Help with a Geometric progression.

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Homework Help Overview

The discussion revolves around a problem involving a geometric progression (GP) where the first term is 12 and the fourth term is -3/2. Participants are tasked with finding the sum to n terms, the sum to infinity, and the least value of n for which the difference between the sum to infinity and the sum to n terms is less than 0.001.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to express the GP and calculate the common ratio, but questions arise regarding the correctness of their calculations. Some participants suggest re-evaluating the relationship between the terms and the sums involved.

Discussion Status

Participants are actively engaging with the problem, with some clarifying their understanding of the series and others providing hints about how to approach the estimation of the difference between the sums. There is no explicit consensus on the method to find n, but guidance has been offered regarding the relationship between the sums.

Contextual Notes

Some participants express uncertainty about the series and its properties, indicating a potential gap in foundational knowledge that may affect their ability to solve the problem. The original poster has made some progress but is struggling with the final part of the question.

misogynisticfeminist
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I need a little help with this problem.

In a geometric progession, the first term is 12 and the fourth term is -3/2. Find the sum to n terms and the sum to infinity. Find also, the least value of n for which the magnitude of the difference between the sum to infinity and to n terms are less than 0.001.

I have first expressed the GP as,

[tex]12, T_2, T_3, -3/2[/tex]

I see that the ratio between the 4th and 1st terms is [tex]-\frac{1}{8}[/tex] and this is 3 times the common ration r, which is -1/24. To find the sum to n terms, i get,

[tex]S_n =\frac {12 ( - \frac {1}{24} ^n -1 )}{-1/24-1}[/tex]

and the sum to infinity is 11.52. However the sum to infinity is given as 8 in the answer.

To find the last part of the question, i did,

[tex]11.52- \frac {12 ( - \frac {1}{24} ^n -1 )}{-1/24-1} = 0.001[/tex] but it didn't work out to get the answer or n=13.

Thanks a lot for your help.
 
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misogynisticfeminist said:
I have first expressed the GP as,

[tex]12, T_2, T_3, -3/2[/tex]

I see that the ratio between the 4th and 1st terms is [tex]-\frac{1}{8}[/tex] and this is 3 times the common ration r, which is -1/24. To find the sum to n terms, i get,

Are you sure that's correct?
 
OHHH ! it should be

[tex]r^3 = -\frac {1}{8}[/tex]. thanks a lot. that should settle it.

edit:

I have found the sum to infinity already and got 8. But have difficulty in the last part where they asked me to find the value of n where the difference between [tex]S_n[/tex] and [tex]S_\infty[/tex] is 0.001

can someone help?
 
Last edited:
I don't remember series very well, but I'm surethey offer a great explanation in your textbook. I remember ours had 3 pages to this cause alone.

but as far as I can remember, you set [itex]S_n[/itex] to an errorestimation variable [itex]\epsilon[/itex], then set [itex]S_{\infty} -\epsilon < 0.001 [/tex] and I think you try solvin for n or something like that. Someone else probably has a better answer.[/itex]
 
The equation for [tex]S_\infty[/tex] comes from the equation for [tex]S_n[/tex] by taking the limit as n goes to infinity. Take the difference between the equations for [tex]S_\infty[/tex] and [tex]S_n[/tex] and set it equal to 0.001
 

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