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

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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  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

  2. -----------------------------------------------------
    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)
    -----------------------------------------------------

  3. [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]

  4. 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.

  5. 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 -

  6. 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]

  7. 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)

  8. [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]

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