hvvrq $ulwkphwlf 6htxhqfhv dqg 6hulhv - NCPN

More documents

Recommendations

Info

$hvvrq 6\vwhpv ri &rqlf 6hfwlrqv - NCPN$

50 ∑ 3 2( −1) = 1 34 ∑3( + 2) ∑7 − 5 = 1 53 ∑8 −1 5 + 2 = 1 82 ∑ = 1 62 = 1 25 ∑ = 1 Reteaching 9.2 (CRB) NAME CLASS DATE AND SERIES RETEACHING 11.2 ARITHMETIC SEQUENCES An arithmetic sequence is a sequence in which the difference between consecutive terms is the same. The difference between consecutive terms is the common difference of the sequence. The arithmetic mean of two numbers is the average of the two numbers. The arithmetic mean can be used to find a missing term of an arithmetic sequence. A series is a mathematical expression for the sum of the terms of a sequence. A series can have a finite number of terms that can be counted individually such as 1 + 2 + 3, or it can have an infinite number of terms such as 2 + 4 + 6 + 8 + …. An arithmetic series is a series whose terms form an arithmetic sequence. The sum Sn of an arithmetic sequence with a finite number of terms, such as a1 + a2 + a3 + … + an, is given by the formula: Sn = 2 n (a1 + an), where a1 is the first term, an is the last term, and n is the number of terms. Summation notation is used to write the terms of a series in a compact way. 5 ∑ = 3(1) + 3(2) + 3(3) + 3(4) + 3(5) = 45 n= 1 3n Under the Σ summation symbol is the lower limit of the series and an index indicated by n = 1. Above the symbol is the upper limit, which is the greatest value of n. To the right of the symbol is an explicit formula for the terms of the sequence. EXERCISES Determine if each sequence is arithmetic. If so, state the common difference of the sequence. 1. 3, 7, 11, 15, 19, … yes; 4 2. 1, 3, 6, 10, 15, 21, … no Use the formula Sn = 2 n (a1 + an) to evaluate each series. 3. 12 ∑ 4. + 3n 2 −1 5n 156 25,150 2n 1,749 n= 1 100 ∑ 5. ∑ n= 1 Copyright © CORD 310 >Algebra 2 Chapter Resource Book 33 n= 1 Extra Practice 9.2 (CRB) Enrichment 9.2 (CRB) NAME CLASS DATE AND SERIES ENRICHMENT 11.2 ARITHMETIC SEQUENCES NAME CLASS DATE AND SERIES EXTRA PRACTICE 11.2 ARITHMETIC SEQUENCES Determine if each sequence is arithmetic. If so, state the common difference of the sequence. 1. –1.4, –1.3, –1.2, –1.1, … yes; 0.1 2. 22, 25, 28, 31, 34, … yes; 3 3. 7, 3, 0, –3, –6, … no 4. 4.2, 5, 5.8, 6.6, 7.4, … yes; 0.8 5. 30, 24, 18, 12, 6, … yes; –6 6. 0.1, 0.01, 0.001, 0.0001, … no 7. 1, 2, 4, 8, 16, 32, … no 8. –25, –14, –3, 8, 19, … yes; 11 Use the arithmetic mean to find the missing term in each arithmetic sequence. 9. 5, 11, 17, ⁪, 29, … 23 10. 24, ⁪, 16, 12, 8, … 20 11. 99, 90, ⁪, 72, 63, … 81 12. –0.4, 1.3, 3, ⁪, 6.4, … 4.7 13. 8, 13, 18, ⁪, 28, … 23 14. 56, ⁪, 24, 8, –8, … 40 15. 6, ⁪, –8, –15, –22, … –1 16. 4.3, 4.9, ⁪, 6.1, 6.7, … 5.5 Use the formula Sn = 2 n (a1 + an) to evaluate each series. 17. 20. 17 ∑ 18. n + 6 8n −3 3n 255 2,525 315 n= 1 21 n= 1 25 ∑ 19. ∑ n= 1 n + 3 n + 3 n 1 2 147 527 –1,225 ∑ 21. 19 ∑ 24. 31 ∑ 22. n= 1 Copyright © CORD n −5 23. n − 2 5n + 10 314 2 >Algebra 2 Chapter Resource Book 152 110.5 1,035 n= 1 26 n= 1 14 n= 1 50 ∑ − + n= 1 ∑ 25. ∑ 18 n= 1 312 >Algebra 2 Chapter Resource Book Copyright © CORD 9.2 Arithmetic Sequences and Series 401
Answers to Math Applications Math Applications for this chapter are on pages 422–429. The notes and solutions shown below accompany the suggested applications to assign with this lesson. 5. a. 500(190 + 7(0)) = 500(190) = 95,000 b. 500(190 + 7(6)) = 500(190 + 42) = 500(232) = 116,000 c. Year 2 = 500(190 + 7) = 500(197) = 98,500 Year 3 = 500(190 + 14) = 500(204) = 102,000 Year 4 = 500(190 + 21) = 500(211) = 105,500 Year 5 = 500(190 + 28) = 500(218) = 109,000 Year 6 = 500(190 + 35) = 500(225) = 112,500 95,000 + 98,500 + 102,000 + 105,500 + 109,000 + 112,500 + 116,000 = 738,500 738,500 > 735,800, so Yoshi can expect to produce enough Sheet of seaweed over the next 7 years to be profitable. 9. a. S n = n ( a + a n ) 2 1 b. S 9 = 9 10 66 2 ( + ) = 9(38) = 342 c. 335 = 9 ( 2 25 + a9 ) 335 = 112.5 + 4.5a 9 222.5 = 4.5a 9 49.4 = a 9 50 signatures d. The funding will increase because Alberto gathered more signatures. 402 Chapter 9 Sequences and Series 11. a. arithmetic sequence b. Each trial, total miles decreases by 39 so Yvonne’s mileage should be 388 – 39 = 299 miles for a 7th trial. c. S n = n 533 260 2 ( + ) 3,172 = n 2 ( 793) 3,172 = 396.5n 8 = n d. The number of mpg decrease by 3 miles in each trial. The number of mpg multiplied by the number of gallons the tank holds gives the total miles traveled. 5 ∑ n= 0 13( 41− 3n) 12. a. 22 – 13 = 9; 40 – 31 = 9 The common difference is 9. b. 4 + 9n c. 4 + 9(7) = 67
Page 1 and 2: LESSON OBJECTIVES 9.1 Patterns •
Page 3 and 4: LESSON PLANNING Vocabulary sequence
Page 5 and 6: INSTRUCTION Example 2 Return to the
Page 7 and 8: Answers to Math Applications Math A
Page 9 and 10: INSTRUCTION Tell students that the
Page 11: INSTRUCTION Ongoing Assessment Show
Page 15 and 16: INSTRUCTION Tell students that the
Page 17 and 18: Problem Solving Before solving this
Page 19 and 20: Answers to Math Applications Math A
Page 21 and 22: INSTRUCTION Students might incorrec
Page 23 and 24: WRAP UP To ensure mastery of object
Page 25 and 26: Mixed Review Additional Answers 35.
Page 27 and 28: INSTRUCTION Ask students to express
Page 29 and 30: WRAP UP To ensure mastery of object
Page 31 and 32: MATH LAB Activity 1 PREPARE • An
Page 33 and 34: Math Applications Solutions and Not
Page 41 and 42: Vocabulary Review arithmetic mean (
Page 43 and 44: 1 2 10 , 922 1 2 ∞ 1 ∑

hvvrq $ulwkphwlf 6htxhqfhv dqg 6hulhv - NCPN

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?