If A, g and H are respectively arithmetic ,
geometric and harmomic means between a and b
both being unequal and positive, then
`A = (a + b)/(2) rArr a + b = 2A`
`G = sqrt(ab) rArr G^(2) = ab `
`H = (2ab)/(a+ b ) rArr G^(2) = AH`
On the basis of above information answer the following questions .
The numbers whose A.M. and G.M. are A and G is

Options

A. `A pm (A^(2) - G^(2))`
B. `sqrt(A) pm sqrt(A^(2) - G^(2))`
C. `A pm sqrt(A^(2) - G^(2))`
D. `(A pm sqrt(A^(2) - G^(2)))/(2)`

Answer

To find the numbers whose Arithmetic Mean (A.M.) and Geometric Mean (G.M.) are A and G respectively, we will follow these steps:

### Step 1: Understand the definitions
- The Arithmetic Mean (A) of two numbers \( x \) and \( y \) is given by:
\[ A = \frac{x + y}{2} \]
- The Geometric Mean (G) of two numbers \( x \) and \( y \) is given by:
\[ G = \sqrt{xy} \]

### Step 2: Set up the equations
From the definitions, we can express \( x + y \) and \( xy \) in terms of A and G:
1. From the A.M.:
\[ x + y = 2A \]
2. From the G.M.:
\[ xy = G^2 \]

### Step 3: Formulate the quadratic equation
The numbers \( x \) and \( y \) can be considered as the roots of the quadratic equation:
\[ t^2 - (x+y)t + xy = 0 \]
Substituting the values from above:
\[ t^2 - (2A)t + G^2 = 0 \]

### Step 4: Use the quadratic formula
To find the roots \( x \) and \( y \), we can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2A \), and \( c = G^2 \):
\[ t = \frac{2A \pm \sqrt{(2A)^2 - 4 \cdot 1 \cdot G^2}}{2 \cdot 1} \]
This simplifies to:
\[ t = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2} \]
\[ t = A \pm \sqrt{A^2 - G^2} \]

### Step 5: Identify the numbers
Thus, the two numbers \( x \) and \( y \) are:
\[ x = A + \sqrt{A^2 - G^2} \]
\[ y = A - \sqrt{A^2 - G^2} \]

### Final Result
The numbers whose A.M. and G.M. are A and G respectively are:
\[ x = A + \sqrt{A^2 - G^2}, \quad y = A - \sqrt{A^2 - G^2} \]

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