The correct option is A (i) and (ii) both correct
Given a2,b2.c2 are in A.P.
⇒a2+(ab+bc+ca),b2+(ab+bc+ca),c2+(ab+bc+ca) are in A.P.
⇒(a+b)(a+c),(b+c)(b+a),(c+a)(c+b) are in A.P.
⇒1b+c,1c+a,1a+b are in A.P.
[Divide by (a+b)(b+c)(c+a)]
Again, a2,b2,c2 are in A.P.
⇒1b+c,1c+a,1a+b are in A.P.
⇒a+b+cb+c,a+b+cc+a,a+b+ca+b are in A.P.
⇒ab+c+1,bc+a+1,ca+b+1 are in A.P.
⇒ab+c,bc+a,ca+b are in A.P.