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 - Page 615: 41


$a_1=-\dfrac{1}{27}$ $a_2=\dfrac{1}{9}$

Work Step by Step

To find the first and second 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: $a_3=-\dfrac{1}{3} \\a_6 = 9$ 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 $a_6=9$ and $a_3=-\dfrac{1}{3}$, substituting these values gives: $a_6=a_3 \cdot r^3 \\9 = -\dfrac{1}{3} \cdot r^3 \\-3 \cdot 9 = -3 \cdot (-\dfrac{1}{3} \cdot r^3) \\-27 = r^3 \\(-3)^3 = r^3$ Take the cube root of both sides to obtain: $-3=r$ With $r=-3$, the second term can be computed by dividing $a_2$ by the common ratio to obtain: $a_2 = \dfrac{a_3}{r} \\a_2 = \dfrac{-\frac{1}{3}}{-3} \\a_2=-\dfrac{1}{3} \cdot \dfrac{1}{-3} \\a_2=\dfrac{1}{9}$ The first term can be found by dividing the second term by the common ratio: $a_1 = \dfrac{a_2}{r} \\a_1=\dfrac{\frac{1}{9}}{-3} \\a_1=\dfrac{1}{9} \cdot \dfrac{1}{-3} \\a_1=-\dfrac{1}{27}$
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