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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 smile )

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

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,403

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.

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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 dizzy

Try this instead:

Last edited by NullRoot (2007-12-07 00:30:32)


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.

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