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Geometric Progression

How would you go about solving:

108=51+r+5(1+r)2+5(1+r)3+....+105(1+r)6 108 = \frac{5}{1+r} + \frac{5}{{(1+r)}^2} + \frac{5}{{(1+r)}^3}+....+\frac{105}{{(1+r)}^6}

for r?

This is what I have done so far:

108=5(1−1(1+r)71−11+r)+100(1+r)6 108 = 5 \left(\dfrac{1-\frac{1}{{(1+r)}^7}}{1 - \frac{1}{1+r}} \right) + \dfrac{100}{{(1+r)}^6}

108=5r((1+r)7−1)+100r(1+r)6r108 = \dfrac{5r((1+r)^7-1) + 100r}{(1+r)^6r}

Bit stuck. Any light? Or is it even possible.

Thanks
(edited 9 years ago)
Original post by 2710


Bit stuck. Any light? Or is it even possible.

Thanks


Is that the whole question - or something you've derived?
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Original post by ghostwalker
Is that the whole question - or something you've derived?


Something I have partially derived. It's just a simple solve for r. No extra information. Was just wondering if my math was deteriorating or that it's actually a hard question xD
Original post by 2710
Something I have partially derived. It's just a simple solve for r. No extra information. Was just wondering if my math was deteriorating or that it's actually a hard question xD


It's certainly hard to get all the possible values for r, I think.

You have some errors in your working as well. The first term is 1/(1+r) rather than 1, once you've taken out the factor of 5, and there are only 6 terms, not 7.
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Original post by ghostwalker
It's certainly hard to get all the possible values for r, I think.

You have some errors in your working as well. The first term is 1/(1+r) rather than 1, once you've taken out the factor of 5, and there are only 6 terms, not 7.


Ah yes.

So essentially it is the case of solving a very high polynomial isn't it. I guess it may not be solvable after all except for trial and error.
Original post by 2710
Ah yes.

So essentially it is the case of solving a very high polynomial isn't it. I guess it may not be solvable after all except for trial and error.


Or software, yes.

As you've derived it, is there any way you can bypass that equation entirely. It's rather unusual to have a last term that doesn't follow the GP format.
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Original post by ghostwalker
Or software, yes.

As you've derived it, is there any way you can bypass that equation entirely. It's rather unusual to have a last term that doesn't follow the GP format.


It is unusual. But even if it were the same we would still be stuck solving a polynomial of size 6 or greater right.

And I guess this goes for any geometric progression sums. As normally in simple maths you are given the r and asked to work out the summation. If you flip it and ask for the summation, it becomes much harder, since you begin dealing with very big polynomials. Is my conclusion
Original post by 2710
It is unusual. But even if it were the same we would still be stuck solving a polynomial of size 6 or greater right.


Yep


And I guess this goes for any geometric progression sums. As normally in simple maths you are given the r and asked to work out the summation. If you flip it and ask for the summation, it becomes much harder, since you begin dealing with very big polynomials. Is my conclusion


Yes, you're being asked to find the roots of a polynomial of degree n+1.
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Original post by 2710
Something I have partially derived. It's just a simple solve for r. No extra information. Was just wondering if my math was deteriorating or that it's actually a hard question xD


If you're working on a real question you might want to post it in case you have missed something. As people have said, what you're trying to isn't easy, or even possible in an exact form.
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Original post by davros
If you're working on a real question you might want to post it in case you have missed something. As people have said, what you're trying to isn't easy, or even possible in an exact form.


It was to do with an economics question that my friend was doing regarding bond yields, she just posted something similar to me and her working and that got me thinking. If I am curious enough and I have the time I will ask her to send me the exact question.

But what I really wanted to know is what I posted originally, to do with GPs and trying to find the r rather than the sum.

But thanks anyways

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