Math Question #1. ICTM R 96 AA. Math Team.?

Question

Let a,b,x,y be positive real numbers such that (a^3)/b,x,y,(b^3)/a, are four consecutive terms of a geometric sequence. Find the ordered pair (x,y). Express your answer using positive rational exponents.

Answer 1

(a^3)/b,x,y,(b^3)/a
geometric progression (Note, it doesn't matter where you start in a prgression, you can always get these numbers)
c,cr, cr^2,cr^3

therefore (a^3)/b=c, x=cr, y=cr^2, (b^3)/a=cr^3

((a^3)/b)r^3=(b^3)/a

r=(b/a)^4/3

therefore

x=(a^3)/b*(b/a)^4/3
=(a^(3-4/3))(b^(4/3-1))
=(a^5/3)*(b^1/3)

y=(a^3)/b*(b/a)^8/3
=(a^(3-8/3))(b^(8/3-1)
=(a^1/3)*(b^5/3))

Wow, this is pretty elegant and seem obvious in hindsight. Nice problem

Answer 2

I assumed that the question was asking for specific numbers to represent x and y.

First, I called the common ratio "z".

So that means:

a^3/b * z^3 = b^3/a
and
z^3 = b^4/a^4

If you substitute some numbers in you will find that what works best is:

z= 16 b= 8 a=1

plug that in and the series is;

1/8, 2, 32, 512

(x,y) = (2,32)

In retrospect, I don't think that's what the question asked for, but ohh well.