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Thread: Help with finding the formula to find the nth term of a geometric sequence.

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    Help with finding the formula to find the nth term of a geometric sequence.

    Geometric sequence is 3/2, -3/4, and 3/8 are the first three terms. what represents the nth term of the sequence?


    I know a is the first term, 3/2.
    ratio is just a term divided by the term proceeding it.
    ar^(n-1) is how you find a specific term.


    I get all the way up to here: a(n) = ar^(n-1) = (3/2)(-1/2)^(n-1)


    the final answer the book says, and this website I got help from: http://www.algebra.com/algebra/homew...on.391759.html


    says the final answer is:
    a(n) = 3*(-1)^(n-1)/2^(n)




    My question is, what happens to the other 2 in the denominator, and where does that n exponent in the denominator come from?

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    Elite Member stapel's Avatar
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    Quote Originally Posted by Johnny_L View Post
    I get all the way up to here: a(n) = ar^(n-1) = (3/2)(-1/2)^(n-1)

    the final answer the book says...is:
    a(n) = 3*(-1)^(n-1)/2^(n)

    My question is, what happens to the other 2 in the denominator, and where does that n exponent in the denominator come from?
    What did you get when you multiplied things out? You split the 3 from the /2 to get (3)(1/2), you noted that (-1/2)^(n-1) = (-1)^(n-1) / 2^(n-1) = (-1)^(n-1) * (1/2)^(n-1), and then you noted that you've now got a (1/2) = (1/2)^1 and a (1/2)^(n-1), so....

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    Quote Originally Posted by Johnny_L View Post
    I get all the way up to here: a(n) = ar^(n-1) = (3/2)(-1/2)^(n-1)

    the final answer the book says, and this website I got help from: http://www.algebra.com/algebra/homew...on.391759.html

    says the final answer is:
    a(n) = 3*(-1)^(n-1)/2^(n)
    What you got is correct; final answer also correct:
    simply another way to get same results.

    Standard formula is ar^(n-1) which you got...
    I'm just an imagination of your figment !

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    Quote Originally Posted by stapel View Post
    What did you get when you multiplied things out? You split the 3 from the /2 to get (3)(1/2), you noted that (-1/2)^(n-1) = (-1)^(n-1) / 2^(n-1) = (-1)^(n-1) * (1/2)^(n-1), and then you noted that you've now got a (1/2) = (1/2)^1 and a (1/2)^(n-1), so....
    (3/2)(-1/2)^(n-1)

    3 X-1 = -3; 2 X 2 =4

    3/4^(n-1)

    why aren't we multiplying numerator by numerator and denominator by denominator?

    I don't understand how we get to this

    a(n) = 3*(-1)^(n-1)/2^(n)

    form at all...



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    Quote Originally Posted by Johnny_L View Post
    (3/2)(-1/2)^(n-1)

    3 X-1 = -3; 2 X 2 =4

    3/4^(n-1)
    (Use * for multiplication symbol...now standard)

    Nooooooo: (3/2) * [(-1/2)^(n-1)] ; exponent first, then multiplication...
    I'm just an imagination of your figment !

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    Quote Originally Posted by Johnny_L View Post
    Geometric sequence is 3/2, -3/4, and 3/8 are the first three terms. what represents the nth term of the sequence?


    I know a is the first term, 3/2.
    ratio is just a term divided by the term proceeding it.
    ar^(n-1) is how you find a specific term.


    I get all the way up to here: a(n) = ar^(n-1) = (3/2)(-1/2)^(n-1)


    the final answer the book says, and this website I got help from: http://www.algebra.com/algebra/homew...on.391759.html


    says the final answer is:
    a(n) = 3*(-1)^(n-1)/2^(n)




    My question is, what happens to the other 2 in the denominator, and where does that n exponent in the denominator come from?
    This is not hard. Just ignore for now that the sign is changing.. You have 3/2=3/2^1, then 3/4=3/2^2. then 3/8=3/2^3. So the nth term will be 3/2^n. How do you get the sign to change?
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

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    Quote Originally Posted by Johnny_L View Post
    (3/2)(-1/2)^(n-1)

    3 X-1 = -3; 2 X 2 =4

    3/4^(n-1)

    why aren't we multiplying numerator by numerator and denominator by denominator?

    I don't understand how we get to this

    a(n) = 3*(-1)^(n-1)/2^(n)

    form at all...
    stapel showed you how but possibly this is more clear
    (3/2)(-1/2)^(n-1)
    =[tex]\frac{3}{2} * (-\frac{1}{2})^{n-1}[/tex]

    =[tex]3 * \frac{1}{2} * (-\frac{1}{2})^{n-1}[/tex]
    =[tex]3 * \frac{1}{2} * (-1)^{n-1} * (\frac{1}{2})^{n-1}[/tex]
    =[tex]3 * (-1)^{n-1} * \frac{1}{2} * (\frac{1}{2})^{n-1}[/tex]
    =[tex]3 * (-1)^{n-1} * (\frac{1}{2})^{n}[/tex]
    =[tex]3 * (-1)^{n-1} * (\frac{1}{2^{n}})[/tex]
    = 3 (-1)^(n-1)/2^n

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    Quote Originally Posted by Ishuda View Post
    = 3 (-1)^(n-1)/2^n
    Agree...however, 3(-1)^(n+1) / 2^n is also correct.

    Seems real strange that "the book" gives that
    "complicated looking" solution.
    Why not simply leave it at:
    a[r^(n-1)] where a = 3/2 and r = -1/2
    I'm just an imagination of your figment !

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    Quote Originally Posted by Ishuda View Post
    stapel showed you how but possibly this is more clear
    (3/2)(-1/2)^(n-1)
    =[tex]\frac{3}{2} * (-\frac{1}{2})^{n-1}[/tex]

    =[tex]3 * \frac{1}{2} * (-\frac{1}{2})^{n-1}[/tex]
    =[tex]3 * \frac{1}{2} * (-1)^{n-1} * (\frac{1}{2})^{n-1}[/tex]
    =[tex]3 * (-1)^{n-1} * \frac{1}{2} * (\frac{1}{2})^{n-1}[/tex]
    =[tex]3 * (-1)^{n-1} * (\frac{1}{2})^{n}[/tex]
    =[tex]3 * (-1)^{n-1} * (\frac{1}{2^{n}})[/tex]
    = 3 (-1)^(n-1)/2^n
    ok, im starting to get it but how do we get from here

    =[tex]3 * (-1)^{n-1} * \frac{1}{2} * (\frac{1}{2})^{n-1}[/tex]
    to here
    =[tex]3 * (-1)^{n-1} * (\frac{1}{2})^{n}[/tex]

    One of the 1/2 disappears and the n-1 turns to n. I don't understand

  10. #10
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by Johnny_L View Post
    ok, im starting to get it but how do we get from here

    =[tex]3 * (-1)^{n-1} * \frac{1}{2} * (\frac{1}{2})^{n-1}[/tex]
    to here
    =[tex]3 * (-1)^{n-1} * (\frac{1}{2})^{n}[/tex]

    One of the 1/2 disappears and the n-1 turns to n. I don't understand
    As pointed out earlier: What do you get when you combine the (1/2)^1 and the (1/2)^(n-1)?

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