College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 9 - Section 9.3 - Geometric Sequences; Geometric Series - 9.3 Assess Your Understanding: 78

Answer

The sequence is geometric. $r=\dfrac{5}{4}$ $S_{50} \approx 350,319.62$

Work Step by Step

$\bf\text{RECALL:}$ $\bf\text{(1) Arithmetic Sequence }$ A sequence is arithmetic if there exists a common difference $d$ among consecutive terms. $d=a_n-a_{n-1}$ The sum of the first $n$ terms of an arithmetic sequence is given by the formulas: $S_n=\frac{n}{2}(a_1 +a_n)$ or $S_n=\frac{n}{2}\left(2 a_1 + (n-1)d\right)$ $\bf\text{(2) Geometric Sequence }$ A sequence is geometric if there exists a common ratio $r$ among consecutive terms. $r=\dfrac{a_n}{a_{n-1}}$ The sum of the first $n$ terms of a geometric sequence is given by the formula: $S_{n}=a_1 \cdot \dfrac{1-r^n}{1-r}$ In the formulas listed above, $d$ = common difference $r$ = common ratio $a_1$ = first term $a_n$ = nth term $n$ = number of terms $\bf\text{List the first few terms of the sequence.}$ $\bf\text{Identify the sequence as arithmetic or geometric.}$ Substitute $1, 2, 3$ for $n$ to list the first three terms: $a_1 =(\frac{5}{4})^1=\frac{5}{4}$ $a_2 = (\frac{5}{4})^2=\frac{25}{4}$ $a_3 = (\frac{5}{4})^3=\frac{125}{64}$ There is no common difference, so the sequence is not arithmetic. Solve for the ratio of pairs of consecutive terms to obtain: $\require{cancel}\dfrac{a_2}{a_1} = \dfrac{\frac{25}{16}}{\frac{5}{4}}=\dfrac{25}{16} \cdot \dfrac{5}{4} = \dfrac{\cancel{25}5}{\cancel{16}4} \cdot \dfrac{\cancel{5}}{\cancel{4}} = \dfrac{5}{4}$ $\require{cancel}\dfrac{a_3}{a_2} = \dfrac{\frac{125}{64}}{\frac{25}{16}}=\dfrac{125}{64} \cdot \dfrac{16}{25} = \dfrac{\cancel{125}5}{\cancel{64}4} \cdot \dfrac{\cancel{16}}{\cancel{25}}=\dfrac{5}{4}$ The sequence has a common ratio of $\dfrac{5}{4}$. Thus, the sequence is geometric with $r=\frac{5}{4}$. $\bf\text{Find the sum of the first 50 terms}:$ With $a_1=\dfrac{5}{4}$ and $r=\frac{5}{4}$, solve for the sum of the first 50 terms using the formula in (2) above to obtain: $S_n = a_1 \cdot \dfrac{1-r^n}{1-r} \\S_{50} = \dfrac{5}{4} \cdot \left(\dfrac{1-\cdot(\frac{5}{4})^{50}}{1-\frac{5}{4}}\right) \\S_{50} \approx 350,319.62$
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