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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 
Try this instead:
Last edited by NullRoot (2007-12-07 23:30:32)
Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.