Question

if a²,b²,c² are in A.P prove that a/b+c,b/c+a,c/a+b are in A.P​

Answer 1

Answer:

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

Answer 2

If a^2 , b^2 , c^2 are in A.P then

b^2 - a^2 = c^2 - b^2

i.e, 2b^2 = c^2 + a^2

If a/b+c , b/c+a , c/a+b are in A.P then,

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

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

b^2+bc -(ac+ ac^2)/(c+a)(b+c) = c^2+ac -(ba+b^2)/(a+b)(c+a)

b^2+bc-ac-ac^2/(b+c) = c^2+ac-ba-b^2/(a+b)

then put2b^2=c^2+a^2 or c^2=2b^2-a or a^2=2b^2-ca^2

then it will be equal....

(a+b)(b^2+bc-ac-ac^2) = (b+c)(c^2+ac-ba-b^2)

ab^2+abc-a^2c-a^2c^2+b^3+b^2c-abc-abc^2=

bc^2+abc-b^2a-b^3+c^3+ac^2-abc-b^2c