# Homework Help: Help with a Geometric progression.

1. Apr 17, 2005

### misogynisticfeminist

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.

2. Apr 17, 2005

### whozum

Are you sure thats correct?

3. Apr 17, 2005

### misogynisticfeminist

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?

Last edited: Apr 17, 2005
4. Apr 17, 2005

### whozum

I dont 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 [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.

5. Apr 17, 2005

### OlderDan

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