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Math Help - definite series!

  1. #1
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    definite series!

    three numbers of a geometric progression whose sum is 28 and whose product is 512. find the first three numbers

    the problem above, i have attempted it and arrived at 8,16,32

    am i correct? thanks
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  2. #2
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    Re: definite series!

    Quote Originally Posted by lawochekel View Post
    three numbers of a geometric progression whose sum is 28 and whose product is 512. find the first three numbers the problem above, i have attempted it and arrived at 8,16,32
    Well 8+16+32\ne 28. What do you think?
    Thanks from lawochekel
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  3. #3
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    Re: definite series!

    Quote Originally Posted by lawochekel View Post
    three numbers of a geometric progression whose sum is 28 and whose product is 512. find the first three numbers

    the problem above, i have attempted it and arrived at 8,16,32

    am i correct? thanks
    As Plato has shown, your solution can not possibly be right.

    You have

    \displaystyle \begin{align*} t_n + r\,t_n + r^2t_n &= 28 \\ (r^2 + r + 1)t_n &= 28 \end{align*}

    and

    \displaystyle \begin{align*} t_n \cdot r\,t_n \cdot r^2t_n &= 512 \\ r^3t_n &= 513 \end{align*}

    Try to solve these two equations simultaneously to find \displaystyle \begin{align*} r \end{align*} and \displaystyle \begin{align*} t_n \end{align*}. From there you can find the following two terms.
    Thanks from lawochekel
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  4. #4
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    Re: definite series!

    Try this

    3 terms a,ar.ar^2 Product = 512 = a^3 r^3 a = 8/r second term =8 third term = 8r
    add the 3 terms whose sum =28 . This will produce a quadratic equation giving the r value
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