Terms of a geometric series (1 Viewer)

[Roobs]

I've got a series question that has me really stumped:

"find the value of n if the nth term of the series : -2+ 3/2 -9/8 + .... is equal to -81/128"


heres my working:

Tn=ar^(n-1)

where a=-2
r= -3/4

-81/128 = -2*(-3/4)^(n-1)

81/256= (-3/4)^(n-1)

now taking logs of both sides:

Ln(81/256)= (n-1)* Ln(-3/4) !!!!!!!!!!!!!!!!!!!!!!!!

and this is where i hit a snag, of trying to take the log of a negative number.....

i tried just ignoring the negative and taking r as =2 and i got the correct answer (n=5)... but for some reason it seems like this sholdent be done..

any help?

 

[Mountain.Dew]

the easy way to do it i reckon is simply do this:

1st term: -2

2nd term: 3/2

3rd term: -9/8

4th term: 27/32

5th term: -81/128 !!!!!!!!!!!!!!!!!!!! voila!

 

[insert-username]

Since multiplying a negative number by itself doesn't change its magnitude (only the sign), it's fine to take the absolute value of these types of questions as long as you remember the signs.


I_F

 

[Roobs]

cheers....yeah justifying it like that makes more sence than just saying "ignore the negative"