# common ratio of the seqence

• Mar 2nd 2011, 07:20 PM
jam2011
common ratio of the seqence
"the length of the sides in a right triangle form 3 consecutive terms of a geometric sequence. Find the common ratio of the sequence"

i don't know how to start this problem. please give me an idea how to answer this problem.

• Mar 2nd 2011, 08:17 PM
Wilmer
Quote:

Originally Posted by jam2011
"the length of the sides in a right triangle form 3 consecutive terms of a geometric sequence. Find the common ratio of the sequence"

a = short side of triangle (1st term)
r = ratio

a^2 + (ar)^2 = (ar^2)^2 (pythagorean theorem)
a^2 + a^2 r^2 = a^2 r^4
a^2 r^4 - a^2 r^2 - a^2 = 0
a^2(r^4 - r^2 - 1) = 0

Take over...
• Mar 2nd 2011, 08:42 PM
rcs
does (r^4 - r^2 - 1) factorable? because (r^2 - 1 ) (r^2 + 1) does not fit to it... how is it done?
• Mar 3rd 2011, 05:16 AM
Wilmer
• Mar 3rd 2011, 06:49 AM
scounged
I guess you substitute r^2 for t(or whatever you'd like to call it), and then solve it like any other quadratic.
• Mar 3rd 2011, 03:52 PM
rcs
i got it thanks alot
• Mar 4th 2011, 06:08 AM
jam2011
thanks a lot...Bless You all
• Mar 4th 2011, 08:00 AM
Wilmer
Quote:

Originally Posted by scounged
I guess you substitute r^2 for t(or whatever you'd like to call it), and then solve it like any other quadratic.

Yes; or this way:
(r^2)^2 - r^2 - 1 = 0

r^2 = [1 +- SQRT(1 + 4)] / 2

r = SQRT{[1 +- SQRT(5)] / 2}