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**Five-Minute Check (over Lesson 10-2)**Then/Now New Vocabulary Example 1: Geometric Sequences Key Concept: The nth Term of a Geometric Sequence Example 2: Explicit and Recursive Formulas Example 3: nth Terms Example 4: Real-World Example: nth Term of a Geometric Sequence Example 5: Geometric Means Key Concept: Sum of a Finite Geometric Series Example 6: Sums of Geometric Series Example 7: Geometric Sum in Sigma Notation Key Concept: The Sum of an Infinite Geometric Series Example 8: Real-World Example: Sums of Infinite Geometric Series Lesson Menu**Determine the common difference and find thenext four terms**of the arithmetic sequence5, –1, –7, –13, … . A.d = –6; –19, –25, –31, –37 B.d = –5; –18, –23, –28, –33 C.d = –5; –19, –25, –31, –37 D.d = 6; –19, –25, –31, –37 5–Minute Check 1**Write an explicit formula and a recursive formulafor finding**the nth term of the arithmeticsequence 2, 7, 12, 17, … . A.an= 5n – 3; a1= 2, an= an – 1+ 5 B.an= 5n; a1 = 2, an= an – 1 + 5 C.an= 5n + 2; a1 = 7, an= an – 1 + 5 D.an= 7n – 2; a1 = 7, an= an – 1 + 5 5–Minute Check 2**Find a1 for the arithmetic sequence for which a25= 42 and d**= –2. A.a1= 92 B.a1 = 90 C.a1 = 88 D.a1 = –8 5–Minute Check 3**Find n for the arithmetic sequence for which an= 127, a1=**83, and d = 4. A.n= 10 B.n = 11 C.n = 12 D.n = 13 5–Minute Check 4**Find .**A.101 B.959.5 C.1010 D.1080 5–Minute Check 5**You found terms and means of arithmetic sequences and sums**of arithmetic series. (Lesson 10-2) • Find nth terms and geometric means of geometric sequences. • Find sums of n terms of geometric series and sums of infinite geometric series. Then/Now**geometric sequence**• common ratio • geometric means • geometric series Vocabulary**Geometric Sequences**A. Determine the common ratio and find the next three terms of the geometric sequence –6, 9, –13.5, … . First, find the common ratio. a2 ÷ a1 = 9 ÷ (–6) or –1.5 Find the ratio between two pairs of consecutive terms to determine the common ratio. a3 ÷ a2 = –13.5 ÷ 9 or –1.5 The common ratio is –1.5. Multiply the third term by –1.5 to find the fourth term, and so on. Example 1**Geometric Sequences**a4 = (–13.5)(–1.5) or 20.25 a5 = 20.25(–1.5) or –30.375 a6 = (–30.375)(–1.5) or 45.5625 The next three terms are 20.25, –30.375, and 45.5625. Answer:–1.5; 20.25, –30.375, 45.5625 Example 1**a2 = –81n + 243**a1 = 243n – 729 Factor. Simplify. Geometric Sequences B. Determine the common ratio and find the next three terms of the geometric sequence 243n – 729, –81n + 243, 27n – 81, … . First, find the common ratio. Example 1**a3 = 27n – 81**a2 = –81n + 243 Factor. Simplify. The common ratio is . Multiply the third term by to find the fourth term, and so on. Geometric Sequences Example 1**Answer: ; –9n + 27, 3n – 9, –n + 3**Geometric Sequences The next three terms are –9n + 27, 3n – 9, and –n + 3. Example 1**Determine the common ratio and find the next three terms of**the geometric sequence 24, 84, 294, … . A. –3; –882, 2646, –7938 B. 3.5; 297.5, 301, 304.5 C. 3.5; 1029, 3601.5, 12,605.25 D. 60; 354, 414, 474 Example 1**Explicit and Recursive Formulas**Write an explicit formula and a recursive formula for finding the nth term in the sequence –1, 2, –4, … . First, find a common ratio. a2 ÷ a1 = 2 ÷ (–1) or –2 Find the ratio between two pairs of consecutive terms to determine the common ratio. a3 ÷ a2 = –4 ÷ 2 or –2 The common ratio is –2. Example 2**Explicit and Recursive Formulas**For an explicit formula, substitute a1 = –1 and r = –2 in the nth term formula. an = a1rn – 1 nth term formula an=–1(–2)n –1a1 = –1 and r = –2 The explicit formula is an= –1(–2)n –1. For a recursive formula, state the first term a1. Then indicate that the next term is the product of the previous