Question

a,b,c are in Ap then prove a(b+c)/bc,b(c+a)/ca,c(a+b)/ab are in ap

Answer 1

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...