 # 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.

11 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

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...

11 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, 