If `a,b,c` are in G.P.then `(b-a)/(b-c)+(b+a)/(b+c)=`

If a,b,c are in H.P., then (b+a)/(b-a)+(b+c)/(b-c)

If a,b,c are in H.P., then (a)/(b+c),(b)/(c+a),(c)/(a+b) will be in

If a,b,c are in G.P., then (A) (a-b)/(b-c)=a/a (B) (a-b)/(b-c)=a/b (C) (a-b)/(b-c)=a/c (D) (a-b)/(b-c)=b/a

If a,b,c are in A.P.the (a)/(bc),(1)/(c),(2)/(b) will be in a. A.P b.G.P.c.H.P.d.none of these

If a,b,c are in H.P.,then prove that (a)/(b+c-a),(b)/(a+b-c),(c)/(a+b-c) are in H.P.

If a,b,c,d are in G.P.then (a^(2)-b^(2)),(b^(2)-c^(2)),(c^(2)-d^(2)) are in

If a,b,c are in G.P. then prove that (1)/(a+b),(1)/(2b),(1)/(b+c) are also in A.P.

If a,b,c,d,e are in H.P., then a/(b+c+d+e), b/(a+c+d+e),c/(a+b+d+e), d/(a+b+c+e), e/(a+b+c+d) are in (A) A.P. (B) G.P. (C) H.P. (D) none of these

If a,b,c are in G.P., then show that a(b-c)^2=c(a-b)^2