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In this guide, we explore the evaluation of recursive rules and derive the first six terms of various arithmetic and geometric sequences. The sequences presented include specific initial values, and we will analyze their corresponding recursive relationships. For instance, given an initial value and common difference or ratio, we will demonstrate how to calculate subsequent terms step by step. This guide aims to enhance understanding of how to establish recursive formulas and calculate terms systematically.
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EXAMPLE 1 Evaluate recursive rules Write the first six terms of the sequence. a. a0 = 1, an= an – 1 + 4 b. a1 = 1, an= 3an – 1 SOLUTION a. a0 = 1 b. a1 = 1 a1 = a0 + 4 = 1 + 4 = 5 a2 = 3a1 = 3(1) = 3 a2 = a1 + 4 = 5 + 4 = 9 a3 = 3a2 = 3(3) = 9 a3 = a2 + 4 = 9 + 4 = 13 a4 = 3a3 = 3(9) = 27 a5 = 3a4 = 3(27) = 81 a4 = a3 + 4 = 13 + 4 = 17 a5 = a4 + 4 = 17 + 4 = 21 a6 = 3a5 = 3(81) = 243
ANSWER So, a recursive rule for the sequence isa1 = 3, an= an– 1 + 10. EXAMPLE 2 Write recursive rules Write the first six terms of the sequence. a. 3, 13, 23, 33, 43, . . . b. 16, 40, 100, 250, 625, . . . SOLUTION The sequence is arithmetic with first term a1 = 3 and common difference d = 13 – 3 = 10. an= an – 1 + d General recursive equation for an = an – 1 + 10 Substitute 10 for d.
b. The sequence is geometric with first term a1 = 16 and common ratio r = = 2.5. an= ran– 1 40 16 ANSWER So, a recursive rule for the sequence is a1 = 16,an= 2.5an – 1. EXAMPLE 2 Write recursive rules General recursive equation for an = 2.5an – 1 Substitute 2.5 for r.
– 1. a1 = 3, an= an – 1 7 – – – a2 = a1 7 = 3 7 = 4 ANSWER 3, –4, –11, –18, –25 for Examples 1 and 2 GUIDED PRACTICE Write the first five terms of the sequence. SOLUTION a1 = 3 a3 = a2– 7 a3 = – 4 – 7 = – 11 a4 = a3– 7 = – 11 – 7 = – 18 a5 = a4– 7 = – 18 – 7 = – 25
ANSWER 162, 81, 40.5, 20.25, 10.125 for Examples 1 and 2 GUIDED PRACTICE Write the first five terms of the sequence. 2. a0 = 162, an= 0.5an – 1 SOLUTION a0 = 162 a1 = 0.5a0 = 0.5 (162) = 81 a2 = 0.5a1 = 0.5 (81) = 40.5 a3 = 0.5a2 = 0.5 (40.5) = 20.25 a4 = 0.5a3 = 0.5 (20.25) = 10.125
ANSWER 1, 2, 4, 7, 11 for Examples 1 and 2 GUIDED PRACTICE Write the first five terms of the sequence. 3. a0 = 1, an= an – 1 + n SOLUTION a0 = 1 a1 = a0+ 1 = 1 + 1 = 2 a2 = a1+ 1 = 2+ 2 = 4 a3 = a2+ 3 = 4 + 3 = 7 a4 = a3+ 4 = 7 + 4 = 11
a2 = 2a1– 1 = (2 4) – 1 = 8 – 1 = 7 a3 = 2a2– 1 = (2 7) – 1 = 14 – 1 = 13 a4 = 2a3– 1 = (2 13) – 1 = 26 – 1 = 25 ANSWER 4, 7, 13 25, 49 a5 = 2a4– 1 = (2 25) – 1 = 49 for Examples 1 and 2 GUIDED PRACTICE Write the first five terms of the sequence. 4. a1 = 4, an= 2an – 1– 1 SOLUTION a1 = 4
an= ran– 1 a2 r = = 7 a1 ANSWER So, a recursive rule for the sequence is a1 = 2, an= 7an – 1 for Examples 1 and 2 GUIDED PRACTICE Write a recursive rule for the sequence. 5. 2, 14, 98, 686, 4802, . . . SOLUTION The sequence is geometric with first term a1 = 2 and common ratio = 7 ·an – 1
ANSWER So, a recursive rule for the sequence is a1 = 19, and an= an – 1 – 6. for Examples 1 and 2 GUIDED PRACTICE Write a recursive rule for the sequence. 6. 19, 13, 7, 1, – 5, . . . SOLUTION The sequence is arithmetic with first term a = 19 and common difference a2– a1= – 6. an= an – 1 + d = an – 1 + (– 6)
ANSWER So, a recursive rule for the sequence is a1 = 11, and an= an – 1 + 11. for Examples 1 and 2 GUIDED PRACTICE Write a recursive rule for the sequence. 7. 11, 22, 33, 44, 55, . . . SOLUTION The sequence is arithmetic with first term a = 11 and common difference a2– a1= 11. an= an – 1 + d = an – 1 + 11
1 a2 108 = 3 = a1 324 a = 324 = an= ran– 1 ANSWER 1 So, a recursive rule for the sequence is an = an – 1 3 1 a = 324, and an= an – 1 3 for Examples 1 and 2 GUIDED PRACTICE Write a recursive rule for the sequence. 8. 324, 108, 36, 12, 4, . . . SOLUTION The sequence is geometric with first term and common ratio
