If 1bāa+1bāc=1a+1c, then a,b,c, are in
1b−a+1b−c=1a+1c 2b−a−c(b−a)(b−c)=a+cac 2abc−a2c−ac2=(b2−ab−bc+ac)(a+c) 2abc−a2c−ac2=b2a−a2b−abc+a2c+cb2−abc−bc2+ac2 ∴ 4abc−2a2c−2ac2−b2a+a2b−cb2+bc2=0 ⇒(2b−1a−1c)=0 2b=1a+1c answer H.P
If a,b,c are in A.P., then abc,1c,2b are in
If 1b−a+1b−c=1a+1c, then a,b,c, are in
If the roots of a(b−c)x2+b(c−a)x+c(a−b) = 0 be equal then a,b,c are in
if a,b,c ϵ R,a,b,c ≠ 0, ∑ a2 = ∑ab, then a,b,c are in