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Math Help - sum of GP (2)

  1. #1
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    sum of GP (2)

    Find the sum of the following GP

    4-6+9-...-30.375

    a=4 , r=-3/2

    T_n=4(-3/2)^{n-1}=-30.375 ... how to solve this part if i take the logs ??
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  2. #2
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    T_n=4 \left(-\frac32\right)^{n-1}=-30.375

    \left(-\frac32\right)^{n-1}=-7.59375

    If n-1 was even then \left(-\frac32\right)^{n-1} would be positive ; therefore n-1 is odd. We can let n-1 = 2p+1

    \left(-\frac32\right)^{2p+1}=-7.59375

    (-1)^{2p+1}\left(\frac32\right)^{2p+1}=-7.59375

    -\left(\frac32\right)^{2p+1}=-7.59375

    \left(\frac32\right)^{2p+1}=7.59375

    Taking logs

    (2p+1) \ln\left(\frac32\right)=\ln(7.59375)

    2p+1 =\frac{\ln(7.59375)}{\ln\left(\frac32\right)}

    2p+1 = n-1 = 5

    n = 6
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  3. #3
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    Quote Originally Posted by running-gag View Post
    T_n=4 \left(-\frac32\right)^{n-1}=-30.375


    If n-1 was even then \left(-\frac32\right)^{n-1} would be positive ; therefore n-1 is odd. We can let n-1 = 2p+1

    Did you assume that \left(-\frac32\right)^{n-1} is negative ? How do i know that ?
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  4. #4
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    Hi thereddevils

    It's not an assumption. \left(-\frac32\right)^{n-1} must be negative
    because the RHS is negative
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