Q 9. Show that each sequence is geome... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 12: Q 9. (page 824)

Show that each sequence is geometric. Then find the common ratio and write out the first four terms $\{a_{n}\} = \{- 3 {(\frac{1}{2})}^{n}\}$

Short Answer

Expert verified

The common ratio is $\frac{1}{2}$ and the first four terms are $\frac{- 3}{2}, \frac{- 3}{4}, \frac{- 3}{8}, \frac{- 3}{16}$ .

Step by step solution

Step 1. Given information.

We have given sequence.

$\{a_{n}\} = \{- 3 {(\frac{1}{2})}^{n}\}$

The $n$ th term and $(n - 1)$ th term are

role="math" localid="1646849849630" $a_{n} = - 3 {(\frac{1}{2})}^{n} a_{n - 1} = - 3 {(\frac{1}{2})}^{n - 1}$

Step 2. To find the common ratio.

We have,

$r = \frac{a_{n}}{a_{n - 1}} r = \frac{- 3 {(\frac{1}{2})}^{n}}{- 3 {(\frac{1}{2})}^{n - 1}} r = {(\frac{1}{2})}^{n - n + 1} r = \frac{1}{2}$

Since the ratio of the successive term is the non zero constant $\frac{1}{2}$ therefore $(s_{n})$ is a geometric sequence with common ratio is $\frac{1}{2}$ .

Step 3. To find the terms.

We have,

$a_{1} = - 3 {(\frac{1}{2})}^{1} a_{1} = \frac{- 3}{2} a_{2} = - 3 {(\frac{1}{2})}^{2} a_{2} = \frac{- 3}{4} a_{3} = - 3 {(\frac{1}{2})}^{3} a_{3} = \frac{- 3}{8} a_{4} = - 3 {(\frac{1}{2})}^{4} a_{4} = \frac{- 3}{16}$

Thus, the first four terms of the sequence are $\frac{- 3}{2}, \frac{- 3}{4}, \frac{- 3}{8}, \frac{- 3}{16}$ .

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91影视

Show that each sequence is geometric. Then find the common ratio and write out the first four terms $\{a_{n}\} = \{- 3 {(\frac{1}{2})}^{n}\}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. To find the common ratio.

Step 3. To find the terms.

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Step by step solution

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Show that each sequence is geometric. Then find the common ratio and write out the first four terms $\{a_{n}\} = \{- 3 {(\frac{1}{2})}^{n}\}$