Question

If a2 , b2 , c2 are in AP then prove that 1 b+c , 1 c+a , 1 a+b are in AP.

Answer 1

Step-by-step explanation:

If a2, b2, c2 are in AP, then

2b2 = a2 + c2

b2 + b2 = a2 + c2

b2 - a2 = c2 - b2

(b - a)(b + a) = (c - b)(c + b)

(b - 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.