College Algebra 7th Edition

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

Chapter 8, Sequences and Series - Section 8.2 - Arithmetic Sequences - 8.2 Exercises: 38

Answer

$d=-3$ $a_5=-1$ The $n^{th}$ term is given by: $a_n=11-3(n-1)$ $a_{100} = -286$

Work Step by Step

The sequence is arithmetic so the terms have a common difference. The common difference $d$ can be found by subtracting any term to the next term in the sequence. Thus, $d=8-11 \\d=-3$ The fifth term $a_5$ can be found by adding the common difference $-3$ to the fourth term. The fourth term of the sequence is $2$. Thus, $a_5 = 2+(-3) \\a_5=-1$ The $n^{th}$ term $a_n$ of an arithmetic sequence is given by the formula $a_n = a+d(n-1)$ where $a$ = first term and $d$ = common difference. The sequence has $a=11$ and $d=-3$. Thus, the $n^{th}$ term is given by: $a_n = 11+(-3)(n-1) \\a_n=11-3(n-1)$ Substituting 100 to $n$ gives the 100th term as: $a_{100} = 11-3(100-1) \\a_{100} = 11-3(99) \\a_{100} = 11-297 \\a_{100} = -286$
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