Let p be the first of the n arithmetic...

MathematicsRelation Between A.M., G.M., and H.M.JEE Advanced 1991Moderate

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 (Hindi)

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

Let the two numbers be $a$ and $b$ .
Let $A_{1}, A_{2}, ..., A_{n}$ be $n$ arithmetic means between $a$ and $b$ .
The common difference is $d = \frac{b - a}{n + 1}$ .
The first arithmetic mean $p = A_{1} = a + d = a + \frac{b - a}{n + 1}$ .
Simplifying, we get $p = \frac{a ( n + 1 ) + b - a}{n + 1} = \frac{na + b}{n + 1}$ .

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

Let $H_{1}, H_{2}, ..., H_{n}$ be $n$ harmonic means between $a$ and $b$ .
The reciprocals $\frac{1}{a}, \frac{1}{H _{1}}, ..., \frac{1}{b}$ are in A.P.
Common difference $D = \frac{\frac{1}{b} - \frac{1}{a}}{n + 1} = \frac{a - b}{ab ( n + 1 )}$ .
$\frac{1}{q} = \frac{1}{H _{1}} = \frac{1}{a} + D = \frac{1}{a} + \frac{a - b}{ab ( n + 1 )}$ .
$\frac{1}{q} = \frac{b ( n + 1 ) + a - b}{ab ( n + 1 )} = \frac{nb + a}{ab ( n + 1 )} \Rightarrow q = \frac{ab ( n + 1 )}{nb + a}$ .

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

Consider the ratio $\frac{p}{q} = \frac{na + b}{n + 1} \div \frac{ab ( n + 1 )}{nb + a}$ .
$\frac{p}{q} = \frac{( na + b ) ( nb + a )}{ab ( n + 1 ) ^{2}}$ .
Expanding the numerator: $n^{2} ab + n a^{2} + n b^{2} + ab$ .
Grouping terms: $ab (n^{2} + 1) + n (a^{2} + b^{2})$ .
So, $\frac{p}{q} = \frac{ab ( n ^{2} + 1 ) + n ( a ^{2} + b ^{2} )}{ab ( n + 1 ) ^{2}}$ .

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

Divide the numerator and denominator by $ab$ :
$\frac{p}{q} = \frac{n ^{2} + 1 + n ( \frac{a}{b} + \frac{b}{a} )}{( n + 1 ) ^{2}}$ .
Let $k = \frac{a}{b} + \frac{b}{a}$ .
If $a, b$ have the same sign, $k \geq 2$ .
If $a, b$ have opposite signs, $k \leq - 2$ .

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

Case 1: $a, b$ have the same sign $\Rightarrow k \geq 2$ .
$\frac{p}{q} \geq \frac{n ^{2} + 1 + n ( 2 )}{( n + 1 ) ^{2}} = \frac{( n + 1 ) ^{2}}{( n + 1 ) ^{2}} = 1$ .
Therefore, $\frac{p}{q} \geq 1 \Rightarrow p \geq q$ (or $q \leq p$ ).

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

Case 2: $a, b$ have opposite signs $\Rightarrow k \leq - 2$ .
$\frac{p}{q} \leq \frac{n ^{2} + 1 + n ( - 2 )}{( n + 1 ) ^{2}} = \frac{( n - 1 ) ^{2}}{( n + 1 ) ^{2}}$ .
Rearranging for $q$ : $q \geq p \cdot \frac{( n + 1 ) ^{2}}{( n - 1 ) ^{2}}$ .

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

From Case 1: $q \leq p$ .
From Case 2: $q \geq p (\frac{n + 1}{n - 1})^{2}$ .
In both scenarios, $q$ does not lie in the open interval $(p, p (\frac{n + 1}{n - 1})^{2})$ .
Hence, $q$ does not lie between $p$ and $(\frac{n + 1}{n - 1})^{2} p$ .
Q.E.D.

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Comprehension Passage

Let

A_{1}, G_{1}, H_{1}

denote the arithmetic, geometric and harmonic means, respectively, of two distinct positive numbers. For

n \geq 2

, Let

A_{n - 1}

and

H_{n - 1}

have arithmetic, geometric and harmonic means as

A_{n}, G_{n}, H_{n}

respectively.

Question 1:

Which one of the following statements is correct ?

(A)

G_{1} > G_{2} > G_{3} > \dots

(B)

G_{1} < G_{2} < G_{3} < \dots

(C)

G_{1} = G_{2} = G_{3} = \dots

(D)

G_{1} < G_{3} < G_{5} < \dots

and

G_{2} > G_{4} > G_{6} > \dots

Answer:C

Question 2:

Which one of the following statements is correct ?

(A)

A_{1} > A_{2} > A_{3} > \dots

(B)

A_{1} < A_{2} < A_{3} < \dots

(C)

A_{1} > A_{2} > A_{5} > \dots

and

A_{2} < A_{4} < A_{6} < \dots

(D)

A_{1} < A_{3} < A_{5} < \dots

and

A_{2} > A_{4} > A_{6} > \dots

Answer:A

Question 3:

Which one of the following statements is correct?

(A)

H_{1} > H_{2} > H_{3} > \dots

(B)

H_{1} < H_{2} < H_{3} < \dots

(C)

H_{1} > H_{3} > H_{5} > \dots

and

H_{2} < H_{4} < H_{6} < \dots

(D)

H_{1} < H_{3} < H_{5} < \dots

and

H_{2} > H_{4} > H_{6} > \dots

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