- #1

chwala

Gold Member

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- 287

- Homework Statement:
- See attached.

- Relevant Equations:
- sequences

*kindly note that i do not have the solutions ...

I was looking at this, not quite sure on what they mean by exact fractions, anyway my approach is as follows;

##\dfrac {a}{243}=\dfrac{a(1-r^3)}{240}##

##\dfrac {1}{243}=\dfrac{1-r^3}{240}##

##\dfrac {240}{243}=1-r^3##

##r^3=1-\dfrac{240}{243}##

##r^3=\dfrac{3}{243}##

##r=\dfrac{1}{\sqrt [3]81}=\dfrac{1}{3\sqrt [3]3}##

My alternative approach,

##(243-243r)(1-r^3)=240-240r##

##243r^4-243r^3-3r+3=0## using ti- nspire i realized same value as first approach...

I am not sure of my answer, I will need to check the r value and see whether i will realize the sum of ##240##. On quick checking i seem to have the correct terms i.e ##[184.4..., 43.6333..., 9.85468..., 2.2813...]## whose ##S_{4}≅240.##

I think i got it...let me post my working...

From

##r=\dfrac{1}{\sqrt [3]81}=\dfrac{1}{3\sqrt [3]3}##,

We shall have;

##r_{1}=\dfrac{1}{3}## and ##r_{2}=\dfrac{1}{\sqrt [3]3}##

On checking with, ##r=\dfrac{1}{3}## , we have the terms of the sequence as ##[162,54,18,6 ]## which add up to ##240##. Cheers guys.

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