You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**100'****Member**- Registered: 2007-12-06
- Posts: 8

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 )

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,397

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.

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.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

Offline

**NullRoot****Member**- Registered: 2007-11-19
- Posts: 162

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.

Offline

Pages: **1**