Question

if a square,b square,c square is in A.P Then proof that 1/b+c,1/c+a,1/a+b will also in A.P plz solve anybody else ......plzzzz
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Answer 1

The answer to this problem is very simple. You start with what is given.

  a2 b2 c2 are in A.P.

To Prove: 1b+c, 1c+a, 1a+b1b+c, 1c+a, 1a+b are in A.P.

Proof: 2b2 = a2 + c2 [as a2 b2 c2 are in A.P.]

=> b2 + b2 = a2 + c2 => b2 - a2 = c2 - b2

 => (b-a) (b+a) = (c-b) (c+b)

=> (b−a)(c+b)=(c−b)(b+a)(b−a)(c+b)=(c−b)(b+a)

Divide both side by 1(c+a)1(c+a),

We get

=> (b−a)(c+b)×(c+a)=(c−b)(b+a)×(c+a)(b−a)(c+b)×(c+a)=(c−b)(b+a)×(c+a)

=> (b−a+c−c)(c+b)×(c+a)=(c−b+a−a)(b+a)×(c+a)(b−a+c−c)(c+b)×(c+a)=(c−b+a−a)(b+a)×(c+a)

=> (b+c)−(c+a)(c+b)×(c+a)=(c+a)−(a+b))(b+a)×(c+a)(b+c)−(c+a)(c+b)×(c+a)=(c+a)−(a+b))(b+a)×(c+a)

=> 1(c+a)−1(c+b)=1(a+b)−1(c+a)1(c+a)−1(c+b)=1(a+b)−1(c+a)

=> 2(c+a)=1(a+b)+1(c+b)2(c+a)=1(a+b)+1(c+b)

Hence by this equation we, can say that 1b+c, 1c+a, 1a+b1b+c, 1c+a, 1a+bare in A.P.