Find an equation for the nth term of the arithmetic sequence.

-15, -6, 3, 12,

an = -15 + 9(n + 1)

an = -15 x 9(n - 1)

an = -15 + 9(n + 2)

an = -15 + 9(n - 1)

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Find an equation for the nth term of the arithmetic sequence.

-15, -6, 3, 12,

an = -15 + 9(n + 1)

an = -15 x 9(n - 1)

an = -15 + 9(n + 2)

an = -15 + 9(n - 1)

These are 14 questions and 14 answers.

Since this exceeds the limit and I had to delete the last questions and I copied all the answers to a file that is attache. See the attachment with all the answers.

1) Question 1. Find the first six terms of the sequence: a1 = -6, an = 4 • an-1

option D) -6, -24, -96, -384, -1536, -6144

Explanation:

A(1) = - 6

A(n) = 4 * A(n-1)

n A(n)

1 - 6

2 4 * (-6) = - 24

3 4 * (-24) = - 96

4 4 * (-96) = - 384

5 4 * (-384) = - 1536

6 4 * ( -1536) = -6144

So, the first six terms are: -6, - 24, - 96, - 384, - 1536, - 6144.

2) Question 2: Find an equation for the nth term of the arithmetic sequence.

-15, -6, 3, 12, ...

option D) - 15 + 9(n - 1)

Explanation:

1. find the difference between the consecutive terms:

-6 - (-15) = -6 + 15 = 9

3 - (-6) = 3 + 6 = 9

12 - 3 = 9

So, the difference is 9, and you can find any term adding 9 to the previous.

2. Since the first term is - 15, you have:

First term, A1 = - 15 + 9(0) = - 15

Second term, A2 = - 15 + 9(1) = - 6

Third term, A3 = -15 + 9(2) = - 15 + 18 = 3

Fourth term, A4 = - 15 + 9(3) = - 15 + 27 = 12

3. So, the general formula is An = - 15 + 9 (n - 1), which is the option D)

3) Question 3. Find an equation for the nth term of the arithmetic sequence A14 = - 33, A15 = 9.

option B) An = - 579 + 42(n - 1)

Explanation:

1) Find the difference: 9 - (-33) = 9 + 33 = 42

2) A15 = A1 + 42 * (15 - 1)

=> A1 = A15 - 42(15 - 1)

A1 = A15 - 42(14)

A1 = 9 - 588 = - 579

Therefore, the formula es An = - 579 + 42(n - 1)

4) Question 4. Determine whether the sequence converges or diverges. If it converges, give the limit.

48, 8, 4/3, 2/9, ...

the sequence converges to 288/5

Explanation:

That is a geometric sequence.

The ratio is 1/6: 8/48 = 1/6; (4/3) / 8 = 4/24 = 1/6; (2/9)/(4/3) = 6/36 = 1/6.

The convergence criterium is that if |ratio| < 1 then the series, this is the sum of all the terms, converge to: A1 / (1 - ratio)

Then the limit 48 / (1 - 1/6) = 48 / (5/6) = 48*6 / 5 = 288/5

5) Question 5. Find an equation for the nth term of the sequence.

-3, -12, - 48, -192

- 3 * (4)^(n-1)

Explanation: clearly any term (from the second) is the previous term multiplied by 4.

The first term is -3

The second term is -3(4) = - 12

The third term is -3(4)(4)= - 48

The fourth term is - 3 (4)(4)(4) = - 192

So, the general formula for the nth term is -3 * 4^ (n-1)

6) Question 6. Find an equation for the nth term of a geometric sequence where the second and fifth terms are -21 and 567, respectively.

An =7 * (-3)^(n-1)

Explanation:

1) The fith term is the second term * (ratio)^3: A5 = A3 * (r)^3

2) A5 = 567, A2 = - 21 => r^3 = A5 / A2 = - 567 / 21 = - 27

=> r = ∛(-27) = - 3

3) So the first term is A1 = A2 / r = -21 / -3 = 7

4) The general formula is

An =7 * (-3)^(n-1)

7) Question 7. Write the sum using summation notation, assuming the suggested pattern continues.

5 - 15 + 45 - 135 + ...

option B) summation of five times negative three to the power of n from n equals zero to infinity

Explanation:

5 = 5

-15 = 5 (-3)

45 = 5(-3)^2

-135 = 5(-3)^3

=> 5 + 5(-3) + 5(-3)^2 + 5(-3)^3+

Using the summation notation that is:

∞

∑ (5)(-3)^n

n=0

Which means summation of five times negative three to the power of n from n equals zero to infinity

8) Question 8. Write the sum using summation notation, assuming the suggested pattern continues.

-9 - 3 + 3 + 9 + ... + 81

option A) summation of the quantity negative nine plus six n from n equals zero to fifteen

Explanation:

Find the difference:

-3 - (-9) = - 3 + 9 = 6

3 - (-3) = 3 + 3 = 6

9 - 3 = 6

First term: - 9

Second term: - 9 + 6(1)

Third term: - 9 + 6(2)

nth term = - 9 + (n -1)

Summation = [- 9] + [- 9 + 6(1) + [-9 + 6(2)] + [-9 + 6(3) ]+ [-9 + 6(15) ]

Using summation notation:

15

∑ [-9 + 6n]

n=0

which means summation of the quantity negative nine plus six n from n equals zero to fifteen.

