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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 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
    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
    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

  7. #7
    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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    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
    When you multiply two factors with the same base you keep the base and add the exponents. Just need to know that 1/2 = (1/2)^1 and that 1+ (n-1) is n.
    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 Jomo View Post
    When you multiply two factors with the same base you keep the base and add the exponents. Just need to know that 1/2 = (1/2)^1 and that 1+ (n-1) is n.
    mind = blown

    i've only done that when the variables were bases.

    but it works with regular numbers too. makes sense.

    thanks!

  10. #10
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    Quote Originally Posted by Johnny_L View Post
    mind = blown

    i've only done that when the variables were bases.

    but it works with regular numbers too. makes sense.

    thanks!
    Variables are place holders for numbers. So yeah, if something works for variables it must work for numbers!
    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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