SOLUTION: If 1/(y+ax),1/(2y),1/(y+az) are in arithmetic progression.prove that k^2x,y, z are in geometric progression

This is true if and only if it is also true that a = k².  You MUST
be given that, even though you didn't state it.  If "a" is not equal
to k², then your proposition is not true.

So we will assume that a=k²

,, are in arithmetic progression.

Then (2nd term)-(1st term) = (3rd term)-(2nd term)



Add  to both sides





Get LCD = y(y+ax)(y+az)

(y+ax)(y+az) - y(y+az) = y(y+ax)

Factor out (y+az) on the left side:

      (y+az)[(y+ax)-y] = y(y+ax)

        (y+az)(y+ax-y) = y²+axy

            (y+az)(ax) = y²+axy

               axy+axz = y²+axy

                   axz = y² 


To show that k^2x, y, z are in geometric progression,

we must show that this is true:



So we have to show that 

We have shown above that

                   axz = y²

Since we assume that a = k², we substitute k² for a

                  k²xz = y²

We divide both sides by k²xy

                 

                                  

                 

So it is a geometric progression if (but only if) a = k²  
                
Edwin