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: 52

Answer

The first four terms of the sequence are: $a_1 = \$500,000$ $a_2=\$575,000$ $a_3=\$661,250$ $a_4=\$760,437.50$ The value of the property at the end of three years is $a_4=\$760,437.50$. Refer to the step-by-step part for the computations and explanation.

Work Step by Step

RECALL: The $n^{th}$ term ($a_n$) of a geometric sequence can be found using the formula: $a_n=a_1 \cdot r^{n-1}$ where $a_1$ is the first term and $r$ is the common ratio. The given geometric sequence has $a_1=\$500,000$ and $r=1.15$. Thus, the first four terms of the sequence are: $a_1 = \$500,000$ $a_2=\$500,000 \cdot 1.15^{2-1}=\$575,000$ $a_3=\$500,000 \cdot 1,15^{3-1}=\$661,250$ $a_4=\$661,250 \cdot 1.15^{4-1} = 760,437.50$ Note that the value of the property at the end of the third year is actually the fourth term of the sequence. This is because the first term of the sequence is the initial value of the property (year 0). Thus, the value of the property at the end of three years is $a_4=\$760,437.50$.
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