# Thread: Question on geometric progression (2)

1. ## 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!!

2. 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.

3. 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!

4. Hello,

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

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

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

5. Originally Posted by Moo
Hello,

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

6. Oh sorry, I was talking about r, not S !

7. Originally Posted by Moo
Oh sorry, I was talking about r, not S !
oh...I found r = -1/2 and S = 8...

8. Yes, it's ok !

And what did you find for n ? Do you manage to do it ?

9. noooo....the expression I got for Sn was 8[1 - (-1/2)^n] and then i couldn't solve the modulus...

10. 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.

$|S_n-S|=|S_n-8|=|{\color{red}-}8 {\color{red}(-1)^{n-1}}(.5)^{n+1}|=8(.5)^{n+1}$

11. But I have a (-1/2)^n so it is impossible to solve the modulus...?

12. 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.

13. HellO! [sorry this is reeaaaally late.] Thank you for replying! I have solved the question.