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 - Page 647: 55

Answer

The general term $a_n$ is given by the formula: $a_n=\$4000 + \$125(n-1)$ At the end of his 12-month training, his monthly salary would be $\$5,375$.

Work Step by Step

RECALL: An arithmetic sequence is a sequence that has a common difference. The next term of an arithmetic sequence can be found by adding the common difference to the previous term. Since the his starting salary is $\$4,000$ and he is guaranteed a monthly increase of $\$125$, then the given sequence involves a common difference of $\$125$. Thus, this sequence is arithmetic. The given arithmetic sequence has $a_1 = \$4,000$ and a common difference $d=\$125$. The $n^{th}$ term ($a_n$) of an arithmetic sequence is given by the formula $a_n=a_1 +d(n-1)$ where $a_1$ is the first term and $d$ is the common difference. Thus, the general term $a_n$ is given by the formula: $a_n = a_1 +d(n-1) \\a_n=\$4000 + \$125(n-1)$ Thus, at the end of his 12-month training, his monthly salary would be: $a_{12} = a_1 + d(12-1) \\a_{12}=\$4000 + \$125(12-1) \\a_{12}=\$4000+\$125(11) \\a_{12}=\$4000+\$1375 \\a_{12}=\$5,375$
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