## College Algebra (10th Edition)

Published by Pearson

# Chapter 9 - Section 9.3 - Geometric Sequences; Geometric Series - 9.3 Assess Your Understanding: 72

#### Answer

The sequence is neither arithmetic nor geometric.

#### 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 $\bf\text{List the first few terms of the sequence.}$ $\bf\text{Identify sequence as arithmetic or geometric.}$ Substitute $1, 2, 3$ for $n$ to list the first three terms: $a_1 =5(1^2)+1=5(1)+1=6 \\a_2 = 5(2^2)+1=5(4)+1=21 \\a_3 = 5(3^2)+1=5(9)+1=46$ The sequence has no common difference. To check if a common ratio exists, solve for $r$ for a few pairs of consecutive terms: $r=\dfrac{a_2}{a_1} = \dfrac{21}{6} = \dfrac{7}{2} \\r=\dfrac{a_3}{a_2} = \dfrac{46}{21}$ The ratios are different, so the sequence is not geometric.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.