Let p be the first of the n arithmetic means between two numbers and q the first of n harmonic means

Q: Why do we define $k = \frac{a}{b} + \frac{b}{a}$? Is this just a random trick?

It is not random! It is the hallmark of symmetric expressions. Whenever you see $a^2+b^2$ alongside $ab$, dividing by $ab$ is the most powerful tool to simplify the expression into a form involving the ratio $\frac{a}{b}$. This transformation is a standard technique in JEE problems to handle symmetric variables.

Q: Can I use the AM-HM inequality directly to solve this?

The AM-HM inequality states $AM \ge HM$ for the *entire set* of numbers. Here, we are dealing with the *first* mean, which is a specific weighted average. The standard inequality doesn't apply directly to $p$ and $q$ without accounting for the weights $n$. We must derive the specific relationship for the first mean.

Q: Why does the sign of $a$ and $b$ change the inequality direction?

Because the expression $k = \frac{a}{b} + \frac{b}{a}$ behaves differently. If $a, b > 0$, $k$ is positive and at least 2. If $a, b < 0$, $k$ is still positive and at least 2. If they have opposite signs, $k$ becomes negative and at most -2. This shift in the range of $k$ is exactly what flips the inequality.

Question

Let $p$ be the first of the $n$ arithmetic means between two numbers and $q$ the first of $n$ harmonic means between the same numbers. Show that $q$ does not lie between $p$ and $\left(\frac{n+1}{n-1}\right)^2 p$.

Sigma Percentile · Accepted Answer

### The Dance of the Means

Welcome, student. Today, we are going to explore the elegant world of arithmetic and harmonic means. Often, these concepts are presented as dry formulas, but they are actually a beautiful dance of symmetry.

We are tasked with proving that the first harmonic mean $q$ never dares to step into the interval between the first arithmetic mean $p$ and a specific scaled version of $p$. Let us embark on this journey.

### Phase 1: Defining the Players

Imagine two numbers, $a$ and $b$. We are inserting $n$ arithmetic means between them. The common difference for this arithmetic progression is:

$$d = \frac{b-a}{n+1}$$

The first arithmetic mean, which the problem calls $p$, is simply the first term $a$ plus this common difference. When we substitute and simplify, we get:

$$p = a + \frac{b-a}{n+1} = \frac{na+b}{n+1}$$

This is our first key equation.

Now, let us turn to the harmonic means. We insert $n$ harmonic means between $a$ and $b$. Remember, the reciprocals of these terms form an arithmetic progression. So, $\frac{1}{a}, \frac{1}{H_1}, \dots, \frac{1}{b}$ are in A.P.

The common difference here, let us call it $D$, is:

$$D = \frac{\frac{1}{b} - \frac{1}{a}}{n+1} = \frac{a-b}{ab(n+1)}$$

The first harmonic mean is $q$. So, $\frac{1}{q} = \frac{1}{a} + D$. Simplifying this carefully, we find:

$$q = \frac{ab(n+1)}{nb+a}$$

### Phase 2: The Algebraic Bridge

We have expressions for both $p$ and $q$. The problem asks us to compare them, so let us look at their ratio, $\frac{p}{q}$. We take our expression for $p$ and divide it by our expression for $q$:

$$\frac{p}{q} = \left( \frac{na+b}{n+1} \right) \div \left( \frac{ab(n+1)}{nb+a} \right) = \frac{(na+b)(nb+a)}{ab(n+1)^2}$$

Expanding the numerator, we get $n^2ab + na^2 + nb^2 + ab$. We can group the terms with $ab$ together and the terms with $n$ together. This gives us:

$$\frac{ab(n^2+1) + n(a^2+b^2)}{ab(n+1)^2}$$

### Phase 3: The $k$ Substitution

To make this ratio easier to analyze, let us divide both the numerator and the denominator by $ab$. The ratio $\frac{p}{q}$ simplifies to:

$$\frac{p}{q} = \frac{n^2+1 + n(\frac{a}{b} + \frac{b}{a})}{(n+1)^2}$$

Notice that interesting term, $\frac{a}{b} + \frac{b}{a}$? Let us call it $k$. This is a classic algebraic form.

