Question

if a2, b2,c2 are in a A.P .then prove that the following are also in A.P (i) 1/b+c ,1/c+a ,1/a+b (ii) a/b+c , b/a+c ,c/b+a

Answer 1

 - a)/(c + b) = (c - b)/(b + a)

Dividing both sides by (c + a),

(b - a)/{(c + b)(c + a)} = (c - b)/{(b + a)(c + a)}

{(b + c) - (c + a)}/{(b + c)(c + a) = {(c + a) - (a + b)}/{(a + b)(c + a)}

{1/(c + a)} - {1/(b + c)} = {1/(a + b)} - {1/(c + a)}

{2/(c + a)} = {1/(a + b)} - {1/(b + c)}

Therefore 1/(a + b), 1/(b + c), 1/(c + a) are in AP.