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Supplemental Material. Arithmetic Sequences. Motivating Examples If you make $27,000 and get $1,800 raises each year, what will your salary be in 10 years?

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  1. Supplemental Material Arithmetic Sequences

  2. Motivating Examples • If you make $27,000 and get $1,800 raises each year, what will your salary be in 10 years? • A new company loses $2,000 in its first month, but its profit increases by $400 in each of the following months for the next year. What is its profit for the year? Problems like these can be solved using arithmetic sequences.

  3. Introductory Example • Suppose you invest $2,000 at 3% simple interest every year for 5 years. Year Interest (I = PRT) Total Amount 1 $2,000(0.03)(1) $2,000 + $60 = $2,060 2 $2,000(0.03)(1) $2,060 + $60 = $2,120 3 $2,000(0.03)(1) $2,120 + $60 = $2,180 4 $2,000(0.03)(1) $2,180 + $60 = $2,240 5 $2,000(0.03)(1) $2,240 + $60 = $2,300

  4. Introductory Example Continued The sequence of numbers $2,000, $2,060, $2,120, $2,180, $2,240, $2,300 is called an arithmetic sequence. Each number, or term, is $60 more than the previous term. An arithmetic sequence is a sequence of numbers, called terms, such that any two consecutive numbers in the sequence are separated by a fixed common number or difference. The common difference is equal to d. (d=$60 in the above sequence)

  5. Arithmetic Sequence Examples • Find the common difference in the sequence 5, 8, 11, 14, . . . d = 3 • Find the common difference in the sequence 24, 20, 16, . . . d = -4 • Find the next 3 terms of the sequence 1, 3, 5, . . . 7, 9, 11 • Find the next 3 terms of the sequence 9, 6, 3, . . . 0, -3, -6

  6. Generalized Way to Write an Arithmetic Sequence • An arithmetic sequence can be written as a1, a2, a3, a4, + . . . a1, a1 + d, a1 +2d, a1 +3d, . . . a2 = a1 +d a3 = a1 +2d a4 = a1 + 3d • The nth term of an arithmetic sequence is an = a1 + (n-1)d a1 is the first term in the sequence d is the common difference

  7. Examples Finding the nth Term • Find the nth term of the sequence 9, 16, 23, 30, 37, . . . d = 7 an = a1 + (n-1)d an = 9 + (n-1)(7) an = 9+ 7n -7 an = 2 + 7n • Find the 20th term of the sequence 9, 16, 23, 30, 37, . . . an = 2 + 7n a20 = 2 + 7x20 = 2 + 140 = 142

  8. Discrete and Continuous Functions • A sequence is a discrete set of points. This discrete set of points forms a line if you connect the points. This line can be represented as a linear function y = mx +b m = slope and b = y-intercept an n

  9. Discrete and Continuous Functions Continued • Line is y = mx +b m = slope = change in height/change in width = d/1 = d b = y-intercept = a1 – d • In the sequence 9, 16, 23, 30, 37 we found that an = 2 + 7n The line that corresponds to this sequence is y = mx +b m = d = 7 b = a1 – d = 9 – 7 = 2. The line becomes y = 7x + 2 Notice that is looks very similar to an = 2 + 7n

  10. Application Example 1 • If you make $27,000 and get $1,800 raises each year, what will your salary be in 10 years? • Solution Year 1 : $27,000 Year 2 : $27,000 + $1,800 Year 3: $27,000 + 2 x $1,800 Find the 10th term in the arithmetic sequence a1 = $27,000 d = $1,800 an = a1 + (n-1)d = $27,000 + (n-1) x $1,800 a10 = $27,000 + (10-1) x $1,800 = $43,200

  11. Application Example 1 Continued • Let’s look at the corresponding linear equation y = mx + b (y = salary, x = years) m = d = $1,800 b = a1 – d = $27,000 - $1,800 = $25,200 y = 1,800 x + 25,200 • The graph of y = mx +b enables us to make future projections of the salary beyond 10 years. salary years

