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: 23

Answer

The first five terms are: $a_1 =6 \\a_2 =18 \\a_3 =54 \\a_4= 162 \\a_5=486$ The sequence is geometric with a common ratio of $3$. The $n^{th}$ term of the geometric sequence is: $a_n = 6\cdot(3^{n-1})$

Work Step by Step

To find the first five terms, substitute 1, 2, 3, 4, and 5 to the given formula to obtain: $a_1 = 2(3^1) = 2(3) = 6 \\a_2 = 2(3^2) = 2(9) = 18 \\a_3 = 2(3^3) = 2(27) = 54 \\a_4=2(3^4) = 2(81) = 162 \\a_5=2(3^5) = 2(243) = 486$ RECALL: A sequence is geometric if there is a common ratio among consecutive terms. Notice that 3 times a term is equal to the value of the next term. This means that the sequence has a common ratio of $3$. Thus, the sequence is geometric with a common ratio of $3$. The $n^{th}$ term of a geometric sequence is given by the formula $a_n = ar^{n-1}$ where $a$ = first term and $r$= common ratio. This means that the $n^{th}$ term of the given geometric sequence, whose $a=6$ and $r=3$, is: $a_n = 6\cdot(3^{n-1})$
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