If 1a+1a−2b+1c+1c−2b=0 and a,b,c are not in A.P then:
a,b,c are in G.P
a,b2,c are in A.P
a,b2,c are in H.P
a,2b,c are in H.P
1a+1a−2b+1c+1c−2b=02(a−b)a(a−2b)+2(c−b)c(c−2b)=0a−ba(a−2b)+c−bc(c−2b)=0 Now, we can verify the result that a,2b,c are in H.P