term an – 1 and r. a1 = –1; an= (–2) an – 1 The recursive formula is a1 = –1; an= (–2) an – 1. Answer:an= –1(–2)n –1; a1 = –1; an= (–2) an – 1 Example 2**Find both an explicit formula and a recursive formula for**the nth term of the geometric sequence 3, 16.5, 90.75, … . A. an = 3(5.5)n – 1; a1 = 3, an = 5.5an – 1 B. an = 5.5(3)n – 1; a1 = 3, an = 5.5an – 1 C. an = 3(5.5)n – 1; a1 = 3, an = 5an – 1 D. an = –2.5 + 5.5n ; a1 = 3, an = 5.5an – 1 Example 2**nth Terms**Find the 11th term of the geometric sequence –122, 115.9, –110.105, … . First, find the common ratio. a2 ÷ a1 = 115.9 ÷ (–122) or –0.95 Find the ratio between two pairs of consecutive terms to determine the common ratio. a3 ÷ a2 = –110.105 ÷ 115.9 or –0.95 Example 3**nth Terms**Use the formula for the nth term of a geometric sequence. an= a1rn– 1 nth term of a geometric sequence a11= –122(–0.95)11– 1n = 11, a1 = –122, and r = –0.95 a11 ≈ 73.05 Simplify. The 11th term of the sequence is about –73.05. Answer:about –73.05 Example 3**Find the 25th term of the geometric sequence 324, 291.6,**262.44, …. A. about 20.93 B. about 23.26 C. about 25.84 D. about 28.72 Example 3**nth Term of a Geometric Sequence**A. REAL ESTATE A couple purchased a home for $225,000. At the end of each year, the value of the home appreciates 3%. Write an explicit formula for finding the value of the home after n years. If the house’s value appreciates at a rate of 3% per year, it will have a value of 100% + 3% or 103% of its original value. Note that the original value given represents the a0 and not the a1 term, so we need to use an adjusted formula for the nth term of this geometric sequence. Example 4**nth Term of a Geometric Sequence**first term a1 = a0r second term a2 = a0r2 nth term an = a0rn Use this adjusted formula to find an explicit formula for the value of the house after n years. an= a0rn Adjusted nth term of a geometric sequence an= 225,000(1.03)na0 = 225,000, r = 1.03 Example 4**nth Term of a Geometric Sequence**An explicit formula for the value of the home after n years is an = 225,000(1.03)n. Answer:an = 225,000(1.03)n Example 4**nth Term of a Geometric Sequence**B. REAL ESTATE A couple purchased a home for $225,000. At the end of each year, the value of the home appreciates 3%. What is the value of the home after the tenth year? Evaluate the formula you found in part A with n = 10. an = 225,000(1.03)n Original formula = 225,000(1.03)10n = 10 ≈ 302,381.19 Simplify. Example 4**nth Term of a Geometric Sequence**The value of the home after the tenth year is about $302,381.19. The value of the home at each year is shown as a point on the graph. The function connecting the points represents exponential growth. Answer:about $302,381.19 Example 4**A. BOAT Jeremy purchased a boat for $12,500. By the end of**each year, the value of the boat depreciates 4%. Write an explicit formula for finding the value of the boat after n years. A. an = 12,500(1.04)n B. an = 12,500(0.96)n C. an = 12,500(0.96)n –1 D. an = 12,000(0.96)n Example 4**B.BOAT Jeremy purchased a boat for $12,500. At the end of**each year, the value of the boat depreciates 4%. What is the value of the boat after 12 years? A. $20,012.90 B. $7658.87 C. $7977.99 D. $7352.52 Example 4**?**? The sequence will resemble 264, _____, _____, _____, 1.03125. ? Geometric Means Write a sequence that has three geometric means between 264 and 1.03125. Note that a1 = 264, n = 5, and a5 = 1.03125. Find the common ratio using the nth term for a geometric sequence formula. a5 = a1rn– 1nth term of a geometric sequence 1.03125= 264r5–1a5 = 1.03125, a1 = 264, and n = 5 Example 5**= r4 Simplify and divide each side by 264.