The correct option is C H.P
Given a1,a2,a3,...,an are in H.P.
⇒1a1,1a2,1a3,....,1an are in A.P
⇒a1+a2+a3+....+ana1,a1+a2+a3+....+ana2,a1+a2+a3+....+ana3,....,a1+a2+a3+....+anan are in A.P
Since we multiply fix quantity in A.P still it remains A.P.
⇒a1+(a2+a3+....+an)a1,a2+(a1+a3+....+an)a2,a3+(a1+a2+....+an)a3,....,an+(a1+a2+a3+....+an−1)an are in A.P
⇒1+(a2+a3+....+an)a1,1+(a1+a3+....+an)a2,1+(a1+a2+....+an)a3,....,1+(a1+a2+a3+....+an−1)an are in A.P
⇒1+{(a2+a3+....+an)a1,(a1+a3+....+an)a2,(a1+a2+....+an)a3,....,(a1+a2+a3+....+an−1)an} are in A.P
⇒(a2+a3+....+an)a1,(a1+a3+....+an)a2,(a1+a2+....+an)a3,....,(a1+a2+a3+....+an−1)an are in A.P.
⇒a1a2+a3+....+an,a2a1+a3+....+an,a3a1+a2+....+an,....,ana1+a2+....+an−1 are in H.P.