Suppose `'a'` is a fixed real number such that `(a-x)/(p x)=(a-y)/(q y)=(a-z)/(r z)` if `p,q, r` are in `AP` then `x,y,z` all are in

Suppose a' is a fixed real number such that (a-x)/(px)=(a-y)/(qy)=(a-z)/(rz) if p,q,r are in AP then x,y,z all are in

Suppose a is a fixed real number such that (a - x)/(px) = (a - y)/(qy) = (a - z)/(rz) If p,q,r, are in A.P., then prove that x,y,z are in H.P.

If (a-x)/(px)=(a-y)/(qy)=(a-z)/(r) andp,q, andr are in A.P.,then prove that x,y,z are in H.P.

Let a(a!=0) is a fixed real number and (a-x)/(px)=(a-y)/(qy)=(a-z)/(rz). If p,q,r are in A.P.show that (1)/(x),(1)/(y),(1)/(z) are in A.P.

If (x+y)/(1-xy),y,(y+z)/(1-yz) be in A.P., " then " x,(1)/(y),z will be in

Let p, q, r epsilon R be such that 2q = p + r and (2017-x)/(px) = (2017-y)/(qy) = (2017-z)/(rz) , then the correct relation between x,y,z is

STATEMENT-1 : If log (x + z) + log (x -2y +z) = 2 log (x -z) then x,v,z are in H.P. STATEMENT-2 : If p , q , r in AP and (a -x)/(px) = (a-y)/(qy) = (a-z)/(rz) , then x, y, z are in A.P. STATEMENT-3 : If (a + b)/(1 - ab), b, (b + c)/(1 - bc) are in A .P. then a, (1)/(b) , c are in H.P.

If reciprocals of (y-x),2(y-a),(y-z) are in A.P then prove that x-a, y-a, z-a are in G.P