Book edition 6th

Author(s) Sullivan

Pages 1200 pages

ISBN 9780321795465

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Short Answer

In Problems 51–66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.
$9 + 12 + 16 + \frac{64}{3} + . . . .$

The given infinite geometric series $9 + 12 + 16 + \frac{64}{3} + . . . .$ is diverges.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Write the given information.

The given geometric series is:

$9 + 12 + 16 + \frac{64}{3} + . . . .$

Step 2: Find the common ratio.

$a_{1} = 9, a_{2} = 12, a_{3} = 16, a_{4} = \frac{64}{3}$

The common ratio is the ratio of successive terms:

$\frac{12}{9} = \frac{4}{3}, \frac{16}{12} = \frac{4}{3} r = \frac{4}{3}$

Step 3. Determine whether the infinite geometric series is converges or diverges.

If $|r| < 1$ then the infinite geometric series $\sum_{k = 1}^{\infty} a_{1} r^{k - 1}$ converges.

As we can see that $r = \frac{4}{3} and \frac{4}{3} > 1$ , therefore the given series is not converges.

The given infinite geometric series is diverges.

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Precalculus Enhanced with Graphing Utilities

Answers without the blur.

Short Answer

In Problems 51–66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.
$9 + 12 + 16 + \frac{64}{3} + . . . .$

Step by Step Solution

Step 1. Write the given information.

Step 2: Find the common ratio.

Step 3. Determine whether the infinite geometric series is converges or diverges.

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Precalculus Enhanced with Graphing Utilities

Answers without the blur.

Short Answer

In Problems 51–66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.9+12+16+643+....

Step by Step Solution

Step 1. Write the given information.

Step 2: Find the common ratio.

Step 3. Determine whether the infinite geometric series is converges or diverges.

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In Problems 51–66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.
$9 + 12 + 16 + \frac{64}{3} + . . . .$