9) Question 9. Write the sum using summation notation, assuming the suggested pattern continues.

64 + 81 + 100 + 121 + ... + n2 + ...

A) summation of n squared from n equals eight to infinity

Explanation:

64 = 8^2

81 = 9^2

100 = 10^2

121 = 11^2

n^2

=>

∞

∑ n^2

n=8

which means summation of n squared from n equals eight to infinity

10) Question 10. Find the sum of the arithmetic sequence.

17, 19, 21, 23, ..., 35

260

Explanation:

The difference is 2:

The sum is: 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35.

You can use the formula for the sum of an arithmetic sequence:

(A1 + An) * n / 2 = (17 + 35)*10/2 = 260

11) Question 11. Find the sum of the geometric sequence.

1, 1/2, 1/4, 1/8, 1/16

option D) 31/16

Explanation:

You can either sum the 5 terms or use the formula for the partial sum of a geometric sequence.

The formula is: Sum = A * ( 1 - r^n) / (1 - r)

Here A = 1, r = 1/2, and n = 5 => Sum = 1 * (1 - (1/2)^5 ) / (1 - 1/2) =

= [ 1 - 1/32] / [1/2] = [31/32] / [1/2] = 31 / 16

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General Formula for arithmetic series :

-----------------------------------------------------

an = a₁ + d(n – 1)

-----------------------------------------------------

Find equation :

-----------------------------------------------------

a₁ = -15 and d = 9,

[tex]a_n = -15 + 9(n-1)[/tex]

-----------------------------------------------------

[tex]a_n = -15 + 9(n-1)[/tex] (Answer D)

-----------------------------------------------------

[tex]-15;-6;\ 3;\ 12;...\\\\a_n=a_1+(n-1)d\\\\a_1=-15;\ a_2=-6\\\\d=a_2-a_1\to d=-6-(-15)=-6+15=9\\\\a_n=-15+(n-1)\cdot9=\boxed{-15+9(n-1)}[/tex]

aₙ = 9n - 24 is the nth term of arithmetic sequence.

Step-by-step explanation:

we have given arithmetic sequence:

-15, -6, 3, 12, ...

we have to find the nth term of the arithmetic sequence.

consider the given arithmetic sequence -15, -6, 3, 12, ...

here a₁= -15 , a₂= -6 ,a₃ = 3,

we first find the common difference between consecutive terms (d).

common difference can be find by finding difference between two successive terms of give arithmetic sequence -15, -6, 3, 12, ...

a₂ - a₁ =( - 6 ) - ( - 15) = 9

a₃ - a₂ = ( 3 ) - ( - 6 ) = 9

hence common difference of given arithmetic sequence is 9.

thus the formula to find the nth term of arithmetic sequence is given below

aₙ = a +( n - 1 ) d

putting a = -15 , d = 9 we get

aₙ = ( -15 ) + ( n-1 ) 9

aₙ = - 15 + 9n - 9

aₙ = 9n - 24 this is the nth term of arithmetic sequence.

A(1) = - 6

A(n) = 4 * A(n-1)

n A(n)

1 - 6

2 4 * (-6) = - 24

3 4 * (-24) = - 96

4 4 * (-96) = - 384

5 4 * (-384) = - 1536

6 4 * ( -1536) = -6144

So, the first six terms are: -6, - 24, - 96, - 384, - 1536, - 6144.

Read more on -

Try a,b,d,a,then b

sorry if its wrong

The nth term of AP is calculated as [tex]a_n=9n-24[/tex]

Step-by-step explanation:

Given : An arithmetic sequence -15, -6, 3, 12, ...

We have to find the nth term of the given arithmetic sequence -15, -6, 3, 12,..

Consider the given arithmetic sequence -15, -6, 3, 12, ...

Here, [tex]a_1=-15\ ,a_2=-6\ ,a_3=3[/tex]

We first find the common difference (d)

Common difference can be find by finding difference between two successive terms of given arithmetic sequence.

[tex]a_2-a_1=-6+15=9\\\\ a_3-a_2=3+6=9[/tex]

Thus, common difference of given AP is 9

Thus, The equation to find the nth term of AP is calculated as,

[tex]a_n=a+(n-1)d[/tex]

Substitute a = -15 , d = 9 , we get,

[tex]a_n=-15+(n-1)9\\\\ a_n=-15+9n-9\\\\ a_n=9n-24[/tex]

Thus, the nth term of AP is calculated as [tex]a_n=9n-24[/tex]

D

Step-by-step explanation:

The n th term of an arithmetic sequence is

[tex]a_{n}[/tex] = a + (n - 1)d

where a is the first term and d the common difference

d = - 6 - (- 15) = - 6 + 15 = 9, and a = - 15

[tex]a_{n}[/tex] = - 15 + 9(n - 1)

[tex]a_n=-15+9(n-1)[/tex]

Step-by-step explanation:

The terms of the arithmetic sequence are...

-15, -6, 3, 12, ...

The first term is

a_1=-15

The common difference is

[tex]d=-6--15=9[/tex]

The equation for the nth term is given by;

[tex]a_n=a_1+(n-1)d[/tex]

We substitute the values into the formula to get;

[tex]a_n=-15+9(n-1)[/tex]