Let a,b,c are the A.M. between two numbers such that `a+b+c=15` and p,q,r be the H.M. between same numbers such that `1/p+1/q+1/r=5/3`, then the numbers are

Options

A. -1,-9
B. -3,-3
C. 3,3
D. 9,1

Answer

To solve the problem, we need to find two numbers \( x \) and \( y \) such that the arithmetic means \( a, b, c \) and harmonic means \( p, q, r \) satisfy the given conditions.

### Step 1: Set up the equations for Arithmetic Mean (A.M.)
Given that \( a + b + c = 15 \), we know that the arithmetic mean of \( x \) and \( y \) can be expressed as:
\[ a + b + c = 3 \cdot \frac{x + y}{2} \]
Thus, we can write:
\[ \frac{3(x + y)}{2} = 15 \]
Multiplying both sides by 2 gives:
\[ 3(x + y) = 30 \]
Dividing by 3:
\[ x + y = 10 \quad \text{(Equation 1)} \]

### Step 2: Set up the equations for Harmonic Mean (H.M.)
The harmonic mean condition is given by:
\[ \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{5}{3} \]
For harmonic means, we can express this in terms of \( x \) and \( y \):
\[ \frac{3}{\frac{1}{x} + \frac{1}{y}} = \frac{5}{3} \]
This implies:
\[ \frac{1}{x} + \frac{1}{y} = \frac{9}{5} \]
Multiplying through by \( xy \):
\[ \frac{y + x}{xy} = \frac{9}{5} \]
Substituting \( x + y = 10 \) from Equation 1:
\[ \frac{10}{xy} = \frac{9}{5} \]
Cross-multiplying gives:
\[ 10 \cdot 5 = 9 \cdot xy \]
Thus:
\[ 50 = 9xy \quad \Rightarrow \quad xy = \frac{50}{9} \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have a system of equations:
1. \( x + y = 10 \)
2. \( xy = \frac{50}{9} \)

We can express \( y \) in terms of \( x \):
\[ y = 10 - x \]
Substituting \( y \) into Equation 2:
\[ x(10 - x) = \frac{50}{9} \]
Expanding this gives:
\[ 10x - x^2 = \frac{50}{9} \]
Rearranging leads to:
\[ x^2 - 10x + \frac{50}{9} = 0 \]
Multiplying through by 9 to eliminate the fraction:
\[ 9x^2 - 90x + 50 = 0 \]

### Step 4: Use the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 9 \), \( b = -90 \), and \( c = 50 \):
\[ x = \frac{90 \pm \sqrt{(-90)^2 - 4 \cdot 9 \cdot 50}}{2 \cdot 9} \]
Calculating the discriminant:
\[ (-90)^2 - 4 \cdot 9 \cdot 50 = 8100 - 1800 = 6300 \]
Thus:
\[ x = \frac{90 \pm \sqrt{6300}}{18} \]
Calculating \( \sqrt{6300} = 30\sqrt{7} \):
\[ x = \frac{90 \pm 30\sqrt{7}}{18} = 5 \pm \frac{5\sqrt{7}}{3} \]

### Step 5: Find the values of \( y \)
Using \( y = 10 - x \):
\[ y = 10 - \left(5 \pm \frac{5\sqrt{7}}{3}\right) = 5 \mp \frac{5\sqrt{7}}{3} \]

### Conclusion
The two numbers are:
\[ \left(5 + \frac{5\sqrt{7}}{3}, 5 - \frac{5\sqrt{7}}{3}\right) \]