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If the harmonic mean between two positive numbers is to their GM as \(12: 13\), the numbers are in the ratio (a) \(12: 13\) (b) \(\frac{1}{12}: \frac{1}{13}\) (c) \(4: 9\) (d) \(2: 3\)

Short Answer

Expert verified
The ratio of the two numbers is 12:13.

Step by step solution

01

Recall the formula for the harmonic mean.

The harmonic mean (HM) of two positive numbers \(a\) and \(b\) is given by the formula: \(\text{HM} = \frac{2ab}{a + b}\).
02

Recall the formula for the geometric mean.

The geometric mean (GM) of two positive numbers \(a\) and \(b\) is given by the formula: \(\text{GM} = \sqrt{ab}\).
03

Set up the relationship between the two means.

We are given that the ratio of HM to GM is \(12:13\). Therefore, we can write the relationship as: \(\frac{\text{HM}}{\text{GM}} = \frac{12}{13}\).
04

Substitute the formulas for the means into the relationship.

Now we substitute the formulas for the harmonic mean \(\frac{2ab}{a + b}\) and the geometric mean \(\sqrt{ab}\) into the relationship: \(\frac{\frac{2ab}{a + b}}{\sqrt{ab}} = \frac{12}{13}\).
05

Solve the equation for the ratio of a to b.

To do this, first we need to clear the denominator by multiplying both sides of the equation by \((a + b)\sqrt{ab}\): \(2ab = \frac{12}{13} (a + b)\sqrt{ab}\). Next, we can divide both sides by \(ab\): \(\frac{2}{\sqrt{ab}} = \frac{12}{13} \frac{a + b}{\sqrt{ab}}\). Now, we can cross multiply: \(2(a + b)(\sqrt{ab}) = 12(\sqrt{ab})\). Divide both sides by \(2\sqrt{ab}\): \(a + b = 6\). Let \(\frac{a}{b} = r\), then we have \(a = rb\), and our equation becomes: \(r + 1 = \frac{6}{b}\).
06

Substitute the expression of a in terms of harmonic mean.

Now, from Step 1, we know harmonic mean is \(\text{HM} = \frac{2ab}{a + b}\), we can substitute \(a = rb\): \(\frac{2rb^2}{(r+1)b} = \frac{24}{13}\).
07

Solve the equation for r.

We can now solve for \(r\) by canceling \(b\) in the equation and multiplying both sides by 13: \(r = \frac{24}{26}= \frac{12}{13}\). So, the ratio of the numbers is: \(\boxed{\text{(a) }12:13}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Mean
The **Geometric Mean** (GM) offers a way to find the central tendency of a list of numbers, mainly when comparing ratios or log-scaled data. For two positive numbers, the geometric mean is represented as the square root of the product of the numbers. This can be written mathematically as \[ \text{GM} = \sqrt{ab} \]Where \(a\) and \(b\) are the numbers being compared. It is an ideal method to calculate average rates of growth, such as investment returns or population growth rates.
The geometric mean is less affected by extreme values than the arithmetic mean, which makes it a preferred method in logarithmically distributed data. Its concept is deeply intertwined with the harmonic mean when it comes to understanding complex mathematical problems involving ratio comparisons. By using the harmonic and geometric means together, as in the original exercise, you expose new insightful relationships between numbers.
Ratio of Numbers
The concept of the **Ratio of Numbers** is fundamental in understanding how two numbers compare directly in magnitude. This is expressed as either a fraction or a ratio, such as \(a:b\), where \(a\) and \(b\) are the numbers being compared. Ratios offer a straightforward method to express and solve problems and are omnipresent in problems involving proportions and direct comparisons.
Connecting this to the exercise, the relationship between harmonic mean and geometric mean was expressed in terms of a ratio \(12:13\). Understanding ratios is crucial in identifying how the harmonic mean differs proportionally from the geometric mean. It allows us to represent complex relationships in more approachable terms, enabling easier solving of mathematical problems. By setting up and manipulating ratios, as done in the solution, one can systematically resolve the mysterious link between means and their advantages over one another.
Problem Solving in Mathematics
**Problem Solving in Mathematics** merges strategic thinking and logical reasoning. It involves navigating through multiple mathematical frameworks to unveil solutions. When faced with exercises like the original problem, a systematic approach is valuable. This typically involves several key steps:
  • Understanding the problem and determining what is being asked.
  • Identifying and applying the relevant formulas or equations, such as the harmonic or geometric means in this context.
  • Solving the equations, which may involve algebraic manipulations or simplifications, to derive an answer.

In the original exercise, setting up the equations using harmonic and geometric means allowed for a structured approach to finding the desired ratio of the numbers. Problem solving is greatly enhanced by the ability to see not just numbers but how they relate to each other through different means and mathematical constructs. This reasoning process strengthens one’s capacity to tackle similar questions of varying complexity.

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Most popular questions from this chapter

If \(a, b, c\) are in AP; \(p, q, r\) are in \(H P\) and ap, bq, cr are in GP, then \(\frac{p}{r}+\frac{r}{p}\) is equal to (a) \(\frac{\mathrm{a}}{\mathrm{c}}-\frac{\mathrm{c}}{\mathrm{a}}\) (b) \(\frac{\mathrm{a}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}}\) (c) \(\frac{b}{q}-\frac{q}{b}\) (d) \(\frac{b}{q}+\frac{q}{p}\)

If the sides \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) of \(\triangle \mathrm{ABC}\) are in GP where \(\log \mathrm{a}-\log 2 \mathrm{~b}, \log 2 \mathrm{~b}-\log 3 \mathrm{c}, \log 3 \mathrm{c}-\log \mathrm{a}\) are in \(\mathrm{AP}\), then the sides \(\mathrm{a}\), b, c are in the ratio (a) \(4: 6: 9\) (b) \(9: 4: 6\) (c) \(9: 6: 4\) (d) \(3: 6: 8\)

If \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\) are in GP then (a) \(\mathrm{a}+\mathrm{b}, \mathrm{b}+c_{2} \mathrm{c}+\mathrm{d}\) are in \(\mathrm{GP}\) (b) \(a x^{2}+c\) is a factor of \(a x^{3}+b x^{2}+c x+d\) (c) \(a^{2}+b^{2}+c^{2}, a b+b c+c d, b^{2}+c^{2}+d^{2}\) are in GP (d) \(a x+c\) is a factor of \(a x^{3}+b x^{2}+c x+d\)

If the second number of three numbers in an increasing \(\mathrm{GP}\) is doubled, we get an \(\mathrm{AP}\), then (a) common ratio of the GP is 1 or \(-1\) (b) common ratio of the GP and common difference of AP are equal. (c) common difference of AP is equal to both first term and last term. (d) common ratio of the GP is \(2+\sqrt{3}\).

The sum of an infinite GP with common ratio \(r(r<1)\) is 4 . The sum of the infinite GP obtained by squaring the terms of this GP is \(\frac{16}{3}\). Then the first term and common ratio of the given GP is (a) \(-2, \frac{1}{2}\) (b) \(-2, \frac{1}{3}\) (c) \(2, \frac{1}{2}\) (d) \(3, \frac{-1}{2}\)

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