Question 251659
Q1.
    a,b,c,d are in H.P
  Then 1/a, 1/b, 1/c, 1/d will be in A.P
Therefore  2(1/b-1/a) = (1/c-1/a)  and 2(1/d-1/c) = (1/d-1/b) 
           2(a-b)/ba  = (a-c)/ca   and 2(c-d)/dc  = (b-d)/db
Multiplying these two equations, we get
          4(a-b)(c-d)/abcd = (a-c)(b-d)/abcd
          4(a-b)(c-d) = (a-c)(b-d)

Q2.  
 Given a+b,b+c,c+a are in H.P.
Therefore, 1/(a+b) + 1/(c+a)      = 2/(b+c)
             (c+a+a+b)/(a+b)(c+a) = 2/(b+c)
             (c+2a+b)(b+c)        = 2(a+b)(c+a)
      bc+c^2+2ab+2ac+b^2+bc       = 2ac+2a^2+2bc+2ab
                 c^2+b^2          = 2a^2
                   (b^2+c^2)/2    =a^2
Therefore b^2,a^2,c^2 are in A.P

Q3 Given, x,y,z are in A.P. 
Therefore, x+z = 2y  ......(1)
Also, x,xy,z are in G.P
Therefore, (xy)^2 = xz
           x^2y^2 = xz
             xy^2 = z  ....(2)
Next, we will prove x,x^2y,z are in H.P
Now, 1/x+1/z = (z+x)/xz
             = 2y/x.xy^2
             = 2/x^2y
Therefore, 1/x+1/z =2/x^2y
Therefore,x,x^2y,z are in H.P