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# Section 5.7

Chapter 5. Section 5.7. Exercise #1. Find each. the first term the common difference the twelfth term the sum of the first 12 terms. 5 , 13 , 21 , 29 , 37 …. a). the first term. b). the common difference. 5. = 8. 13  5. 5 , 13 , 21 , 29 , 37 …. c). Télécharger la présentation ## Section 5.7

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1. Chapter 5 Section 5.7 Exercise #1

2. Find each. the first term the common difference the twelfth term the sum of the first 12 terms

3. 5, 13, 21, 29, 37… a) the first term b) the common difference 5 = 8 13  5

4. 5, 13, 21, 29, 37… c) the twelfth term an = a1 + (n  1)d a12 = 5+ (12  1)(8) = 5+ (11)(8) = 5+ 88 = 93

5. 5, 13, 21, 29, 37… d) the sum of the first 12 terms + = + = = = 588

6. Chapter 5 Section 5.7 Exercise #9

7. Find each. the first term the common ratio the twelfth term the sum of the first 12 terms

8. 4, 12, 36, 108, 324… a) the first term b) the common ratio 4 = 3

9. 4, 12, 36, 108, 324… c) the twelfth term an = a1rn  1 a12 = 4(3)12  1 = 4(311) = 708,588

10. 4, 12, 36, 108, 324… d) the sum of the first 12 terms = = = = 1,062,880

11. Chapter 5 Section 5.7 Exercise #19

12. 9, a1= 9 d= 3

13. a1= 9 d= 3 9, a2  a1 = d a2  (9) = 3 a2 + 9 = 3 a2 = 3 9 a2 = 12 12,

14. a1= 9 d= 3 a3  a2 = d 9, 12, a3  (12) = 3 a2 + 12 = 3 a2 = 3 12 a2 = 15 15,

15. a1= 9 d= 3 a4  a3 = d 9, 12, 15, a4  (15) = 3 a4 + 15 = 3 a4 = 3 15 a4 = 18 18,

16. a1= 9 d= 3 a5  a4 = d 9, 12, 15, 18, a5  (18) = 3 a5 + 18 = 3 a5 = 3 18 a5 = 21 21…

17. Chapter 5 Section 5.7 Exercise #23

18. 12, a1= 12 r= 2

19. a1= 12 r= 2 a2 = r a1 a2 = 2 12 a2 = 2 12 12, a1= 12 r= 2 = 24 24,

20. a1= 12 r= 2 a3 = r a2 a3 = 2 24 a3 = 2 24 12, 24, a1= 12 r= 2 = 48 48,

21. a1= 12 r= 2 a4 = r a3 a4 = 2 48 a4 = 2 48 12, 24, 48, a1= 12 r= 2 = 96 96,

22. a1= 12 r= 2 a5 = r a4 a5 = 2 96 a4 = 2 96 12, 24, 48, 96, a1= 12 r= 2 = 192 192…

23. Chapter 5 Section 5.7 Exercise #31

24. Determine whether the sequence is an arithmetic sequence or a geometric sequence.

25. 6, 2,2,6,10… 2  6 = 4 2  2 = 4 6 (2) = 4 10 (6) = 4 There is a common difference between successive terms so the sequence is arithmetic.

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