# Thread: sum of GP (2)

1. ## 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 ??

2. $\displaystyle T_n=4 \left(-\frac32\right)^{n-1}=-30.375$

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

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

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

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

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

$\displaystyle n = 6$

3. Originally Posted by running-gag
$\displaystyle T_n=4 \left(-\frac32\right)^{n-1}=-30.375$

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 ?

4. Hi thereddevils

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