If A, G & H are respectively the A.M., G.M. & H.M. of three positive numbers a, b, & c, then equation whose roots are a, b, & c is given by

Options

A. `a^(2)=AH`
B. A is an integer if `altbltclt4`
C. A=H iff a=b=c
D. `AgtGgtH,ifa!=b!=c`

Explanation

To find the equation whose roots are \( a, b, \) and \( c \), where \( A, G, H \) are the arithmetic mean (A.M.), geometric mean (G.M.), and harmonic mean (H.M.) of the three positive numbers \( a, b, \) and \( c \), we can follow these steps:

### Step 1: Identify the relationships
1. **Arithmetic Mean (A.M.)**:
\[ A = \frac{a + b + c}{3} \]
Therefore, we have:
\[ a + b + c = 3A \]

2. **Geometric Mean (G.M.)**:
\[ G = \sqrt[3]{abc} \]
Hence,
\[ abc = G^3 \]

3. **Harmonic Mean (H.M.)**:
\[ H = \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \]
This can be rearranged to find:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{3}{H} \]
Multiplying through by \( abc \) gives:
\[ bc + ca + ab = \frac{3abc}{H} \]
Substituting \( abc = G^3 \):
\[ ab + ac + bc = \frac{3G^3}{H} \]

### Step 2: Form the cubic equation
Using Vieta's formulas, the cubic equation with roots \( a, b, c \) can be expressed as:
\[ x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = 0 \]

Substituting the values we found:
1. **Sum of roots**:
\[ a + b + c = 3A \]

2. **Sum of roots taken two at a time**:
\[ ab + ac + bc = \frac{3G^3}{H} \]

3. **Product of roots**:
\[ abc = G^3 \]

### Step 3: Substitute into the cubic equation
Now, substituting these into the cubic equation gives:
\[ x^3 - (3A)x^2 + \left(\frac{3G^3}{H}\right)x - G^3 = 0 \]

### Final Form of the Equation
Thus, the equation whose roots are \( a, b, c \) is:
\[ x^3 - 3Ax^2 + \frac{3G^3}{H}x - G^3 = 0 \]