The two numbers whose arithmetic mean is 34 and their geometric mean is 16 are:

Question

Prepp Team · Accepted Answer

Finding Two Numbers Using Arithmetic and Geometric Means

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

The definitions of Arithmetic Mean and Geometric Mean for two numbers \(a\) and \(b\) are:

Arithmetic Mean (AM) = \(\frac{a+b}{2}\)
Geometric Mean (GM) = \(\sqrt{ab}\)

We are given the following information:

The arithmetic mean is 34.
The geometric mean is 16.

We can write these as two equations:

Equation 1 (from AM): \(\frac{a+b}{2} = 34\)

Equation 2 (from GM): \(\sqrt{ab} = 16\)

Let's simplify these equations.

From Equation 1: \(\frac{a+b}{2} = 34\)

Multiplying both sides by 2, we get:

\(a+b = 34 \times 2\)

\(a+b = 68\)

From Equation 2: \(\sqrt{ab} = 16\)

Squaring both sides to eliminate the square root:

\((\sqrt{ab})^2 = 16^2\)

\(ab = 256\)

Now we have a system of two equations with two variables:

\(a+b = 68\)
\(ab = 256\)

We need to find the two numbers \(a\) and \(b\) that satisfy both these equations. We can solve this system. From the first equation, we can express \(b\) in terms of \(a\):

\(b = 68 - a\)

Substitute this expression for \(b\) into the second equation:

\(a(68 - a) = 256\)

Distribute \(a\) on the left side:

\(68a - a^2 = 256\)

Rearrange the terms to form a standard quadratic equation \(Ax^2 + Bx + C = 0\):

\(a^2 - 68a + 256 = 0\)

We can solve this quadratic equation for \(a\). We are looking for two numbers whose product is 256 and whose sum is 68. Let's consider the options provided or try to factor the quadratic equation.

Alternatively, we can check the given options by calculating their AM and GM to see which pair satisfies the conditions.

Checking the Options

Let's examine each option:

Option	Numbers (a, b)	Arithmetic Mean \(\left(\frac{a+b}{2}\right)\)	Geometric Mean \((\sqrt{ab})\)	Match AM=34?	Match GM=16?
(a)	60, 8	\(\frac{60+8}{2} = \frac{68}{2} = 34\)	\(\sqrt{60 \times 8} = \sqrt{480}\)	Yes	No (\(\sqrt{480} \neq 16\))
(b)	34, 34	\(\frac{34+34}{2} = \frac{68}{2} = 34\)	\(\sqrt{34 \times 34} = \sqrt{34^2} = 34\)	Yes	No (\(34 \neq 16\))
(c)	64, 4	\(\frac{64+4}{2} = \frac{68}{2} = 34\)	\(\sqrt{64 \times 4} = \sqrt{256} = 16\)	Yes	Yes
(d)	48, 28	\(\frac{48+28}{2} = \frac{76}{2} = 38\)	\(\sqrt{48 \times 28} = \sqrt{1344}\)	No (\(38 \neq 34\))	No (\(\sqrt{1344} \neq 16\))

Option (c) with the numbers 64 and 4 satisfies both conditions: their arithmetic mean is 34, and their geometric mean is 16.

Solving the Quadratic Equation

Let's verify this by solving the quadratic equation \(a^2 - 68a + 256 = 0\).

We can factor this equation. We need two numbers that multiply to 256 and add up to -68. The numbers -64 and -4 fit this description since \((-64) \times (-4) = 256\) and \((-64) + (-4) = -68\).

So, the equation can be factored as:

\((a - 64)(a - 4) = 0\)

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

\(a - 64 = 0 \implies a = 64\)
\(a - 4 = 0 \implies a = 4\)

If \(a = 64\), then using \(b = 68 - a\), we get \(b = 68 - 64 = 4\). The pair of numbers is (64, 4).

If \(a = 4\), then using \(b = 68 - a\), we get \(b = 68 - 4 = 64\). The pair of numbers is (4, 64).

In both cases, the two numbers are 64 and 4. These are the numbers whose arithmetic mean is 34 and geometric mean is 16.

Conclusion

The two numbers are 64 and 4. This matches option (c).

Revision Table: Arithmetic and Geometric Means

Concept	Formula (for two numbers a, b)	Properties	Example
Arithmetic Mean (AM)	\(\frac{a+b}{2}\)	Always \(\ge\) GM for non-negative numbers. Represents average value.	AM of 10 and 20 is \(\frac{10+20}{2} = 15\).
Geometric Mean (GM)	\(\sqrt{ab}\)	Used for growth rates, ratios, scaling. Requires non-negative numbers.	GM of 9 and 4 is \(\sqrt{9 \times 4} = \sqrt{36} = 6\).

Additional Information: Relation Between AM and GM

For any two non-negative numbers \(a\) and \(b\), the arithmetic mean is always greater than or equal to the geometric mean (\(AM \ge GM\)). The equality holds only when \(a = b\). In this problem, AM (34) is greater than GM (16), which is consistent with the fact that the two numbers found (64 and 4) are not equal.

The problem of finding two numbers given their sum \(S = a+b\) and product \(P = ab\) is related to solving a quadratic equation \(x^2 - Sx + P = 0\). In our case, the sum \(S = 68\) and the product \(P = 256\), leading to the equation \(x^2 - 68x + 256 = 0\), whose roots are the two numbers.

Understanding the concepts of arithmetic mean and geometric mean is fundamental in sequences, series, and various areas of mathematics and statistics.