# Geometric Progression

• September 10th 2009, 10:22 AM
archa1c
Geometric Progression
Hi guys,
I am a little stuck on where to begin here and any help would be great.

This is the question I am faced with:

Quote:

The sum of the first three terms of a geometric series is http://mathbin.net/equations/31707_1.png. The sum of the first six terms is http://mathbin.net/equations/31707_2.png. Find the first term and common ratio.
Now, the formula I have been using recently for working out the sum of a geometric series to n places is:

$S_{n} = \frac{a(1-r^n)}{1-r}$

But I am unsure how to use it in this context.

Thanks for any help!
• September 10th 2009, 12:33 PM
running-gag
Hi

You know

$S_{6} = \frac{a(1-r^6)}{1-r} = \frac{3367}{512}$

$S_{3} = \frac{a(1-r^3)}{1-r} = \frac{37}{8}$

Dividing gives

$\frac{1-r^6}{1-r^3} = \frac{91}{64}$

Using $R = r^3$ you will find a quadratic whose unknown is $R$

You will get R then r and finally a
• September 10th 2009, 04:26 PM
luobo
Quote:

Originally Posted by running-gag
Hi

You know

$S_{6} = \frac{a(1-r^6)}{1-r} = \frac{3367}{512}$

$S_{3} = \frac{a(1-r^3)}{1-r} = \frac{37}{8}$

Dividing gives

$\frac{1-r^6}{1-r^3} = \frac{91}{64}$

Using $R = r^3$ you will find a quadratic whose unknown is $R$

You will get R then r and finally a

There is no need to solve a quadratic since

$\frac{1-r^6}{1-r^3} =1+r^3= \frac{91}{64} \Rightarrow r^3=\frac{27}{64} \Rightarrow r=\frac{3}{4}$
• September 11th 2009, 03:42 AM
archa1c
Quote:

Originally Posted by running-gag
Hi

You know

$S_{6} = \frac{a(1-r^6)}{1-r} = \frac{3367}{512}$

$S_{3} = \frac{a(1-r^3)}{1-r} = \frac{37}{8}$

Dividing gives

$\frac{1-r^6}{1-r^3} = \frac{91}{64}$

Using $R = r^3$ you will find a quadratic whose unknown is $R$

You will get R then r and finally a

Could you please explain the steps you took to divide those equations?

Thanks for you help
• September 11th 2009, 11:48 AM
running-gag
Quote:

Originally Posted by luobo
There is no need to solve a quadratic since

$\frac{1-r^6}{1-r^3} =1+r^3= \frac{91}{64} \Rightarrow r^3=\frac{27}{64} \Rightarrow r=\frac{3}{4}$

Yes

I realized this after switching off my computer ! (Wink)
• September 11th 2009, 11:52 AM
running-gag
$\frac{a(1-r^6)}{1-r} = \frac{3367}{512}$

$\frac{a(1-r^3)}{1-r} = \frac{37}{8}$

Dividing is the same than multiplying the first equation by the inverse of the second one

$\frac{a(1-r^6)}{1-r} \: \frac{1-r}{a(1-r^3)} = \frac{3367}{512}\:\frac{8}{37}$

a and (1-r) are both at the numerator and the denominator, therefore they disappear

$\frac{1-r^6}{1-r^3} = \frac{91}{64}$
• September 13th 2009, 01:24 AM
archa1c
Thanks for the help guys, much appreciated