Geometric progression: first term 12, fourth term -3/2

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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