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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.

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.


2 Answers

vikas askiitian expert
509 Points
10 years ago

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


               = 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...


mycroft holmes
272 Points
10 years ago

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,


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