Help with a Geometric progression.

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In a geometric progression with the first term of 12 and the fourth term of -3/2, the common ratio is determined to be -1/24. The sum to infinity is correctly calculated as 8, resolving an earlier miscalculation of 11.52. To find 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, the difference should be set as an error estimation variable, ε. The approach involves taking the limit of the sum to n terms as n approaches infinity and solving for n. This method ensures that the conditions of the problem are met accurately.
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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,

12, T_2, T_3, -3/2

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

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

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,

11.52- \frac {12 ( - \frac {1}{24} ^n -1 )}{-1/24-1} = 0.001 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,

12, T_2, T_3, -3/2

I see that the ratio between the 4th and 1st terms is -\frac{1}{8} 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

r^3 = -\frac {1}{8}. thanks alot. 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 S_n and S_\infty is 0.001

can someone help?
 
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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 S_n to an errorestimation variable \epsilon, then set 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.
 
The equation for S_\infty comes from the equation for S_n by taking the limit as n goes to infinity. Take the difference between the equations for S_\infty and S_n and set it equal to 0.001
 
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