If a,b,c be in A.P. and b,c,d be in H.P., then
ab = cd
ad = bc
bc = ad
abcd = 1
Given that a,b,c in A.P. and b,c,d in H.P.
So, 2b = a + c and c = 2bdb+d
⇒ c(b + d) = 2bd = (a + c)d ⇒ bc = ad.