Question

If A, G and H are respectively the AM< GM and HM of three positive numbers a, b and c; find the equations whose roots are a, b and c.
\[\begin{align}
  & (a)G{{x}^{3}}-3AG{{x}^{2}}+3{{A}^{3}}x-{{A}^{3}}{{G}^{3}}=0 \\
 & (b)H{{x}^{3}}-3AH{{x}^{2}}+3{{G}^{3}}x-H{{G}^{3}}=0 \\
 & (c)A{{x}^{3}}-3AG{{x}^{2}}+3{{G}^{2}}H-{{H}^{3}}=0 \\
 & (d)A{{x}^{3}}+HG{{x}^{2}}-GHx+G=0 \\
\end{align}\]

Answer 1

Hint: Assume the equations as $ {{x}^{3}}-\left( a+b+c \right){{x}^{2}}+\left( ab+bc+ca \right)x-abc $ . Now apply the formula $ A=\dfrac{a+b+c}{3} $ and substitute the values of (a+b+c) in the assumed equation. Similarly, calculate the value of GM, given as $ G={{\left( abc \right)}^{\dfrac{1}{3}}} $ and HM given as $ \dfrac{1}{H}=\dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{3} $ . Substitute all the values of (ab+bc+ca) and abc in terms of H and G in the equation and get the answer.

Complete step by step answer:
Here, we have been given that A, G, and H are representing the arithmetic, geometric and harmonic means of three numbers a, b and c. We have to find an equation whose roots are a, b and c.
Since there are three roots of the equation to be determined, the equation must be a cubic equation. We know that the equations of a cubic polynomial is given in general form as $ {{x}^{3}}-\left( a+b+c \right){{x}^{2}}+\left( ab+bc+ca \right)x-abc $ , where a, b and c are the roots of the equation. In the above relation, we have
(a+b+c) = Sum of roots
(ab+bc+ca) = Sum of the product of roots taken two at a time
Abc = Product of roots
Now, let us replace a, b and c with the given values of AM, GM and HM.
We know that the arithmetic mean (AM) of three numbers a, b and c is given as
  $ \begin{align}
  & A=\dfrac{a+b+c}{3} \\
 & \Rightarrow a+b+c=3A\ldots \ldots \ldots \left( i \right) \\
\end{align} $
Now, the geometric mean (GM) of three numbers a, b and c is given as
  $ G={{\left( abc \right)}^{\dfrac{1}{3}}} $
Cubing both sides, we get
  $ abc={{G}^{3}}\ldots \ldots \ldots \left( ii \right) $
At last, we know that the harmonic mean (HM) of three numbers a, b and c is given as
  $ \begin{align}
  & \dfrac{1}{H}=\dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{3} \\
 & \Rightarrow \dfrac{1}{H}=\dfrac{ab+bc+ac}{3abc} \\
\end{align} $
Using equation (ii), we get
  $ \begin{align}
  & \Rightarrow \dfrac{1}{H}=\dfrac{ab+bc+ac}{3{{G}^{3}}} \\
 & \Rightarrow ab+bc+ac=\dfrac{3{{G}^{3}}}{H}\ldots \ldots \ldots \left( iii \right) \\
\end{align} $
So, substituting the values of (a+b+c), (ab+bc+ac) and abc in the assumed equation of cubic polynomial, we get
  $ \Rightarrow {{x}^{3}}-3A{{x}^{2}}+\dfrac{3{{G}^{3}}}{H}x-{{G}^{3}}=0 $
Multiplying both sides with H, we get
  $ \Rightarrow H{{x}^{3}}-3AH{{x}^{2}}+3{{G}^{3}}x-{{G}^{3}}H=0 $
Hence, option (b) is the correct answer.

Note:
One must remember the formula of Arithmetic, Geometric, Harmonic means of any number of terms. You may see that the reciprocal of HM is nothing but the AM of reciprocals of a, b and c. This is because Harmonic progression is a series of terms in which its terms when taken in reciprocal forms Arithmetic progression. One must remember the general form of an equation having three roots given as (x-a)(x-b)(x-c)=0, whose simplified form is $ {{x}^{3}}-\left( a+b+c \right){{x}^{2}}+\left( ab+bc+ca \right)x-abc $ .