# Question on geometric progression (2)

• Apr 26th 2008, 06:02 PM
Tangera
Question on geometric progression (2)
Hello! Please show how to solve this question! I don't know how to solve the second part of the question...

In a geometric progression, the first term is 12, and the fourth term is (-3/2).

i)Find the sum of the first n terms of the progression, Sn.

ii) Find the sum to infinity, S, of the progression [I found this to be 8] and the least value of n for which the magnitude of the difference between Sn and S is less than 0.001.

Thank you very much!!
• Apr 26th 2008, 06:06 PM
icemanfan
Quote:

Originally Posted by Tangera
Hello! Please show how to solve this question! I don't know how to solve the second part of the question...

In a geometric progression, the first term is 12, and the fourth term is (-3/2).

i)Find the sum of the first n terms of the progression, Sn.

ii) Find the sum to infinity, S, of the progression [I found this to be 8] and the least value of n for which the magnitude of the difference between Sn and S is less than 0.001.

Thank you very much!!

Since this is a geometric progression, our terms will be $12, 12r, 12r^2, 12r^3, ...$ so the fourth term of the series is $12r^3= -\frac{3}{2}$.

The sum to infinity is $\frac{12}{1-r}$. So you can find this after solving the previous equation for r.
• Apr 27th 2008, 01:47 AM
Tangera
Hello! Thank you for replying! Could you explain how to solve part (ii) of the question too? The part regarding the least value of n....Thank you!
• Apr 27th 2008, 01:54 AM
Moo
Hello,

How much do you get for the progression ? Because it's not 8 :p

$S_n=12 \times \frac{1-r^{n+1}}{1-r}$

So $S_n-S=\frac{12}{1-r} ((1-r^{n+1})-1)$

$\Longleftrightarrow S_n-S=\frac{12}{1-r} (-r^{n+1})$

You're looking for n such as $|S_n-S|=\frac{12 |r|^{n+1}}{1-r} < 0.001$

Now plugin r in this expression and solve for n :)
• Apr 27th 2008, 02:10 AM
Tangera
Quote:

Originally Posted by Moo
Hello,

How much do you get for the progression ? Because it's not 8 :p

Hello! I found the Sum to infinity to be 8...and that's also the answer that is in the answer key, so it should be correct. (Thinking)
• Apr 27th 2008, 02:21 AM
Moo
Oh sorry, I was talking about r, not S !
• Apr 27th 2008, 02:23 AM
Tangera
Quote:

Originally Posted by Moo
Oh sorry, I was talking about r, not S !

oh...I found r = -1/2 and S = 8...
• Apr 27th 2008, 02:27 AM
Moo
Yes, it's ok !

And what did you find for n ? Do you manage to do it ? :)
• Apr 27th 2008, 03:06 AM
Tangera
noooo....the expression I got for Sn was 8[1 - (-1/2)^n] and then i couldn't solve the modulus...
• Apr 27th 2008, 03:23 AM
Moo
Quote:

S_n=8[1 - (-1/2)^n]
It's ok, it's just n+1 instead of n because there are n+1 terms in $S_n$ :)

$S_n=8-8(-1)^{n-1}(.5)^{n+1}$

You have to take the absolute value of the difference, because S is constant and the sign always change for the terms.

$|S_n-S|=|S_n-8|=|{\color{red}-}8 {\color{red}(-1)^{n-1}}(.5)^{n+1}|=8(.5)^{n+1}$
• Apr 27th 2008, 05:15 AM
Tangera
But I have a (-1/2)^n so it is impossible to solve the modulus...?
• Apr 27th 2008, 08:04 AM
Moo
Take the absolute value of -1/2, that is to say 1/2. You want to care about the magnitude, that is to say the absolute difference between the two.
• May 5th 2008, 04:20 AM
Tangera
HellO! [sorry this is reeaaaally late.] Thank you for replying! I have solved the question. :)