Results 1 to 4 of 4

Thread: sum of GP (2)

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    381

    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 ??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    $\displaystyle T_n=4 \left(-\frac32\right)^{n-1}=-30.375$

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

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

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

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

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

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

    Taking logs

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

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

    $\displaystyle 2p+1 = n-1 = 5$

    $\displaystyle n = 6$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2009
    Posts
    381
    Quote Originally Posted by running-gag View Post
    $\displaystyle T_n=4 \left(-\frac32\right)^{n-1}=-30.375$


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

    Did you assume that $\displaystyle \left(-\frac32\right)^{n-1}$ is negative ? How do i know that ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jul 2009
    Posts
    397
    Hi thereddevils

    It's not an assumption. $\displaystyle \left(-\frac32\right)^{n-1}$ must be negative
    because the RHS is negative
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum