Math Is Fun Forum

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 )

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,422

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.

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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.

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

NullRoot

Member

Registered: 2007-11-19

Posts: 162

Re: Geometric Progression Problem

Edit: Oops. forgot about the Geometric Progression

Try this instead:

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

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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.