Stuck on a couple geometric series issue

[bobbyc]

Hi,

I've been trying to figure out the following and got stuck as cannot recall the rules (mental block) and looking for some assistance.

I have three numbers, one is a geometric series total (eg: 500), another is the number of iterations (eg: 5) and the other is the ratio (eg: 2).
I need to work out the starting number and have tried the following:

My normal understanding is from: https://en.wikipedia.org/wiki/Geometric_series
a + ar + ar2 + ar3 + ...
so I know from guessing my for my example the starting value is approximately 7.93 with 5 series
starting value = 7.933 then 23.799 then 55.531 then 118.995 then 245.923 then finally 499.779 (this is the total value I have and I know the ration & # of iterations)
so start = 7.933
1st iteration is ( [last which is the starting value here which is 7.933] )+ [start which is 7.933] ✖ [ratio which is 2] = 23.799
2nd iteration is ( [last which is 23.799] )+ [start which is 7.933] ✖ [ratio which is 2]^2 = 55.531
3rd iteration is ( [last which is 55.531] )+ [start which is 7.933] ✖ [ratio which is 2]^3 =118.995
4th iteration is ( [last which is 118.995] )+ [start which is 7.933] ✖ [ratio which is 2]^4 =245.923
5th iteration is ( [last which is 245.923] )+ [start which is 7.933] ✖ [ratio which is 2]^5 =499.80 (close enough for the example)

so this looks like t = n + ((n ✖ 2) + (n ✖ 2^2) + (n ✖ 2^3) + (n ✖ 2^4) + (n ✖ 2^5))

I may be completely wrong here to work out 'x' from many past days and due to some lost brain cells, I vaguely recall as n is being multiplied on the right side so to work it out it's a division.

would that then be:
500 = n + ((n ✖ r) + (n ✖ r^2) + (n ✖ r^3) + (n ✖ r^4) + (n ✖ r^5))

total = n + (n ✖ r) + (n ✖ r^2) + (n ✖ r^3) + (n ✖ r^4) + (n ✖ r^5)
................................500

Every example I've seen so far knows two of the numbers somewhere in the sequence so can work out the ratio so can determine the entire series but I've not seen any examples (looking on and off for 3 days now) when having the last number of the series. I think this may be where I'm getting things wrong as its' the 6th number so the series is in fact 5 and not 6 iterations? There is also the case where the 1st number can be different and not included in the series.
eg: 10 then the series (of 5 iterations) starts at 15 with ratio of say 1.5.

500 = o + ((n ✖ 1.5) + (n ✖ 1.5^2) + (n ✖ 1.5^3) + (n ✖ 1.5^4) + (n ✖ 1.5^5))

total = o (n ✖ r) + (n ✖ r^2) + (n ✖ r^3) + (n ✖ r^4) + (n ✖ r^5)

What part of the puzzle have I missed?

Thank you.

Bobby with a sore brain attempting to work this out as I've done only binary & octal math for the last 20 years.

 

[Steven G]

I have three numbers, one is a geometric series total (eg: 500), another is the number of iterations (eg: 5) and the other is the ratio (eg: 2).
What is the number of iterations??
ratio = 2 means r=2.
a/(1-r) = a/(1-2) = -a = 500. So a=-500
Now you have the whole series. Is this what you want? What was the problem getting a?

 

[Dr.Peterson]

Hi,

I've been trying to figure out the following and got stuck as cannot recall the rules (mental block) and looking for some assistance.

I have three numbers, one is a geometric series total (eg: 500), another is the number of iterations (eg: 5) and the other is the ratio (eg: 2).
I need to work out the starting number and have tried the following:

Jomo assumed the total is for an infinite series. I think you are asking about a finite sum, for which your link contains the formula here.

The sum of n terms, starting at [MATH]a[/MATH] with common ratio [MATH]r[/MATH], is [MATH]S_n=a\frac{1-r^{n}}{1-r}[/MATH]. I think you are using the term "iterations" to mean terms after the first, so that your [MATH]n[/MATH] is 1 less than the number of terms. Taking it that way, [MATH]S_{n+1}=a\frac{1-r^{n+1}}{1-r}[/MATH], which is the formula Wikipedia gave.

For example, if [MATH]a=2[/MATH], [MATH]n=3[/MATH], and [MATH]r=3[/MATH], there are 4 terms, and the sum [MATH]2+6+18+54=80[/MATH], and [MATH]S_4=2\frac{1-3^{4}}{1-3}=2\frac{-80}{-2} = 80[/MATH].

You want, I think, to find the first term, [MATH]a[/MATH], given [MATH]n, r,[/MATH] and [MATH]S_n[/MATH]. For your example, given [MATH]n=5, r=2, S_n=500[/MATH], the formula says that [MATH]500=a\frac{1-2^{6}}{1-2}[/MATH], so [MATH]63a = 500[/MATH], and [MATH]a=\frac{500}{63} = 7.9365...[/MATH]
The same link shows you how this formula is derived, which is the method you are trying to reconstruct.

 

[bobbyc]

Brilliant. thank you. so part of my mistake was using wrong 'definitions'. Yes I now realize the 1st 'term' is what I have been meaning.

Got it. Thank God for mathematicians.

Jomo assumed the total is for an infinite series. I think you are asking about a finite sum, for which your link contains the formula here.

The sum of n terms, starting at [MATH]a[/MATH] with common ratio [MATH]r[/MATH], is [MATH]S_n=a\frac{1-r^{n}}{1-r}[/MATH]. I think you are using the term "iterations" to mean terms after the first, so that your [MATH]n[/MATH] is 1 less than the number of terms. Taking it that way, [MATH]S_n=a\frac{1-r^{n+1}}{1-r}[/MATH], which is the formula Wikipedia gave.

For example, if [MATH]a=2[/MATH], [MATH]n=3[/MATH], and [MATH]r=3[/MATH], there are 4 terms, and the sum [MATH]2+6+18+54=80[/MATH], and [MATH]S_4=2\frac{1-3^{4}}{1-3}=2\frac{-80}{-2} = 80[/MATH].

You want, I think, to find the first term, [MATH]a[/MATH], given [MATH]n, r,[/MATH] and [MATH]S_n[/MATH]. For your example, given [MATH]n=5, r=2, S_n=500[/MATH], the formula says that [MATH]500=a\frac{1-2^{6}}{1-2}[/MATH], so [MATH]63a = 500[/MATH], and [MATH]a=\frac{500}{63} = 7.9365...[/MATH]
The same link shows you how this formula is derived, which is the method you are trying to reconstruct.