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