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#1 2006-03-24 17:08:10
- RickyOswaldIOW
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- Registered: 2005-11-18
- Posts: 212
Geometric Progression
In a GP, u3 = 18 and u5 = 162. Find u1.
I know that in a GP, ux / u(x-1) = ratio (r). I deduce from that;
u5 / n == n / u3.
so 162 / n = n / 18
I then try to balance this - I think that's where I am going wrong.
162 = 2n/18
2916 = 36n
n = 81
Which is of course incorrect. Where am I going wrong?
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#2 2006-03-24 17:10:33
- RickyOswaldIOW
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Re: Geometric Progression
162 = n² / 18n
Is this along the right track?
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#3 2006-03-24 17:37:20
- Jai Ganesh
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- Registered: 2005-06-28
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Re: Geometric Progression
In a GP, u3 = 18 and u5 = 162. Find u1.
rickyoswaldiow,
u3*n=u4, u3*n²=u5
That is, n²=u5/u3=162/18=9, n=3.
u1*n²=u3. Therefore, u1*9=18, u1=2.
The Geometric Progression is 2, 6, 18, 54, 162, 486.....
The first term is 2 and the common ratio is 3. 
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#4 2006-03-25 01:25:58
- RickyOswaldIOW
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- Registered: 2005-11-18
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Re: Geometric Progression
Thanks Ganesh. Will have to study this a bit more 
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#5 2006-03-29 03:59:45
- RickyOswaldIOW
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- Registered: 2005-11-18
- Posts: 212
Re: Geometric Progression
ux / u(x-1) = ratio
Is this correct?
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#6 2006-03-29 06:10:03
- Ikcelaks
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- Registered: 2006-03-13
- Posts: 8
Re: Geometric Progression
You setup the equation right, but your algebra is incorrect.
162 / n = n / 18
162 = n^2 / 18 (this is where you went wrong)
162 * 18 = n^2
2916 = n^2
54 = n
You should be able to work out the rest for yourself.
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