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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 12: Q 152. (page 1198)

In the following exercises, find the indicated term of the given sequence. Find the general term for the sequence.
Find $a_{12}$ of the sequence, $5, 15, 45, 135, 405, 1215, . . .$

Short Answer

Expert verified

The value of $a_{12}$ is $885, 735$ and the general term of the sequence is $a_{n} = (5) {(3)}^{n - 1}$ .

Step by step solution

01

Step 1. Write the given information, determine the first term and common ratio.

The given sequence is:

$5, 15, 45, 135, 405, 1215, . . .$

First term is:

$a_{1} = 5$

Common ratio:

$\frac{15}{5} = 3, \frac{45}{15} = 3, \frac{135}{45} = 3, \frac{405}{135} = 3, \frac{1215}{405} = 3 r = 3$

02

Step 2. Use the formula of general term to find the twelfth term

$a_{n} = a_{1} r^{n - 1}$

Substitute the values $a_{1} = 5, r = 3$ in the formula,

role="math" localid="1645348479546" $a_{12} = (5) {(3)}^{12 - 1} = (5) {(3)}^{11} = (5) (177147) = 885, 735 a_{12} = 885, 735$

03

Step 3. Find the general term.

Substitute $a_{1} = 5$ and $r = 3$ in the formula,

$a_{n} = a_{1} r^{n - 1} a_{n} = (5) {(3)}^{n - 1}$

The general term of the sequence is $a_{n} = 5 {(3)}^{n - 1}$

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