Given that `log_(x)y,log_(z)x,log_(y)z` are in GP, `xyz=64" and "x^(3),y^(3),z^(3)` are in A.P.
Which one of the following is correct ?
x,y and z are

Options

A. in AP only
B. in GP only
C. in both AP and GP
D. neither in AP nor in GP

Answer

C

Explanation

Given `log_(x)y,log_(z)x,log_(y)z" are in G.P."`
`:.(log_(z)x)^(2)=(log_(x)y)(log_(y)z)`
`((logx)/(logz))^(2)=((logy)/(logx))((logz)/(logy))=(logz)/(logx)`
`implies((logx)/(logz))^(3)=1`
`implieslogx=logzimpliesx=z`
`becausexyz=64`
`y=(64)/(x^(2))`
Also given `x^(3),y^(3)` and `z^(3)` are in A.P.
`:.y^(3)=(x^(3)+z^(3))/(2)=(x^(3)+x^(3))/(2)`
`y^(3)=x^(3)impliesy=x`
`impliesx=y=z`
`x.y.z=64`
`x=y=z=4`
Thus x, y, z are in A.P. and G.P. both.