Conditional identity consisting of AP and GP

[rama]

Homework Statement

x,y,z are three terms in GP and a,b,c are three terms in AP
prove that (x^b÷x^c)(y^c÷y^a)(z^a÷z^b)=1

Homework Equations

The Attempt at a Solution

(x^b-c)(y^c-a)(z^a-b)

since x y z are in GP
x^b-c÷y^c-a=y^c-a÷z^a-b
(x^{b- c})(z^a-b)=y^c-a(y^c-a)

 

[Mark44]

Homework Statement

x,y,z are three terms in GP and a,b,c are three terms in AP
prove that (x^b÷x^c)(y^c÷y^a)(z^a÷z^b)=1

Homework Equations

The Attempt at a Solution

(x^b-c)(y^c-a)(z^a-b)

since x y z are in GP
x^b-c÷y^c-a=y^c-a÷z^a-b
(x^{b- c})(z^a-b)=y^c-a(y^c-a)

You are given that x, y, and z are in geometric progression. Did you use that fact in your work?

Also, a, b, and c are in arithmetic progression. Did you use that fact in your work?

 

[rama]

got it thank you, I seem to be posting here without thinking hard
next time I won't post without thinking out all options sorry