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 )
Let the numbers be 1/a, 1, a, and a².
1/a + 1 + a + a²=13.
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.
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.
Edit: Oops. forgot about the Geometric Progression
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 cant cope with is therefore your own problem.