Question Number 101563 by harckinwunmy last updated on 03/Jul/20

Answered by bramlex last updated on 03/Jul/20

(2) (1/(y+z)) , (1/(z+x)) , (1/(x+y)) →AP ⇒(2/(z+x)) = (1/(y+z)) + (1/(x+y)) (2/(z+x)) = ((x+2y+z)/((y+z)(x+y))) 2(y^2 +yx+zx+zy)=(z+x)(x+2y+z) 2y^2 +2yx+2zx+2zy = zx+2zy+z^2 +x^2 +2xy+xz 2y^2 = x^2 +z^2 ⇒so x^2 , y^2 ,z^2 →AP

Answered by PRITHWISH SEN 2 last updated on 03/Jul/20

∵ a,b and c are the three cosecutive terms of a linear sequence ∴ a+c=2b now a((1/b)+(1/c))+c((1/a)+(1/b)) =((b(a^2 +c^2 )+ac(a+c))/(abc)) =((b(a^2 +c^2 )+2abc)/(abc)) =(((a+c)^2 )/(ac)) = ((2b(a+c))/(ac)) =2b((1/a)+(1/c)) ∴ the terms are also the 3 consecutive terms of the same linear sequence proved