n^{th}$ term of a geometric sequence, $a_n$, is represented by the formula: $a_n=a_1 \cdot r^{n-1}$ where $a_1$=first term $r$ = common ratio $n$ = term number (2) The next term of a geometric sequence can be found by multiplying the common ratio $r$ to the present term. $a_n=a_{n-1} \cdot r$ The population of the sparrow can be represented by a geometric sequence with a first term of $800$ and a common ratio of $\frac{1}{2}$ or $0.5$. The geometric sequence has $a_1=800$ and $r=0.5$. Note that the year 2004 represents $n=5$ since $n=1$ represents the year $2000$, Use the formula in (1) above to obtain: $a_n=a_1 \cdot r^{n-1} \\a_5=800 \cdot 0.5^{5-1} \\a_5=800 \cdot 0.5^4 \\a_5=50$ To find the term number when the sparrow will be extinct, just find the succeeding populations to obtain: $2005: 50(0.5)=25 \\2006: 25(0.5) =12.5 \approx 13 \\2007: 12.5(0.5)=1 \approx 6 \\2008: 6.25(0.5)= 3.125 \approx 3 \\2009: 3.125(0.5) = 1.5625\approx 2 \\2010: 1.5625(0.5) = 0.78125$ This number is already less than $1$ so the sparrow will be extinct by this time. Therefore, the sparrow will be extinct by the year 2010.
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