If 1/x,1/y,1/z are AP, show that:

(i) yz,zx,xy are in AP

(ii) xy,zx,yz are in AP

(iii) y+z/x,z+x/y,x+y/z are in AP

I got the first one but how to do the other two?

given: 1/x , 1/y and 1/z are in AP.
multiplying each term with xyz will also produce an AP
$$\begin{gathered} \frac{xyz}{x} ,\frac{xyz}{y},\frac{xyz}{z} are also in AP \\ yz , xz , xy are in AP \end{gathered}$$
since reverse of AP is also in AP.
therefore
xy , xz , yz are also in AP

multiplying each term of given AP by (x+y+z)
​therefore
$$\begin{gathered} \frac{x+y+z}{x} ,\frac{x+y+z}{y} , \frac{x+y+z}{z} are in AP \\ \frac{x+y+z}{x}-1 ,\frac{x+y+z}{y}-1 , \frac{x+y+z}{z}-1 are in AP [subtracting 1 from each term \\ \frac{y+z}{x} ,\frac{x+z}{y},\frac{x+y}{z} are in AP \end{gathered}$$

hope this helps you