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Geometric Sequences

Geometric Sequences. ALGEBRA 1 LESSON 6-4. (For help, go to Lesson 5-6.). Find the common difference of each sequence. 1. 1, 3, 5, 7, ... 2. 19, 17, 15, 13, ... 3. 1.3, 0.1, –1.1, –2.3, ... 4. 18, 21.5, 25, 28.5, ... Use inductive reasoning to find the next two numbers in

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Geometric Sequences

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  1. Geometric Sequences ALGEBRA 1 LESSON 6-4 (For help, go to Lesson 5-6.) Find the common difference of each sequence. 1. 1, 3, 5, 7, ... 2. 19, 17, 15, 13, ... 3. 1.3, 0.1, –1.1, –2.3, ... 4. 18, 21.5, 25, 28.5, ... Use inductive reasoning to find the next two numbers in each pattern. 5. 2, 4, 8, 16, ... 6. 4, 12, 36, ... 7. 0.2, 0.4, 0.8, 1.6, ... 8. 200, 100, 50, 25, ... 6-4

  2. Geometric Sequences ALGEBRA 1 LESSON 6-4 1. 1, 3, 5, 7, ... 2. 19, 17, 15, 13, ... 7 – 5 = 2, 5 – 3 = 2, 3 – 1 = 2 13 – 15 = –2, 15 – 17 = –2, Common difference: 2 17 – 19 = –2 Common difference: –2 3. 1.3, 0.1, –1.1, –2.3, ... 4. 18, 21.5, 25, 28.5, ... –2.3 – (–1.1) = –1.2, –1.1 28.5 – 25 = 3.5, 25 – 21.5 = 3.5, – 0.1 = –1.2, 0.1 – 1.3 = –1.2 21.5 – 18 = 3.5 Common difference: –1.2 Common difference: 3.5 Solutions 6-4

  3. Geometric Sequences ALGEBRA 1 LESSON 6-4 5. 2, 4, 8, 16, ... 6. 4, 12, 36, ... 2(2) = 4, 4(2) = 8, 8(2) = 16, 4(3) = 12, 12(3) = 36, 16(2) = 32, 32(2) = 64 36(3) = 108, 108(3) = 324 Next two numbers: 32, 64 Next two numbers: 108, 324 7. 0.2, 0.4, 0.8, 1.6, ... 8. 200, 100, 50, 25, ... (0.2)2 = 0.4, 0.4(2) = 0.8, 0.8(2) = 200  2 = 100, 100  2 = 50, 1.6, 1.6(2) = 3.2, 3.2(2) = 6.4 50  2 = 25, 25  2 = 12.5, Next two numbers: 3.2, 6.4 12.5  2 = 6.25 Solutions (continued) 6-4

  4. 3 –15 75 –375 (–5) (–5) (–5) 3 2 3 4 3 8 1 2 1 2 1 2 b. 3, , , , ...    1 2 3 2 3 4 3 8 3 The common ratio is . Geometric Sequences ALGEBRA 1 LESSON 6-4 Find the common ratio of each sequence. a. 3, –15, 75, –375, . . . The common ratio is –5. 6-4

  5. 5 –10 20 –40 (–2) (–2) (–2) Geometric Sequences ALGEBRA 1 LESSON 6-4 Find the next three terms of the sequence 5, –10, 20, –40, . . . The common ratio is –2. The next three terms are –40(–2) = 80, 80(–2) = –160, and –160(–2) = 320. 6-4

  6. 62 54 18 6 1 3 1 3 1 3    Geometric Sequences ALGEBRA 1 LESSON 6-4 Determine whether each sequence is arithmetic or geometric. a. 162, 54, 18, 6, . . . The sequence has a common ratio. The sequence is geometric. 6-4

  7. 98 101 104 107 + 3 + 3 + 3 Geometric Sequences ALGEBRA 1 LESSON 6-4 (continued) b. 98, 101, 104, 107, . . . The sequence has a common difference. The sequence is arithmetic. 6-4

  8. Geometric Sequences ALGEBRA 1 LESSON 6-4 Find the first, fifth, and tenth terms of the sequence that has the rule A(n) = –3(2)n – 1. first term: A(1) = –3(2)1– 1 = –3(2)0 = –3(1) = –3 fifth term: A(5) = –3(2)5 – 1 = –3(2)4 = –3(16) = –48 tenth term: A(10) = –3(2)10 – 1 = –3(2)9 = –3(512) = –1536 6-4

  9. The first term is 2 meters, which is 200 cm. Draw a diagram to help understand the problem. Geometric Sequences ALGEBRA 1 LESSON 6-4 Suppose you drop a tennis ball from a height of 2 meters. On each bounce, the ball reaches a height that is 75% of its previous height. Write a rule for the height the ball reaches on each bounce. In centimeters, what height will the ball reach on its third bounce? 6-4

  10. A rule for the sequence is A(n) = 200 • 0.75n– 1. Use the sequence to find the height of the third bounce. A(n) = 200 • 0.75n – 1 Substitute 4 for n to find the height of the third bounce. A(4) = 200 • 0.754 – 1 = 200 • 0.753 Simplify exponents. = 200 • 0.421875 Evaluate powers. = 84.375 Simplify. Geometric Sequences ALGEBRA 1 LESSON 6-4 (continued) The ball drops from an initial height, for which there is no bounce. The initial height is 200 cm, when n = 1. The third bounce is n = 4. The common ratio is 75%, or 0.75. The height of the third bounce is 84.375 cm. 6-4

  11. 1 3 3, 1, Geometric Sequences ALGEBRA 1 LESSON 6-4 1. Find the common ratio of the geometric sequence –3, 6, –12, 24, . . . 2. Find the next three terms of the sequence 243, 81, 27, 9, . . . 3. Determine whether each sequence is arithmetic or geometric. a. 37, 34, 31, 28, . . . b. 8, –4, 2, –1, . . . 4. Find the first, fifth, and ninth terms of the sequence that has the rule A(n) = 4(5)n–1. 5. Suppose you enlarge a photograph that is 4 in. wide and 6 in. long so that its dimensions are 20% larger than its original size. Write a rule for the length of the copies. What will be the length if you enlarge the photograph five times? (Hint: The common ratio is not just 0.2. You must add 20% to 100%.) –2 arithmetic geometric 4, 2500, 1,562,500 A(n) = 6(1.2)n-1; about 14.9 in. 6-4

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