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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 12: Q 54. (page 825)

In Problems 51鈥�66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.
$6 + 2 + \frac{2}{3} + . . . . .$

Short Answer

Expert verified

The given infinite geometric series $6 + 2 + \frac{2}{3} + . . . . .$ is convergent and its sum is 9.

Step by step solution

01

Step 1. Write the given information.

The given geometric series is:

$6 + 2 + \frac{2}{3} + . . . . .$

02

Step 2. Find the common ratio.

$a_{1} = 6, a_{2} = 2, a_{3} = \frac{2}{3}$

The common ratio is the ratio of successive terms:

$\frac{2}{6} = \frac{1}{3}, \frac{\frac{2}{3}}{2} = \frac{1}{3} r = \frac{1}{3}$

03

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

If $|r| < 1$ then the infinite geometric series ${鈭�}_{k = 1}^{鈭�} a_{1} r^{k - 1}$ converges.

As we can see that $r = \frac{1}{3} and \frac{1}{3} < 1$ , therefore the given series is convergent.

04

Step 4. Find the sum of the series.

Use the formula to find the sum of the given geometric series ${鈭�}_{k = 1}^{鈭�} a_{1} r^{k - 1} = \frac{a_{1}}{1 - r}$ ,

${鈭�}_{k = 1}^{鈭�} 6 路 {(\frac{1}{3})}^{k - 1} = \frac{6}{(1 - \frac{1}{3})} = \frac{6}{\frac{2}{3}} = 9$

Therefore, the sum of the infinite geometric series is 9.

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Most popular questions from this chapter

In Problems 51鈥�66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.

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

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In Problems 51鈥�66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.

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