  12. Application Example 2 • If you make $27,000 and get $1,800 raises each year, in how many years will your salary double? • Solution Year 1 : $27,000 Year 2 : $27,000 +$1,800 Year 3: $27,000 + 2 x $1,800 Value of doubled salary = $27,000 + $27,000 = $54,000 We need to find when the nth term is $54,000. an = a1 + (n-1)d $54,000 = $27,000 +(n-1)$1,800 $27,000 = (n-1)$1,800 15 = n-1 n = 16 (Your salary will double in 16 years)

  13. Application Example 3 • A new company loses $2,000 in its first month, but its profit increases by $400 in each of the following months for the next year. What is its profit in the twelfth month? • Solution Month 1 : -$2,000 Month 2 : -$2,000 + $400 Month 3: -$2,000 + 2 x $400 Find the 12th term in the arithmetic sequence a1 = -$2,000 d = $400 an = a1 + (n-1)d = -$2,000 + (n-1) x $400 a12 = -$2,000 + (12-1) x $400 = $2,400

  14. Application Example 3 Continued • Let’s look at the corresponding linear equation y = mx + b (y = profit, x = months) m = d = $400 b = a1 – d = - $2,000 - $400 = - $2,400 y = 400 x – 2,400 • The graph of y = mx +b enables us to see when the company started making a profit and what the profit is for each month. profit months

  15. Sum of the First n Terms of an Arithmetic Sequence • The first n terms of an arithmetic sequence can be written from a1 to an as a1, a1 + d, a1 +2d, a1 +3d, . . . , a1 + (n-1)d • The first n terms of an arithmetic sequence can be written backward from an to a1 an, an - d, an - 2d, an - 3d, . . . , an - (n-1)d • We let Sn represent the sum of the first n terms on the sequence (1) Sn = a1 + (a1 + d) + (a1 +2d) + (a1 +3d) + . . . + (a1 + (n-1)d) or (2) Sn = an + (an – d) + (an - 2d) + (an - 3d) + . . . + (an - (n-1)d)

  16. Sum of the First n Terms of an Arithmetic Sequence Cont (1) Sn = a1 + (a1 + d) + (a1 +2d) + (a1 +3d) + . . . + (a1 + (n-1)d) or (2) Sn = an + (an – d) + (an - 2d) + (an - 3d) + . . . + (an - (n-1)d) If we add equations (1) and (2) together, we obtain 2Sn = (a1 + an) + (a1 + an) + (a1 + an) + . . . + (a1 + an) 2Sn = n(a1 + an) Sn = (n/2)(a1 + an) The sum of the first n terms of an arithmetic sequence is Sn = (n/2)(a1 + an) where a1 is the first term and an is the nth term.

  17. Application Example 4 • A new company loses $2,000 in its first month, but its profit increases by $400 in each of the following months for the next year. What is its profit for the year? • Solution Month 1 : -$2,000 Month 2 : -$2,000 + 400 Month 3: -$2,000 + 2 x $400 Find the sum of the first 12 terms in the arithmetic sequence Sn = (n/2)(a1 + an) S12 = (12/2)(a1 + a12) a1 = -$2,000 and a12 = $2,400 S12 = (12/2)(-$2,000 + $2,400) = 6 x $400 = $2,400

  18. Application Example 5 • In buying a house from their relatives, a couple agrees to pay $3,000 at the end of the first year, $3,500 at the end of the second year, $4,000 at the end of the third year, and so on. How much do they pay for the house if they make 20 payments in all? • Solution Year 1 : $3,000 Year 2 : $3,000 + 500 Year 3: $3,000 + 2 x $500 Find the sum of the first 20 terms in the arithmetic sequence Sn = (n/2)(a1 + an) S20 = (20/2)(a1 + a20) a1 = $3,000 and a20 = ?

  19. Application Example 5 Continued • Solution Find a20 an = a1 + (n-1)d = $3,000 + (n-1) x $500 a20 = $3,000 + (20-1) x $500 = $12,500 S20 = (20/2)(a1 + a20) S20 = (20/2)($3,000 + $12,500) S20 = $155,000 They pay $155,000 for their house.

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