# Geometric Sequence Help

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• Jul 13th 2010, 01:46 PM
fletch
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.
• Jul 13th 2010, 02:14 PM
Plato
Quote:

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 $\displaystyle r$.
• Jul 13th 2010, 03:46 PM
Archie Meade
Quote:

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

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

From your information

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

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

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

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 $\displaystyle r^7$