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## #1 2007-12-06 08:01:48

100'
Member
Registered: 2007-12-06
Posts: 8

### Geometric Progression Problem

Four numbers form a geometric progression. The sum of the four numbers is 13, the sum of their squares is 1261. Find the four numbers.

(note, this had me stumped for ages, have fun )

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## #2 2007-12-06 17:11:32

ganesh
Registered: 2005-06-28
Posts: 24,300

### Re: Geometric Progression Problem

Let the numbers be 1/a, 1, a, and a².
Therefore,
1/a + 1 + a + a²=13.
1/a²+1+a²+a^4=1261.
Square the first equation, eleminate a^4.
Thereafter, you are left with a quadratic equation which, I guess, can be solved easily.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #3 2007-12-06 21:27:21

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Geometric Progression Problem

Those numbers should all be multiplied by an unknown constant to be completely general.
Also, squaring the first equation would get an expression involving every power of x between -2 and 4, which means that after cancelling with the second one you'd have a quintic thingy.

Why did the vector cross the road?
It wanted to be normal.

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## #4 2007-12-06 22:02:34

NullRoot
Member
Registered: 2007-11-19
Posts: 162

### Re: Geometric Progression Problem

Edit: Oops. forgot about the Geometric Progression