If 1/a,1/b,1/c are in a.p show that a(b+c),b(c+a),c (a+b) are in a.p

Answer : Given :If 1/a,1/b,1/c are in a.p

To prove : a(b+c),b(c+a),c (a+b) are also in a.p

 1/a, 1/b, 1/c are in AP

=> (1/a) + (1/c) = 2(1/b)      {x, y, z are in AP iff : x + z = 2y}

=> (c+a) / (ca) = 2/b

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

 (bc+ab) = 2(ca) ................... (1) 

 

now ,  a(b+c) + c(a+b)

= ab + ca + ca + bc

= (bc+ab) + 2(ca)

= (bc+ab) + (bc+ab) ............ from (1)

= 2(bc+ab)

=> a(b+c) + c(a+b) = 2 b(c+a)  {x, y, z are in AP iff : x + z = 2y}    

 a(b+c),b(c+a),c (a+b) are in a.p 

Hence Proved