A) AP
B) HP
C) GP
D) both AP and GP
E) none of these
Correct Answer: B
Solution :
Given expression is \[a(b-c){{x}^{2}}+b(c-a)xy+c(a-b){{y}^{2}}\] Put\[x=1\]in the given expression \[\therefore \] \[a(b-c)+b(c-a)+c(a-b)\] \[=ab-ac+bc-ab+ca-cb\] \[=0\] Since, it is a perfect square, therefore both roots are same i. e., 1,1. \[\therefore \]Sum of roots \[=-\frac{b(c-a)}{a(b-c)}\] \[\Rightarrow \] \[1+1=\frac{-bc+ab}{ab-ac}\] \[\Rightarrow \] \[2ab-2ac=-bc+ab\] \[\Rightarrow \] \[ab+bc=2ac\] \[\Rightarrow \] \[b(a+c)=2ac\] \[\Rightarrow \] \[b=\frac{2ac}{a+c}\] Hence, a, b, c are in HP.You need to login to perform this action.
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