## TrojanPoem one year ago If three consecutive terms for a geometric sequence formed a triangle sides.what is the base for the geometric sequence.

1. TrojanPoem

Assume the terms are x, y, z. As they form a geometric sequence ( x, y, z) zx = y^2 -> (1) Now we need 2 more equations.

2. ganeshie8

Let the terms be $$a, ~ar,~ ar^2$$ For these to be sides of a triangle, they must satisfy the triangle inequality relation : $a+ar \gt ar^2\tag{1}$ $ar+ar^2 \gt a\tag{2}$ $ar^2+a \gt ar\tag{3}$

3. TrojanPoem

I thought about the inequality matter. But we will get periods.

4. ganeshie8

solve the 3 inequalities and take the intersection

5. TrojanPoem

I think according to your assumption , we can take (a) as common factor, can't we ?

6. ganeshie8

divide both sides by $$a$$ so that it disappears

7. ganeshie8

you can do so because $$a$$ is assumed to be side of a triangle, so $$a\gt 0$$

8. TrojanPoem

If inequalities really help, why don't we use GM AM inequality , xy = z^2 and the triangle inequality. $(r-\frac{ 1 - \sqrt(5) }{ 2 })( r - \frac{ 1 + \sqrt(5) }{ 2 }) >$

9. ganeshie8

use it if you find it simpler than triangle inequality

10. TrojanPoem

Not easier at all. Imagine z + y + x > 3root(zyz)

11. ganeshie8

triangle inequlity does the job here

12. ganeshie8
13. TrojanPoem

what about this ? zy = x^2 z + y + x > 3 3root(xyz) from (1) z + y+ x > 3 3root(x^3) z +y + x > 3 x z + y > 2x

14. TrojanPoem

assume x, y , z = a, ar, ar^2 and use this with your inequality and maybe we will reach something.

15. ganeshie8

we're done.

16. TrojanPoem

Really ? Xd

17. TrojanPoem

brb

18. ganeshie8

$\frac{1}{\phi} \lt r\lt \phi$ where $$\phi$$ = $$\href{https:///en.wikipedia.org/wiki/Golden_ratio}{\text{golden ratio}}$$

19. TrojanPoem

Ok, I read about the golden ration but it's over my post level.