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: 69

Answer

The sequence is arithmetic. common difference = $d=1$ Sum of first 50 terms = $S_{50}=1375$

Work Step by Step

RECALL: (1) 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 $S_n=\frac{n}{2}[2a_1+(n−1)d]$ (2) A sequence is geometric if there exists a common ratio $r$ among consecutive terms. $r=\frac{a_n}{a_{n−1}}$ 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 List the first few terms of the sequence. Substitute 1,2,3 to n to list the first three terms: $a_1=1+2=3 \\a_2=2+2=4 \\a_3=3+2=5$ Since the values increase by 1, the sequence is arithmetic with $d=1$. Find the sum of the first 50 terms: With $a_1=3$ and $d=1$, solve for the sum of the first 50 terms using the formula in (1) above to obtain: $Sn=\frac{n}{2}(2a_1+(n−1)d)) \\S_{50}=\frac{50}{2}(2⋅3+1(50−1)) \\S_{50}=25(6+1⋅49) \\S_{50}=25(6+49) \\S_{50}=25(55) \\S_{50}=1375$
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