Question

a/b+c, b/c+a, c/a+b are in AP and a+b+c not equal to zero. Then prove that 1/b+c,1/c+a,1/a+b are in AP

Answer 1

a/(b + c), b/(c + a) , c/(a + b) are in AP
⇒a/(b + c) +1 , b/(c + a) + 1, c/(a + b) +1 are in AP
⇒(a + b + c)/(b + c), (b + c + a)/(c + a) , (c + a + b)/(a + b) are in AP
⇒ (a + b + c)/(b + c) , (a + b + c)/(c + a ) , (a + b + c)/(a + b) are in AP
dividing by (a + b + c) in all terms ,

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