1 / 36

Understanding Geometric Sequences and Series: A Comprehensive Guide

This guide provides an overview of geometric sequences and series, covering key concepts such as the nth term formula, common ratios, and calculations for specific terms and sums. Examples illustrate how to find terms and sums within these sequences, including both finite and infinite series. Learn to derive the total distance traveled by bouncing balls modeled by geometric sequences. Perfect for students seeking clarity on the subject or educators looking to enhance their teaching materials.

navarro-lajara
Télécharger la présentation

Understanding Geometric Sequences and Series: A Comprehensive Guide

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. Geometric Sequences & Series By: Jeffrey Bivin Lake Zurich High School jeff.bivin@lz95.org Last Updated: March 23, 2011

  2. Geometric Sequences 1, 2, 4, 8, 16, 32, … 2n-1, … 3, 9, 27, 81, 243, … 3n, . . . 81, 54, 36, 24, 16, … , . . . Jeff Bivin -- LZHS

  3. nth term of geometric sequence an = a1·r(n-1) Jeff Bivin -- LZHS

  4. Find the nth term of thegeometric sequence First term is 2 Common ratio is 3 an = a1·r(n-1) an = 2(3)(n-1) Jeff Bivin -- LZHS

  5. Find the nth term of a geometric sequence First term is 128 Common ratio is (1/2) an = a1·r(n-1) Jeff Bivin -- LZHS

  6. Find the nth term of the geometric sequence First term is 64 Common ratio is (3/2) an = a1·r(n-1) Jeff Bivin -- LZHS

  7. Finding the 10th term a1 = 3 r = 2 n = 10 3, 6, 12, 24, 48, . . . an = a1·r(n-1) a10 = 3·(2)10-1 a10 = 3·(2)9 a10 = 3·(512) a10 = 1536 Jeff Bivin -- LZHS

  8. Finding the 8th term a1 = 2 r = -5 n = 8 2, -10, 50, -250, 1250, . . . an = a1·r(n-1) a8 = 2·(-5)8-1 a8 = 2·(-5)7 a8 = 2·(-78125) a8 = -156250 Jeff Bivin -- LZHS

  9. Sum it up Jeff Bivin -- LZHS

  10. 1 + 3 + 9 + 27 + 81 + 243 a1 = 1 r = 3 n = 6 Jeff Bivin -- LZHS

  11. 4 - 8 + 16 - 32 + 64 – 128 + 256 a1 = 4 r = -2 n = 7 Jeff Bivin -- LZHS

  12. Alternative Sum Formula We know that: Multiply by r: Simplify: Substitute: Jeff Bivin -- LZHS

  13. Find the sum of the geometric Series Jeff Bivin -- LZHS

  14. Evaluate = 2 + 4 + 8+…+1024 a1 = 2 r = 2 n = 10 an = 1024 Jeff Bivin -- LZHS

  15. Evaluate = 3 + 6 + 12 +…+ 384 a1 = 3 r = 2 n = 8 an = 384 Jeff Bivin -- LZHS

  16. Review -- Geometric Sum of n terms nth term an = a1·r(n-1) Jeff Bivin -- LZHS

  17. Geometric Infinite Series Jeff Bivin -- LZHS

  18. The Magic Flea(magnified for easier viewing) There is no flea like a Magic Flea Jeff Bivin -- LZHS

  19. The Magic Flea(magnified for easier viewing) Jeff Bivin -- LZHS

  20. Sum it up -- Infinity Jeff Bivin -- LZHS

  21. Remember --The Magic Flea Jeff Bivin -- LZHS

  22. Jeff Bivin -- LZHS

  23. A Bouncing Ball rebounds ½ of the distance from which it fell -- What is the total vertical distance that the ball traveled before coming to rest if it fell from the top of a 128 feet tall building? 128 ft 64 ft 32 ft 16 ft 8 ft Jeff Bivin -- LZHS

  24. A Bouncing Ball Downward = 128 + 64 + 32 + 16 + 8 + … 128 ft 64 ft 32 ft 16 ft 8 ft Jeff Bivin -- LZHS

  25. A Bouncing Ball Upward = 64 + 32 + 16 + 8 + … 128 ft 64 ft 32 ft 16 ft 8 ft Jeff Bivin -- LZHS

  26. A Bouncing Ball Downward = 128 + 64 + 32 + 16 + 8 + … = 256 Upward = 64 + 32 + 16 + 8 + … = 128 TOTAL = 384 ft. 128 ft 64 ft 32 ft 16 ft 8 ft Jeff Bivin -- LZHS

  27. A Bouncing Ball rebounds 3/5 of the distance from which it fell -- What is the total vertical distance that the ball traveled before coming to rest if it fell from the top of a 625 feet tall building? 625 ft 375 ft 225 ft 135 ft 81 ft Jeff Bivin -- LZHS

  28. A Bouncing Ball Downward = 625 + 375 + 225 + 135 + 81 + … 625 ft 375 ft 225 ft 135 ft 81 ft Jeff Bivin -- LZHS

  29. A Bouncing Ball Upward = 375 + 225 + 135 + 81 + … 625 ft 375 ft 225 ft 135 ft 81 ft Jeff Bivin -- LZHS

  30. A Bouncing Ball Downward = 625 + 375 + 225 + 135 + 81 + … = 1562.5 Upward = 375 + 225 + 135 + 81 + … = 937.5 TOTAL = 2500 ft. 625 ft 375 ft 225 ft 135 ft 81 ft Jeff Bivin -- LZHS

  31. Find the sum of the series Jeff Bivin -- LZHS

  32. Fractions - Decimals Jeff Bivin -- LZHS

  33. Let’s try again + + Jeff Bivin -- LZHS

  34. One more subtract Jeff Bivin -- LZHS

  35. OK now a series Jeff Bivin -- LZHS

  36. .9 = 1 .9 = 1 That’s All Folks Jeff Bivin -- LZHS

More Related