Intermediate Algebra (6th Edition)

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

Chapter 11 - Section 11.2 - Arithmetic and Geometric Sequences - Exercise Set: 26

Answer

$a_{1} = \frac{4}{9}, r= -3$

Work Step by Step

Given, $a_{3} = 4$ $a_{4} = -12$ In geometric sequence, common ratio, $r = \frac{a_{n}}{a_{n-1}}$ $r = \frac{a_{4}}{a_{3}} = \frac{-12}{4} =-3$ To find $a_{1},$ $a_{3} = a_{1} . r^{3-1} $ using $a_{n} = a_{1} . r^{n-1} $ $a_{3} = a_{1} . r^{2} $ Using $r$ and $a_{3}$ values, $4= a_{1} .(-3)^{2} $ $4 = a_{1} .(9) $ $a_{1} = \frac{4}{9}$
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