**± = r Take the fourth root of each side. The common ratio is ± . Use r to find the geometric means. Geometric Means Example 5**Geometric Means**The sequence is 264, 66, 16.5, 4.125, and 1.03125 or 264, 66, 16.5, 4.125, and 1.03125. Answer:264, 66, 16.5, 4.125, 1.03125 or 264, 66, 16.5, 4.125, 1.03125 Example 5**Write a geometric sequence that has two geometric means**between 20 and 8.4375. A. 20, 15, 11.25, 8.4375 B. 20, 16.15, 12.296, 8.4375 C. 20, 14.2188, 8.4375 D. 15, 11.25 Example 5**Sums of Geometric Series**A. Find the sum of the first eleven terms of the geometric series 4, –6, 9, … . First, find the common ratio. a2 ÷ a1 = –6 ÷ 4 or –1.5 Find the ratio between two pairs of consecutive terms to determine the common ratio. a3 ÷ a2 = 9 ÷ (–6) or –1.5 The common ratio is –1.5. Use Formula 1 to find the sum of the series. Example 6**Sum of a geometric series formula 1.**n = 11, a1 = 4, and r = –1.5 Simplify. Sums of Geometric Series The sum of the geometric series is about 140. Answer:about 140 Example 6**Sum of a geometric series formula 2.**a1 = –4, an= –65,536, and r = 2 Simplify. Sums of Geometric Series B. Find the sum of the first n terms of a geometric series with a1 = –4, an= –65,536, and r = 2. Use Formula 2 for the sum of a finite geometric series. Example 6**Sums of Geometric Series**The sum of the first n terms of the geometric series is –131,068. Answer:–131,068 Example 6**Find the sum of the first 8 terms of the geometric series 8**+ 36 + 162 + … . A. 18,977.75 B. 85,407.875 C. 384,343.4375 D. 2,113,888.906 Example 6**Find .**Geometric Sum in Sigma Notation Method 1 Use Formula 1. Find n, a1, and r. n = 8 – 3 + 1 or 6 Upper bound minus lower bound plus 1 a1 = –2(–2)3–1 or –8 n = 3 r = –2 r is the base of the exponential function. Substitute n = 6, a1 = 8, and r = 2 into Formula 1. Example 7**Formula 1**n = 6, a1 = –8, and r = –2 Simplify. Geometric Sum in Sigma Notation Method 2 Use Formula 2. Find an. an= a1rn 1nth term of a geometric sequence Example 7**Formula 2**n = 6, a1 = 8, an= 256, and r = 2 Simplify. Therefore, = 168. Geometric Sum in Sigma Notation = –8(–2)6–1 a1 = 8, r = 2, and n = 6 = 256 Simplify. Substitute a1 = 8, an = 256, and r = 2 into Formula 2. Answer:168 Example 7**Find .**A. 10,912 B. 32,768 C. 43,680 D. 174,752 Example 7**a2 ÷ a1 = 18 ÷ 24 or . Find the ratio between two**pairs of consecutive terms to determine the common ratio. The common ratio r is , and . This infinite geometric series has a sum. Use the formula for the sum of an infinite geometric series. a3 ÷ a2 = 13.5 ÷ 18 or Sums of Infinite Geometric Series A. If possible, find the sum of the infinite geometric series 24 + 18 + 13.5 + … . First, find the common ratio. Example 8**Sum of an infinite geometric series formula**a1 = 9 and r = Simplify. Sums of Infinite Geometric Series The sum of the infinite series is 96. Answer:96 Example 8**Sums of Infinite Geometric Series**B. If possible, find the sum of the infinite geometric series 0.33 + 0.66 + 1.32 + … . First, find the common ratio. a2 ÷ a1 = 0.66 ÷ 0.33 or 2 Find the ratio between two pairs of consecutive terms to determine the common ratio. a3 ÷ a2 = 1.32 ÷ 0.66 or 2 The common ratio r is 2, and |2| > 1. Therefore, this infinite geometric series has no sum. Answer:does not exist Example 8**C. If possible, find .**Sums of Infinite Geometric Series The common ratio r is 0.65, and |0.65| < 1. Therefore, this infinite geometric series has a sum. Find a1. a1 = 3(0.65)2 – 1 Lower bound = 2 = 1.95 Use the formula for the sum of an infinite geometric series to find the sum. Example 8**Sum of an infinite geometric series formula**a1 = 1.95 and r = 0.65 Simplify. Answer: Sums of Infinite Geometric Series Example 8**If possible, find the sum of the infinite geometric series**. A. 1.5 B. 7.5 C. 30 D. does not exist Example 8

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