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#1 2006-03-24 17:08:10

RickyOswaldIOW
Member
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
Member
Registered: 2005-11-18
Posts: 212

Re: Geometric Progression

162 = n² / 18n

Is this along the right track?


Aloha Nui means Goodbye.

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#3 2006-03-24 17:37:20

Jai Ganesh
Administrator
Registered: 2005-06-28
Posts: 45,835

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. smile


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.

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#4 2006-03-25 01:25:58

RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212

Re: Geometric Progression

Thanks Ganesh.  Will have to study this a bit more neutral


Aloha Nui means Goodbye.

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#5 2006-03-29 03:59:45

RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212

Re: Geometric Progression

ux / u(x-1) = ratio

Is this correct?


Aloha Nui means Goodbye.

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#6 2006-03-29 06:10:03

Ikcelaks
Member
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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