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?

multiplying each term with xyz will also produce an AP

$\frac{xyz}{x},\frac{xyz}{y},\frac{xyz}{z}arealsoinAP\phantom{\rule{0ex}{0ex}}yz,xz,xyareinAP$

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

$\frac{x+y+z}{x},\frac{x+y+z}{y},\frac{x+y+z}{z}areinAP\phantom{\rule{0ex}{0ex}}\frac{x+y+z}{x}-1,\frac{x+y+z}{y}-1,\frac{x+y+z}{z}-1areinAP[subtracting1fromeachterm\phantom{\rule{0ex}{0ex}}\frac{y+z}{x},\frac{x+z}{y},\frac{x+y}{z}areinAP\phantom{\rule{0ex}{0ex}}$

hope this helps you

**
**