Maths: Geometric Series

[Script Coder]

Question wrote:

A Geometric series has a first term of -2 and a common ratio of -3/2 find the smallest number of terms whose sum will be greater than 12.

What I did:
Sn > 12
a(r^n-1)/(r-1) > 12
((-3/2)^n-1)/(-5/2) <-6
(-3/2)^n > 16

And I know that you cannot have a log with a negative base. I also know that summing the first 8 terms will be the first time that the sum is greater than 12. So where do I stand mathematically: Is it undefined or 8?

[ne555]

you could divide your series into positive and negative terms.
a) start at 2, ratio 9/4
1) start at 3, ratio 9/4

then solve for which `n' the difference of their sum is greater than 12
S_{1} - S_{a} > 12


Or more simple, you know that n is even so
(-3/2)^n = (3/2)^n

[helios]

x^n-1 != x^(n-1)
Also, the correct form of the expression is with r^(n+1), not n-1, since the series is of the form [a, ar, ar^2, ...]

n=5: (-3/2)^(n+1) = (-3/2)^6 > 11
n=6: (-3/2)^(n+1) = (-3/2)^7 < -17
n=7: (-3/2)^(n+1) = (-3/2)^10 > 25
So n = 7.

(-3/2)^n > 16 has a positive solution. Let f(n) = (-3/2)^n and g(m) = f(2m). g is a positive (the base of the exponent is positive) and monotonically increasing (the base is > 1) function. For all k, there exists some natural n such that g(n) > k.

[Script Coder]

ne555 wrote:

you could divide your series into positive and negative terms.
a) start at 2, ratio 9/4
1) start at 3, ratio 9/4

then solve for which `n' the difference of their sum is greater than 12
S_{1} - S_{a} > 12

Solving gives you:
(9/4)^n <-14
which is also undefined.

ne555 wrote:

Or more simple, you know that n is even so
(-3/2)^n = (3/2)^n

I know that, but I do not know if it is mathematically correct or not.

helios wrote:

Also, the correct form of the expression is with r^(n+1), not n-1, since the series is of the form [a, ar, ar^2, ...]

The nth term of a GP is ar^(n-1). However I am referring to the sum of a GS which is a((r^n)-1)/(r-1).

Thank both of you for your help thus far.

[ne555]

> Solving gives you:
> (9/4)^n <-14

S = a(r^n-1)/(r-1)

S_{1} = 3 [ (9/4)^n - 1 ] / [ 9/4 - 1]
S_{1} = 3*4 [ (9/4)^n - 1 ] / 5
S_{a} = 2*4 [ (9/4)^n - 1 ] / 5

S_{1} - S_{a} = 4 [ (9/4)^n - 1 ] / 5 > 12
(9/4)^n - 1 > 15
(9/4)^n > 16
note the similarity with (-3/2)^n > 16


> I know that, but I do not know if it is mathematically correct or not.
n \in N
N = {odd} U {even}
n \notin {odd} \iff n \in {even}


> And I know that you cannot have a log with a negative base
ln( -1 ) = j \pi

[Script Coder]

>S_{1} = 3*4 [ (9/4)^n - 1 ] / 5
>S_{a} = 2*4 [ (9/4)^n - 1 ] / 5
How?

Also does S_{a} and S_{1} mean the same as the sum of the first a term and the sum of the first term respectively?

>ln( -1 ) = i \pi
Fair enough.

Thank you thus far.

[ne555]

Sorry, I think I abused the notation I didn't want to depreciate any series

You've got two series
the series called `a' starts at 2, and has a ratio of 9/4
the series called `1' starts at 3, and has a ratio of 9/4

S_{a} means the sum of the series called `a'
S_{1} means the sum of the series called `1'

the series called `a' represents the terms that are negative
the series called `1' represents the terms that are positive
that's why you do S_{1} - S_{a}

[Script Coder]

Oh, thank you very much; I understand now.

[ne555]

>> ln( -1 ) = i \pi
http://xkcd.com/326/