College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.3 - Geometric Sequences - 8.3 Exercises: 47

Answer

$a_6=6$

Work Step by Step

The given geometric sequence has: $a= 1536 \\r=\dfrac{1}{2}$ RECALL: The $n^{th}$ term $a_n$ of a geometric sequence is given by the formula: $a_n=a \cdot r^{n-1}$ where $a$ = first term $r$ = common ratio This means that the $n^{th}$ term of the given sequence is given by the formula: $a_n= 1536 \cdot \left(\dfrac{1}{2}\right)^{n-1}$ To know which term of the sequence is $6$, substitute $6$ to $a_n$ to obtain: $\require{cancel} 6 = 1536 \cdot \left(\dfrac{1}{2}\right)^{n-1} \\\dfrac{6}{1536} = \dfrac{1536 \cdot (\frac{1}{2})^{n-1}}{1536} \\\dfrac{\cancel{6}}{\cancel{6}(256)}=\dfrac{\cancel{1536} \cdot (\frac{1}{2})^{n-1}}{\cancel{1536}} \\\dfrac{1}{256} = \left(\dfrac{1}{2}\right)^{n-1}$ Note that $256 = 2^8$. Thus, the expression above is equivalent to: $\dfrac{1}{2^8} = \left(\dfrac{1}{2}\right)^{n-1}$ Since $\dfrac{1}{2^8} = \left(\dfrac{1}{2}\right)^8$, the expression above is equivalent to: $\left(\dfrac{1}{2}\right)^8=\left(\dfrac{1}{2}\right)^{n-1}$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $8=n-1 \\8+1 = n-1+1 \\9 =n$ Thus, $a_9=6$.
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