1. ## Geometric Sequence Help

Having trouble finding a way of working through this problem:

A geometric sequence has first term 1. The ninth term exceeds the fifth term by 240. Find possible values for the eighth term.

I know that the common ratio has to be 2 because the difference between the two terms is small number, but this was found through trial and error.

Any help would be appreciated.

2. Originally Posted by fletch
Having trouble finding a way of working through this problem:

A geometric sequence has first term 1. The ninth term exceeds the fifth term by 240. Find possible values for the eighth term.

I know that the common ratio has to be 2 because the difference between the two terms is small number, but this was found through trial and error.
I am not sure what you are asking.
You are given that $\displaystyle a_1=1,~a_5=r^4,~\&~a_9=r^8$.
Also $\displaystyle a_9-a_5=240$.
From that find $\displaystyle r$.

3. Originally Posted by fletch
Having trouble finding a way of working through this problem:

A geometric sequence has first term 1. The ninth term exceeds the fifth term by 240. Find possible values for the eighth term.

I know that the common ratio has to be 2 because the difference between the two terms is small number, but this was found through trial and error.

Any help would be appreciated.
The geometric sequence is

$\displaystyle 1,\ r,\ r^2,\ r^3,\ r^4,\ r^5,\ r^6,\ r^7,\ r^8,\ r^9....$

$\displaystyle \left(r^4\right)^2=r^4+240$

$\displaystyle x^2-x-240=0$

$\displaystyle (x-16)(x+15)=0$

x is not -15, so x is 16

r is the 4th root of 16, which is $\displaystyle \pm2$.

The 5th and 9th terms are both positive, regardless of whether r is 2 or -2

The 8th term is the two solutions to $\displaystyle r^7$