Finding Numbers Using Arithmetic Mean and Geometric Mean

The problem asks us to find two numbers given their arithmetic mean (AM) and geometric mean (GM). Let the two numbers be \(a\) and \(b\).

Understanding Arithmetic Mean and Geometric Mean

The arithmetic mean of two numbers \(a\) and \(b\) is calculated as:

$$ AM = \frac{a+b}{2} $$

The geometric mean of two positive numbers \(a\) and \(b\) is calculated as:

$$ GM = \sqrt{ab} $$

Setting Up the Equations

According to the question, the arithmetic mean of the two numbers is 7. So, we have the equation:

$$ \frac{a+b}{2} = 7 $$

Multiplying both sides by 2, we get:

$$ a+b = 14 \quad \text{(Equation 1)} $$

The geometric mean of the two numbers is \(2\sqrt{10}\). So, we have the equation:

$$ \sqrt{ab} = 2\sqrt{10} $$

Squaring both sides of this equation to eliminate the square root:

$$ (\sqrt{ab})^2 = (2\sqrt{10})^2 $$

$$ ab = 2^2 \times (\sqrt{10})^2 $$

$$ ab = 4 \times 10 $$

$$ ab = 40 \quad \text{(Equation 2)} $$

Solving for the Two Numbers

We now have a system of two equations with two variables \(a\) and \(b\):

1. \(a+b = 14\)
2. \(ab = 40\)

We are looking for two numbers whose sum is 14 and whose product is 40.

One way to solve this is to consider \(a\) and \(b\) as the roots of a quadratic equation. A quadratic equation whose roots are \(a\) and \(b\) can be written as:

$$ x^2 - (a+b)x + ab = 0 $$

Substituting the values from our equations \(a+b=14\) and \(ab=40\), we get the quadratic equation:

$$ x^2 - 14x + 40 = 0 $$

We can solve this quadratic equation for \(x\) to find the values of the two numbers. We can factor the quadratic expression. We need two numbers that multiply to 40 and add up to -14. These numbers are -4 and -10. However, the equation is \(x^2 - 14x + 40\), so we need two numbers that multiply to 40 and add to 14. These numbers are 4 and 10.

So, we can factor the equation as:

$$ (x-4)(x-10) = 0 $$

This gives us two possible values for \(x\):

• \(x-4 = 0 \implies x = 4\)
• \(x-10 = 0 \implies x = 10\)

Thus, the two numbers are 4 and 10.

Alternatively, we could use substitution. From Equation 1, \(b = 14-a\). Substitute this into Equation 2:

$$ a(14-a) = 40 $$

$$ 14a - a^2 = 40 $$

Rearrange into a standard quadratic form:

$$ a^2 - 14a + 40 = 0 $$

This is the same quadratic equation as before, which yields the solutions \(a=4\) or \(a=10\). If \(a=4\), then \(b=14-4=10\). If \(a=10\), then \(b=14-10=4\). The pair of numbers is {4, 10}.

Verifying the Numbers

Let's check if the numbers 4 and 10 satisfy the given conditions:

• Arithmetic Mean: \(\frac{4+10}{2} = \frac{14}{2} = 7\). This matches the given AM.
• Geometric Mean: \(\sqrt{4 \times 10} = \sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2\sqrt{10}\). This matches the given GM.

Both conditions are satisfied, so the numbers are indeed 4 and 10.

Reviewing the Options

Let's quickly check the given options against our findings:

The numbers 4 and 10 from Option 3 are the only pair that correctly gives an arithmetic mean of 7 and a geometric mean of \(2\sqrt{10}\).

Revision Table: Arithmetic and Geometric Mean

Additional Information: Finding Numbers from Sum and Product

If the sum of two numbers is \(S\) and their product is \(P\), the numbers are the roots of the quadratic equation \(x^2 - Sx + P = 0\).

In this problem, the sum of the two numbers \(a+b\) is \(2 \times AM = 2 \times 7 = 14\). So \(S=14\).

The product of the two numbers \(ab\) is \((GM)^2 = (2\sqrt{10})^2 = 40\). So \(P=40\).

The quadratic equation is \(x^2 - 14x + 40 = 0\). Solving this equation gives the two numbers. This method is a standard technique for finding two quantities when their sum and product are known.