(a,b,c) in A.P. ⇒ (2b=a+c)

(b,c,d) in H.P. ⇒ (\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d}) in A.P. ⇒ (2\cdot\dfrac{1}{c}=\dfrac{1}{b}+\dfrac{1}{d})

[ \dfrac{2}{c}=\dfrac{d+b}{bd}\implies 2bd=c(b+d) ]

Need (ad=bc). From (2b=a+c) we get (a=2b-c). Substitute and use the H.P. relation to verify (ad=bc).

Proved.