If $A,G$ and $H$ are respectively arithmetic, geometric and harmonic means between $a$ and $b$ both being unequal and positive, then $A=\dfrac{a+b}{2}\Rightarrow a+b=2A, G=\sqrt{ab} \Rightarrow ab={G}^{2}$ and $H=\dfrac{2ab}{a+b}\Rightarrow{G}^{2}=AH$ From the above discussion we can say that $a,b$ are the roots of the equation ${x}^{2}-2Ax+{G}^{2}=0$ Now,quadratic equation, ${x}^{2}-Px+Q=0$ and quadratic equation $a\left(b-c\right){x}^{2}+b\left(c-a\right)x+c\left(a-b\right)=0$ have a root common and satisfy the relation $b=\dfrac{2ac}{a+c},$ where $a,b,c$ are real numbers. On the basis of the above information, answer the following questions: The ratio of A.M, G.M and H.M of the roots of the given quadratic equation is:

The given relation b equals 2ac divided by (a+c) means that a, b, and c are in harmonic progression. For the quadratic equation a(b-c)x^2 + b(c-a)x + c(a-b) = 0, you can verify by substitution that x = 1 is a root. Using the product of the roots formula for the standard quadratic ax^2 + bx + c = 0, the product of the roots here is c(a-b) divided by a(b-c), which simplifies to 1, proving that both roots are equal to 1. The arithmetic mean is (1+1)/2 = 1, the geometric mean is the square root of 1*1 = 1, and the harmonic mean is 2/(1/1 + 1/1) = 1, making their ratio 1:1:1.

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