If `x^2+9y^2+25 z^2=x y z((15)/2+5/y+3/z),t h e n` `x ,y ,a n dz` are in H.P. b. `1/x ,1/y ,1/z` are in A.P. c. `x ,y ,z` are in G.P. d. `1/a+1/d=1/b=1/c`

Options

A. x,y and z are in H.P
B. `1/x, 1/y,1/z ` are in G.P
C. x,y,z are in G.P
D. `1/x,1/y,1/z` are in G.P

Answer

A, C

Explanation

`x^(2)+9y^(2)+25z^(2)=15yz+5zx+3xy`
`rArr(x)^(2)+(3y)^(2)+(5z)^(2)-(x)(3y)-(3y)(5z)-(x)(5z)=0`
`rArr1/2[(x-3y)^(2)+(3y-5z)^(2)+(x-5z)^(2)]=0`
`rArrx-3y=0,3y-5z=0,x-5z=0`
x=3y=5z
`rArrx:y:z=1/1:1/3:1/5`
Therefore,1/x,1/y and 1/z are in A.P. and x,y, and z are in H.P.