Question

If three consecutive terms for a geometric sequence formed a triangle sides.what is the base for the geometric sequence.

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

solve the 3 inequalities and take the intersection

Answer 5

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

Answer 6

divide both sides by \(a\) so that it disappears

Answer 7

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

Answer 8

If inequalities really help, why don't we use GM AM inequality , xy = z^2 and the triangle inequality.

\[(r-\frac{ 1 - \sqrt(5) }{ 2 })( r - \frac{ 1 + \sqrt(5) }{ 2 }) > \]

Answer 9

use it if you find it simpler than triangle inequality

Answer 10

Not easier at all.

Imagine z + y + x > 3root(zyz)

Answer 11

triangle inequlity does the job here

Answer 12

http://www.wolframalpha.com/input/?i=solve+r%5E2%3C1%2Br%2C+1%3Cr%2Br%5E2%2C+r%3C1%2Br%5E2

Answer 13

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

Answer 14

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

Answer 15

we're done.

Answer 16

Really ? Xd

Answer 17

brb

Answer 18

\[\frac{1}{\phi} \lt r\lt \phi\]

where \(\phi\) = \(\href{https:///en.wikipedia.org/wiki/Golden_ratio}{\text{golden ratio}}\)

Answer 19

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