a,b,c are ub A.P. the prove that(i)a^2(b+c), b^2(c+a), c^2(a+b) are in A.P.(ii) (ab+ac)/bc, (bc+ba)/ca, (ca+bc)/ab are in A.P.

1)  if a²(b+c) , b²(a+c), c²(a+b)  are in AP then

           2b²(a+c)  =  a²(b+c) + c²(a+b)        (we have to prove this)        ..............................1

 RHS:  

               = a² (b+c) + c² (a+b)

               = a²b +a²c +c²a +c²b

               =b(a² +c²)   +  ac(a+c)  

since a,b,c are in AP so b=a+c/2

   RHS =     (a+c)(a² +b²)/2 + ac(a+c)

          =(a+c)(a² +b² +2ac)/2

          =(a+c)³ /2           or   2(a+c) b²                 (after putting a+c =b)

 hence proved

2)      (ab+ac)/bc , (bc+ba)/ac ,(ca+bc)/ab are in AP so

                 2(bc+ba)/ac    =    (ab+ac)/bc + (ac+bc)/ab          we have to prove this

         multiplying the equation by abc

      now we get

                  2b²(a+c) = a²(b+c) + c²(a+b)               we have to prove this

 this expression is same as of eq 1 of previous ans ,so now  take RHS and prove as i have done in previous ans...

Approved

Two results to remember:

If k is a constant and a,b,c are in AP, the following are also in AP-

1) ka,kb,kc

2) a+k, b+k, c+k

Using this,