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 - Page 607: 36

Answer

$d=-15$ $a_5=4$ The $n^{th}$ term is given by: $a_n=64-15(n-1)$ $a_{100} = -1421$

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, $d=49-64 \\d=-15$ The fifth term $a_5$ can be found by adding the common difference $-15$ to the fourth term. The fourth term of the sequence is $19$. Thus, $a_5 = 19+(-15) \\a_5=4$ 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 $a=64$ and $d=-15$. Thus, the $n^{th}$ term is given by: $a_n = 64+(-15)(n-1) \\a_n=64-15(n-1)$ Substituting 100 to $n$ gives the 100th term as: $a_{100} = 64-15(100-1) \\a_{100} = 64-15(99) \\a_{100} = 64-1485 \\a_{100} = -1421$
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