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## #1 2007-12-07 07:01:48

100'
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### 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 )

## #2 2007-12-07 16:11:32

ganesh
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### 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.

Character is who you are when no one is looking.

## #3 2007-12-07 20:27:21

mathsyperson
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### 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.

## #4 2007-12-07 21:02:34

NullRoot
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### Re: Geometric Progression Problem

Edit: Oops. forgot about the Geometric Progression