1 / 18

Example

Example. For each geometric sequence, find the common ratio. a) - 2, - 12, - 72, - 432, . . . b) 50, 10, 2, 0.4, 0.08,. Solution. Sequence Common Ratio a) - 2, - 12, - 72, - 432,. r = 6. b) 50, 10, 2, 0.4, 0.08,. r = 0.2. Example. Find the 8 th term of each sequence.

airlia
Download Presentation

Example

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Example For each geometric sequence, find the common ratio. a) -2, -12, -72, -432, . . . b) 50, 10, 2, 0.4, 0.08, . . . Solution Sequence Common Ratio a) -2, -12, -72, -432, . . . r = 6 b) 50, 10, 2, 0.4, 0.08, . . . r = 0.2

  2. Example Find the 8th term of each sequence. a) –2, –12, –72, –432, –2592, . . . b) 50, 10, 2, 0.4, 0.08, . . . Solution a) First, we note that a1 = –2, n = 8, and r = 6. The formula an = a1rn – 1 gives us a8 = –2·68 – 1 = –2·67 = –2(279936) = –559872.

  3. Solution continued b) First, we note that a1 = 50, n = 8, and r = 0.2. The formula an = a1rn – 1 gives us a8 = 50·(0.2)8 – 1 = 50·(0.2)7 = 50(0.0000128) = 0.00064.

  4. Sum of the First n Terms of a Geometric Sequence

  5. Example Find the sum of the first 9 terms of the geometric sequence -1, 4, -16, 64, . . . . Solution First, we note that a1 = -1, n = 9, and Then, substituting in the formula we have

  6. Example Determine whether each series has a limit. If a limit exists, find it. a) -2 – 12 – 72 – 4323 – · · · b) 50 + 10 + 2 + 0.4 + 0.08 + · · · Solution a) Here r = 6, so | r| = | 6 | = 6. Since | r | > 1, the series does not have a limit.

  7. Solution continued b) Here r = 0.2, so | r | = | 0.2 | = 0.2. Since | r | < 1, the series does have a limit. We find the limit by substituting into the formula for S∞:

  8. Example Find the fraction notation for 0.482482482…. Solution We can express this as 0.482 + 0.000482 + 0.000000482 + · · ·. This is an infinite geometric series, where a1 = 0.482 and r = 0.001. Since | r| < 1, this series has a limit: Thus fraction notation for 0.482482482… is

  9. Problem Solving For some problem-solving situations, the translation may involve geometric sequences or series.

More Related