If 3 arithmetic means, 3 geometric means and 3 harmonic means are inserted between 1 and 5, then the cubic equation whose roots are first A.M., second G.M. and third H.M. between 1 and 5, is

Options

A. `x^(3)-((9)/(2)+sqrt(5))x^(2)+((9sqrt(5))/(2)+5)x-5sqrt(5)=0`
B. `x^(3)+((9)/(2)+sqrt(5))x^(2)-((9sqrt(5))/(2)+5)x-5sqrt(5)=0`
C. `x^(3)+((9)/(2)-sqrt(5))x^(2)-((9sqrt(5))/(2)-5)x+5sqrt(5)=0`
D. none of these

Explanation

To solve the problem, we need to find the cubic equation whose roots are the first arithmetic mean (A.M.), the second geometric mean (G.M.), and the third harmonic mean (H.M.) between the numbers 1 and 5.

### Step 1: Find the first arithmetic mean (A.M.)
The first arithmetic mean (A.M.) can be found using the formula for the A.M. of two numbers a and b, which is given by:

\[ A.M. = \frac{a + b}{2} \]

In this case, \(a = 1\) and \(b = 5\):

\[ A.M. = \frac{1 + 5}{2} = \frac{6}{2} = 3 \]

### Step 2: Find the second geometric mean (G.M.)
The geometric means between two numbers can be found using the formula:

\[ G.M. = \sqrt{a \cdot b} \]

For the second geometric mean, we need to find the common ratio \(r\) in the geometric sequence. The first term is \(1\) and the fifth term is \(5\). The fifth term of a geometric sequence can be expressed as:

\[ a \cdot r^4 = 5 \]

Substituting \(a = 1\):

\[ 1 \cdot r^4 = 5 \implies r^4 = 5 \implies r = 5^{1/4} \]

Now, the second geometric mean \(G_2\) is given by:

\[ G_2 = a \cdot r^2 = 1 \cdot (5^{1/4})^2 = 5^{1/2} = \sqrt{5} \]

### Step 3: Find the third harmonic mean (H.M.)
The harmonic mean can be calculated using the formula:

\[ H.M. = \frac{2ab}{a + b} \]

For the harmonic means, we can also find the common difference \(d\) in the harmonic series. The first term is \(1\) and the fifth term is \(5\):

\[ \frac{1}{H.M.} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{e} \]

For three harmonic means, we have:

\[ \frac{1}{H_3} = \frac{1}{1} + \frac{1}{H_1} + \frac{1}{H_2} + \frac{1}{H_3} + \frac{1}{5} \]

We can find \(H_3\) using the formula:

\[ H_3 = \frac{2 \cdot 1 \cdot 5}{1 + 5} = \frac{10}{6} = \frac{5}{3} \]

### Step 4: Form the cubic equation
Now we have the roots:
- First A.M. \(A_1 = 3\)
- Second G.M. \(G_2 = \sqrt{5}\)
- Third H.M. \(H_3 = \frac{5}{3}\)

The cubic equation can be formed using the roots \(r_1, r_2, r_3\):

\[ x^3 - (r_1 + r_2 + r_3)x^2 + (r_1r_2 + r_2r_3 + r_3r_1)x - (r_1r_2r_3) = 0 \]

Calculating the sum of the roots:

\[ r_1 + r_2 + r_3 = 3 + \sqrt{5} + \frac{5}{3} \]

Calculating the product of the roots:

\[ r_1r_2r_3 = 3 \cdot \sqrt{5} \cdot \frac{5}{3} = 5\sqrt{5} \]

Calculating the sum of the products of the roots taken two at a time:

\[ r_1r_2 + r_2r_3 + r_3r_1 = 3\sqrt{5} + \sqrt{5} \cdot \frac{5}{3} + 3 \cdot \frac{5}{3} \]

Now substituting these values into the cubic equation gives us:

\[ x^3 - \left(3 + \sqrt{5} + \frac{5}{3}\right)x^2 + \left(3\sqrt{5} + \sqrt{5} \cdot \frac{5}{3} + 3 \cdot \frac{5}{3}\right)x - 5\sqrt{5} = 0 \]

### Final Equation
After simplifying, we arrive at the final cubic equation.