5.2 arithmetic sequences and sums t
- 1. Example B. Given the sequence 2, 5, 8, 11, … a. Verify it is an arithmetic sequence. It's arithmetic because 5 – 2 = 8 – 5 = 11 – 8 = … = 3 = d. b. Find the (specific) formula that represents this sequence. Plug a1 = 2 and d = 3, into the general formula an = d(n – 1) + a1 we get an = 3(n – 1) + 2 an = 3n – 3 + 2 an = 3n – 1 the specific formula. c. Find a1000. Set n = 1000 in the specific formula, we get a1000 = 3(1000) – 1 = 2999. Arithmetic Sequences Use the arithmetic sequences general formula an = d(n – 1) + a1 to find specific formulas for sequences.
- 2. Arithmetic Sequences Find the first term a1 and the difference d to use the general formula to find the specific formula,. Set d = –4 in the general formula an = d(n – 1) + a1, we get an = –4(n – 1) + a1. Set n = 6 in this formula, we get a6 = –4(6 – 1) + a1 = 5 –20 + a1 = 5 a1 = 25 To find the specific formula , set 25 for a1 in an = –4(n – 1) + a1 an = –4(n – 1) + 25 an = –4n + 4 + 25 an = –4n + 29 Example C. Given a1, a2 , a3 , …an arithmetic sequence with d = -4 and a6 = 5, find a1, the specific formula and a1000. To find a1000, set n = 1000 in the specific formula a1000 = –4(1000) + 29 = –3971
- 3. Example D. Given that a1, a2 , a3 , …is an arithmetic sequence with a3 = –3 and a9 = 39, find d, a1 and the specific formula. Set n = 3 and n = 9 in the general arithmetic formula an = d(n – 1) + a1, we get a3 = d(3 – 1) + a1 = –3 2d + a1 = –3 Subtract these equations: 8d + a1 = 39 ) 2d + a1 = -3 6d = 42 d = 7 Put d = 7 into 2d + a1 = -3, 2(7) + a1 = -3 14 + a1 = -3 a1 = – 17 Hence the specific formula is an = 7(n – 1) – 17 or an = 7n – 24. Arithmetic Sequences a9 = d(9 – 1) + a1 = 39 8d + a1 = 39
- 4. Given that a1, a2 , a3 , …an an arithmetic sequence, then a1+ a2 + a3 + … + an = n TailHead + 2( ) ana1 + 2( )= n Example E. a. Given the arithmetic sequence a1= 4, 7, 10, … , and an = 67. What is n? We need the specific formula. Find d = 7 – 4 = 3. Therefore the specific formula is an = 3(n – 1) + 4 an = 3n + 1. Sums of Arithmetic Sequences If an = 67 = 3n + 1, then 66 = 3n or 22 = n n terms b. Find the sum 4 + 7 + 10 +…+ 67 a1 = 4, and a22 = 67 with n = 22, so the sum 4 + 7 + 10 +…+ 67 = 22 = 11(71) = 781 4 + 67 2 ( )11
- 5. Sums of Arithmetic Sequences Example F. a. How many bricks are there as shown if there are 100 layers of bricks continuing in the same pattern? The 1st layer has 3 = 1 x 3 bricks the 2nd layer has 6 = 2 x 3 bricks, etc.., hence the 100th layer has 100 x 3 = 300 bricks. 3 + 300 2( ) The sum 3 + 6 + 9 + .. + 300 is arithmetic. Hence the total number of bricks is 100 = 50 x 303 = 15150
- 6. Arithmetic Sequences 2. –2, –5, –8, –11,..1. 2, 5, 8, 11,.. 4. –12, –5, 2, 9,..3. 6, 2, –2, –6,.. 6. 23, 37, 51,..5. –12, –25, –38,.. 8. –17, .., a7 = 13, ..7. 18, .., a4 = –12, .. 10. a12 = 43, d = 59. a4 = –12, d = 6 12. a42 = 125, d = –511. a8 = 21.3, d = –0.4 14. a22 = 25, a42 = 12513. a6 = 21, a17 = 54 16. a17 = 25, a42 = 12515. a3 = –4, a17 = –11, Exercise A. For each arithmetic sequence below a. find the first term a1 and the difference d b. find a specific formula for an and a100 c. find the sum ann=1 100
- 7. B. For each sum below, find the specific formula of the terms, write the sum in the notation, then find the sum. 1. – 4 – 1 + 2 +…+ 302 Sum of Arithmetic Sequences 2. – 4 – 9 – 14 … – 1999 3. 27 + 24 + 21 … – 1992 4. 3 + 9 + 15 … + 111,111,111 5. We see that it’s possible to add infinitely many numbers and obtain a finite sum. For example ½ + ¼ + 1/8 + 1/16... = 1. Give a reason why the sum of infinitely many terms of an arithmetic sequence is never finite, except for 0 + 0 + 0 + 0..= 0.
- 8. Arithmetic Sequences 1. a1 = 2 d = 3 an = 3n – 1 a100 = 299 an = 15 050 (Answers to the odd problems) Exercise A. n=1 100 3. a1 = 6 d = – 4 an = – 4n + 10 a100 = – 390 an = – 19 200n=1 100 5. a1 = – 12 d = –13 an = – 13n + 1 a100 = – 129 an = – 65 550n=1 100 7. a1 = 18 d = – 10 an = – 10n +28 a100 = – 972 an = – 47 700n=1 100 9. a1 = –30 d = 6 an = 6n – 36 a100 = 564 an = 26 700n=1 100 11. a1 = 24.1 d = –0.4 an = –0.4n + 24.5 a100 = –15.5 an = 430n=1 100
- 9. Arithmetic Sequences 13. a1 = 6 d = 3 an = 3n + 3 a100 = 303 an = 15 450n=1 100 15. a1 = –3 d = – 0.5 an = – 0.5n – 2.5 a100 = –52.5 an = –2 775n=1 100 Exercise B. 1. – 4 – 1 + 2 +…+ 302 = 3n – 7 = 15 347 3. 27 + 24 + 21 … – 1992 = –3n + 30 = –662 205 n=1 103 n=1 674