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5.7 Arithmetic and Geometric Sequences

5.7 Arithmetic and Geometric Sequences. An arithmetic sequence is a sequence of numbers that is defined by adding a constant term to get from one term to the next term. 3, 7, 11, 15, 19, … is an arithmetic sequence with first term ( a 1 ) 3, and the constant term ( d ) is 4.

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5.7 Arithmetic and Geometric Sequences

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  1. 5.7 Arithmetic and Geometric Sequences

  2. An arithmetic sequence is a sequence of numbers that is defined by adding a constant term to get from one term to the next term. 3, 7, 11, 15, 19, … is an arithmetic sequence with first term (a1) 3, and the constant term (d) is 4. a1 = 3, a2 = 7, a3 = 11, a4 = 15, … 7, 1, 5, 11,  17, … is an arithmetic sequence with first term (a1) 7, and the constant term (d) is 6. a1 = 7, a2 = 1, a3 =  5, a4 =  11, …

  3. 3, 7, 11, 15, 19, … is an arithmetic sequence with first term (a1) 3, and the constant term (d) is 4. a1 = 3 a2 = a1 + (1 × 4) a3 = a1 + (2× 4) a4 = a1 + (3 × 4) a5 = a1 + (4 × 4) a6 = a1 + (5 × 4) Find the thirty first term, a31 , of the sequence. a31 = 3 + (30 × 4) = 123 Thegeneral term for this sequence is an = 3 + (n - 1) × 4

  4. 7, 1, -5, -11, -17, … is an arithmetic sequence with first term (a1) 7, and the constant term (d) is -6. a1 = 7 Thegeneral term for any arithmetic sequence is an = a1 + (n – 1)d a2 = a1 + (1 × ( 6)) a3 = a1 + (2× ( 6)) a4 = a1 + (3 × ( 6)) a5 = a1 + (4 × ( 6)) a6 = a1 + (5 × ( 6)) Find the fifty first term, a51 , of the sequence. a51 = 7 + (50 × ( 6)) =  293 Thegeneral term for this sequence is an = 7 + (n1) × ( 6)

  5. Thegeneral term for any arithmetic sequence is an = a1 + (n – 1)d Find the sixty first term, a61 , of the sequence. 17, 14,  11,  8,  5, … Find the eighty second term, a82 , of the sequence. The common difference, d, can always be found by subtracting. an + 1 – an = d  5 – ( 8) = 3 a82 =  17 + 81(3) =  17 + 243 = 226

  6. Thegeneral term for any arithmetic sequence is an = a1 + (n – 1)d 31, 24, 17, 10, 3, … Find the sixty fifth term, a65 , of the sequence. a65 = 31+ 64( 7) = 31+ (-448) =  417 Find the one hundred and first term, a101 , of the sequence.

  7. A geometric sequence is a sequence of numbers that is defined by multiplying bya constant term to get from one term to the next term. 3, 6, 12, 24, 48, … is a geometric sequence with first term (a1) 3, and the constant term (r) is 2. a1 = 3, a2 = 6, a3 = 12, a4 = 24, … 5, -15, 45, -135, 405, … is a geometric sequence with first term (a1) 5, and the constant term (r) is -3. a1 = 5, a2 = -15, a3 = 45, a4 = -135, …

  8. 5, 10, 20, 40, 80, … is a geometric sequence with first term (a1) 5, and the constant term (r) is 2. a1 = 5 a2 = a1 (2)1 a3 = a1 (2)2 a4 = a1 (2)3 a5 = a1 (2)4 a6 = a1 (2)5 Find the twenty first term, a21 , of the sequence. a21 = 5 (2)20 = 5,242,880 Thegeneral term for this sequence is an = 5 (2)n - 1

  9. 2, -6, 18, -54, 162, … is a geometric sequence with first term (a1) 2, and the constant term (r) is -3. a1 = 2 Thegeneral term for any geometric sequence is an = a1 (r )n-1 a2 = a1 (-3) a3 = a1 (-3)2 a4 = a1 (-3)3 a5 = a1 (-3)4 a6 = a1 (-3)5 Find the fifteenth term, a15 , of the sequence. a15 = 2 (-3)14 = 9,565,938 Thegeneral term for this sequence is an= a1( 3)n-1

  10. Thegeneral term for any geometric sequence is an = a1 (rn – 1) Find the eleventh term, a11 , of the sequence. -3, -12, -48, -192, … Find the thirteenth term, a13 , of the sequence. The common ratio, r, can always be found by dividing. an + 1 ÷ an = r -12 ÷ (-3) = 4 a13 = -3 (4)12 = -50,331,648

  11. Thegeneral term for any geometric sequence is an = a1 (rn – 1) 4, 20, 100, 500, … Find the tenth term, a10 , of the sequence. r = 20÷ 4 = 5 a10 = 4(5)9 =7,812,500 Find the twenty first term, a21 , of the sequence.

  12. Comparing Arithmetic and Geometric Sequences For an arithmetic sequence of a1 = 3 and d = 5, the graph would look like the red line. For a geometric sequence of a1 = 3 and r = 5, the graph would look like the blue curve. Geometric sequences change much faster than arithmetic sequences.

  13. A baseball player is given a contract of $3,000,000 per year with a raise of $500,000 each year. How much will he earn in his fourteenth year of playing? Is the sequence arithmetic or geometric? a1= 3,000,000 d = 500,000 n = 14 a15= 3,000,000 + 13(500,000) = 9,500,000 A baseball player is given a contract of $3,000,000 per year with a raise of 10% each year. How much will he earn in his fourteenth year of playing? Is the sequence arithmetic or geometric? a1= 3,000,000 r= 1.1 n = 14 a15= 3,000,000 (1.1)13 = 10,356,813.64

  14. Two offers of employment are available. The work is the same and all of the benefits are the same. The only difference is the way raises are given. Opportunity A will give a $1000 raise for each six month period. Opportunity B will give an annual raise of $4000. Which opportunity will be most profitable in the long run? Assume that A will start with $25,000 for the first six months and B will start with $50,000 for the first year. During the first year A will pay $25,000 + ($25,000 + $1,000) =$51,000 During the first year B will pay $50,000 During the second year A will pay $27,000 + ($27,000 + $1,000) =$55,000 During the second year B will pay $54,000 During the third year A will pay $29,000 + ($29,000 + $1,000) =$59,000 During the third year B will pay $58,000

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