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## #1 2006-03-25 16:08:10

RickyOswaldIOW
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### 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?

Aloha Nui means Goodbye.

## #2 2006-03-25 16:10:33

RickyOswaldIOW
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### Re: Geometric Progression

162 = n² / 18n

Is this along the right track?

Aloha Nui means Goodbye.

## #3 2006-03-25 16:37:20

ganesh
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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.

Character is who you are when no one is looking.

## #4 2006-03-26 00:25:58

RickyOswaldIOW
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### Re: Geometric Progression

Thanks Ganesh.  Will have to study this a bit more

Aloha Nui means Goodbye.

## #5 2006-03-30 02:59:45

RickyOswaldIOW
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### Re: Geometric Progression

ux / u(x-1) = ratio

Is this correct?

Aloha Nui means Goodbye.

## #6 2006-03-30 05:10:03

Ikcelaks
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### 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.