If $a$ and $b$ have the same sign, $k \ge 2$. But if they have opposite signs, $k \le -2$. This is the crucial turning point in our logic.

### Phase 4: The Two Worlds

Let us analyze the first case, where $a$ and $b$ have the same sign. Here, our value $k$ is greater than or equal to 2. Let us substitute the minimum value, 2, into our ratio equation:

$$\frac{p}{q} \ge \frac{n^2+1 + n(2)}{(n+1)^2} = \frac{(n+1)^2}{(n+1)^2} = 1$$

This simply means $p \ge q$, or in other words, $q \le p$.

Now for the second case. What if $a$ and $b$ have opposite signs? Then $k \le -2$. Substituting $-2$ into our ratio, we get:

$$\frac{p}{q} \le \frac{n^2+1 + n(-2)}{(n+1)^2} = \frac{(n-1)^2}{(n+1)^2}$$

If we rearrange this to isolate $q$, we find that:

$$q \ge p \cdot \left( \frac{n+1}{n-1} \right)^2$$

### Conclusion

Let us bring it all together. In our first case, we found that $q \le p$. In our second case, we found that $q \ge p \left( \frac{n+1}{n-1} \right)^2$.

Think about what this means on a number line. $q$ is either to the left of $p$, or to the right of that larger expression. It never falls in the space between them.

Therefore, we have successfully proven that **$q$ does not lie between $p$ and $p \left( \frac{n+1}{n-1} \right)^2$**. You have just mastered a beautiful piece of algebraic reasoning!

Animated Solution for Mathematics - Sequence and Series: Let $p$ be the first of the $n$ arithmetic means between two numbers and $q$ the first of $n$ harmonic means between the same numbers. Show that $q$ does not lie between $p$ and $(\frac{n + 1}{n - 1})^{2} p$ .

Visualized Solution

Defining the First Arithmetic Mean $p$

Defining the First Harmonic Mean $q$

Forming the Ratio $\frac{p}{q}$

Simplifying the Ratio using $k = \frac{a}{b} + \frac{b}{a}$

Analyzing Case 1: $a, b$ have the same sign

Analyzing Case 2: $a, b$ have opposite signs

Conclusion: $q$ lies outside the interval

The Sigma Insight: Relation Between A.M., G.M., and H.M.

The Dance of the Means

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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Visualized Solution

Defining the First Arithmetic Mean .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Defining the First Harmonic Mean .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Forming the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } qp​

Simplifying the Ratio using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=ba​+ab​

Analyzing Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b have the same sign

Analyzing Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b have opposite signs

Conclusion: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q lies outside the interval

The Sigma Insight: Relation Between A.M., G.M., and H.M.

The Dance of the Means

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

Comprehension Passage

Which one of the following statements is correct ?

Which one of the following statements is correct ?

Which one of the following statements is correct?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a,b,c are in A.P., a2,b2,c2 are in H.P., then prove that either a=b=c or a,b,−c/2 form a G.P.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the arithmetic mean and geometric mean of the pth and qth terms of the sequence −16,8,−4,2,… satisfy the equation 4x2−9x+5=0, then p+q is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the harmonic mean and geometric mean of two positive numbers be the ratio 4:5. Then the two number are in the ratio \dots.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x1​,x2​,…,xn​ are any real numbers and n is any positive integer, then

Defining the First Arithmetic Mean $p$

Defining the First Harmonic Mean $q$

Forming the Ratio $\frac{p}{q}$

Simplifying the Ratio using $k = \frac{a}{b} + \frac{b}{a}$

Analyzing Case 1: $a, b$ have the same sign

Analyzing Case 2: $a, b$ have opposite signs

Conclusion: $q$ lies outside the interval

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If the arithmetic mean and geometric mean of the $p^{t h}$ and $q^{t h}$ terms of the sequence $- 16, 8, - 4, 2, \dots$ satisfy the equation $4 x^{2} - 9 x + 5 = 0$ , then $p + q$ is equal to

Let the harmonic mean and geometric mean of two positive numbers be the ratio $4 : 5$ . Then the two number are in the ratio \dots.

If $x_{1}, x_{2}, \dots, x_{n}$ are any real numbers and $n$ is any positive integer, then