precal question

Question

Find the number of terms of the arithmetic sequence with the given description that must be added to get a value of 3360.

The first term is 5, and the common difference is 2.

Answer 1

TO CHECK:

n a(n) sum

1 5 5

2 7 12

3 9 21

4 11 32

5 13 45

6 15 60

7 17 77

8 19 96

9 21 117

10 23 140

11 25 165

12 27 192

13 29 221

14 31 252

15 33 285

16 35 320

17 37 357

18 39 396

19 41 437

20 43 480

21 45 525

22 47 572

23 49 621

24 51 672

25 53 725

26 55 780

27 57 837

28 59 896

29 61 957

30 63 1020

31 65 1085

32 67 1152

33 69 1221

34 71 1292

35 73 1365

36 75 1440

37 77 1517

38 79 1596

39 81 1677

40 83 1760

41 85 1845

42 87 1932

43 89 2021

44 91 2112

45 93 2205

46 95 2300

47 97 2397

48 99 2496

49 101 2597

50 103 2700

51 105 2805

52 107 2912

53 109 3021

54 111 3132

55 113 3245

56 115 3360

57 117 3477

58 119 3596

59 121 3717

60 123 3840

Answer 2

Hi Viviana,

The formula for the n^th term of an arithmetic sequence is: a_n = a₁ + (n - 1)*d.

The formula for the arithmetic sum is S_n = n(a₁ + a_n)/2.

Substituting the arithmetic sequence a_n into S_n:

S_n = n(a₁ + [a₁ + (n - 1)*d])/2

With a₁ = 5; d = 2; and S_n = 3360:

3360 = n(5 + [5 + (n - 1)*2])/2

3360 = n(10 + 2n - 2)/2

3360 = (2n² + 8n)/2

3360 = n² + 4n

n² + 4n - 3360 = 0, you can solve this.

n = 56 and - 60

So 56 terms must be added.

I hope this helps, Joe.