3 questions

Question

Given t₁ = -23, n = 25, and t_n = 57, find the sum of the arithmetic series.

	A.	125
	B.	250
	C.	325
	D.	425

 

Find a formula for t_n so that the first four terms are 15, 25, 35, 45.

	A.	tn = 15n
	B.	tn = 10n - 5
	C.	tn = 10n + 5
	D.	tn = 15n - 5

Find the first four terms of the given sequence t_n = (-1)ⁿ(n).

	A.	-1, -2, -3, -4
	B.	-1, 2, -3, 4
	C.	1, ½, 2/3, ¾
	D.	1, 2, 3, 4

 

Answer 1

1)

Given t₁ = -23, n = 25, and t_n = 57.

Find the common difference by using the formula:

t_n = t₁ + (n-1)d

Substitute -23 for t₁ , 57 for t_n , and 25 for n.

57 = -23 + (25-1) d

24d = 80

d = 80/24 =10/3

Find the sum of the n terms.

S_n = (n/2)(2t₁ + (n-1)d )

    = (25/2)(2(-23) + (25-1)(10/3) )

   = (25/2)(2(-23) + 80 )

   = (25/2)(34)

    = 425

So Option D is correct

Answer 2

2)

Given 15, 25, 35, 45.

Find the common difference by using the formula:

d =t₂ - t₁

Substitute 15 for t₁ and 25 for t₂.

d = 25 - 15 =10

Find the nth term by using the formula.

t_n = t₁ + (n-1)d

Substitute 15 for t₁ and 10 for d .

t_n = 15 + (n-1)(10)

   = 15 + 10n - 10

    = 5 + 10n

So Option C is correct

 

Answer 3

The given sequence t_n = (-1)ⁿ(n).

Find the first term of the sequence.

Substitute 1 for n in sequence.

t₁ = (-1)¹(1) = -1

Find the second term of the sequence.

Substitute 2 for n in sequence.

t₂ = (-1)²(2) = 2

Find the third term of the sequence.

Substitute 3 for n in sequence.

t₃ = (-1)³(3) = -3

Find the fourth term of the sequence.

Substitute 4 for n in sequence.

t₄ = (-1)⁴(4) = 4

So sequence is -1,2,-3, and 4

So Option B is correct.