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: 43

Answer

$a = -\frac{9}{32}$ The $n^{th}$ term of the given sequence is: $a_n = -\frac{9}{32} \cdot (-8)^{n-1}$

Work Step by Step

To find the first and nth terms, the value of the common ratio $r$ is needed. Note that the next term of a geometric sequence can be found by multiplying the common ratio $r$ to the current term. The known terms are: $a_3=-18 \\a_6 =9216$ Note that using the third term as reference point, the sixth term can be found by multiplying the common ratio $r$ three times to the value of the third term. Thus, $a_6=a_3 \cdot r \cdot r \cdot r \\a_6 = a_3 \cdot r^3$ Since $a_3=-18$ and $a_6=9216$, substituting these values gives: $a_6=a_3 \cdot r^3 \\9216 = -18 \cdot r^3 \\\dfrac{9216}{-18} = \dfrac{-18r^3}{-18} \\-512 = r^3 \\(-8)^3 = r^3$ Take the cube root of both sides to obtain: $-8=r$ With $r=-8$, the first term can be computed by dividing $a_3$ by the common ratio twice to obtain: $a = a_3 \div (r \times r) \\a = a_3 \div r^2 \\a=-18 \div (-8)^2 \\a= -18 \div 64 \\a = -\frac{9}{32}$ RECALL: The $n^{th}$ term $a_n$ of a geometric sequence can be found using the formula $a_n = a \cdot r^{n-1}$ where $a$ = first term $r$ = common ratio Thus, the $n^{th}$ term of the given sequence is: $a_n = -\frac{9}{32} \cdot (-8)^{n-1}$
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