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: 74

Answer

The sequence if arithimetic. $d=-\dfrac{3}{4}$ $S_{50}=-\dfrac{2225}{4}$

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: $a_1 =8-\frac{3}{4}(1)=8-\frac{3}{4} = \frac{32}{4} - \frac{3}{4} =\frac{29}{4}$ $a_2 = 8-\frac{3}{4}(2)=8-\frac{6}{4} = \frac{32}{4} - \frac{6}{4} =\frac{26}{4}=\frac{13}{2}$ $a_3 = 8-\frac{3}{4}(3)=8-\frac{9}{4} = \frac{32}{4} - \frac{9}{4} =\frac{23}{4}$ Notice that the values decrease by $\frac{3}{4}$. Thus, the sequence is arithmetic with $d=-\frac{3}{4}$. $\bf\text{Find the sum of the first 50 terms}:$ With $a_1=\dfrac{29}{4}$ and $d=-\frac{3}{4}$, solve for the sum of the first 50 terms using the formula in (1) above to obtain: $S_n = \dfrac{n}{2}\left(2a_1+(n-1)d)\right) \\S_{50} = \dfrac{50}{2}\left(2\cdot(\frac{29}{4}) + (-\frac{3}{4})(50-1)\right) \\S_{50} = 25(\frac{58}{4}+(-\frac{3}{4}) \cdot 49) \\S_{50} = 25(\frac{58}{4} -\frac{147}{4}) \\S_{50}=25(-\frac{89}{4}) \\S_{50}=-\dfrac{2225}{4}$
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