Let a, b be positive real numbers. If...

MathematicsRelation Between A.M., G.M., and H.M.JEE Advanced 2002Moderate

Let $a, b$ be positive real numbers. If $a, A_{1}, A_{2}, b$ are in arithmetic progression, $a, G_{1}, G_{2}, b$ are in geometric progression and $a, H_{1}, H_{2}, b$ are in harmonic progression, show that $\frac{G _{1} G _{2}}{H _{1} H _{2}} = \frac{A _{1} + A _{2}}{H _{1} + H _{2}} = \frac{( 2 a + b ) ( a + 2 b )}{9 ab}$ .

Visualized Solution (English)

Arithmetic Means .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A_{1}, A_{2}

Given $a, A_{1}, A_{2}, b$ are in AP.
Sum of arithmetic means: $A_{1} + A_{2} = a + b$ .

Geometric Means .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G_{1}, G_{2}

Given $a, G_{1}, G_{2}, b$ are in GP.
Product of geometric means: $G_{1} G_{2} = ab$ .

Harmonic Means .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H_{1}, H_{2}

Given $a, H_{1}, H_{2}, b$ are in HP.
Reciprocals are in AP: $\frac{1}{a}, \frac{1}{H _{1}}, \frac{1}{H _{2}}, \frac{1}{b}$ .
Sum of reciprocals: $\frac{1}{H _{1}} + \frac{1}{H _{2}} = \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$ .
Rearranging: $H_{1} + H_{2} = H_{1} H_{2} \frac{a + b}{ab}$ .

Comparing the Ratios

Ratio 1: $\frac{A _{1} + A _{2}}{H _{1} + H _{2}} = \frac{a + b}{H _{1} H _{2} \frac{a + b}{ab}} = \frac{ab}{H _{1} H _{2}}$ .
Ratio 2: $\frac{G _{1} G _{2}}{H _{1} H _{2}} = \frac{ab}{H _{1} H _{2}}$ .
Thus, $\frac{G _{1} G _{2}}{H _{1} H _{2}} = \frac{A _{1} + A _{2}}{H _{1} + H _{2}}$ .

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H_{1} Explicitly

Common difference of reciprocal AP: $d^{'} = \frac{\frac{1}{b} - \frac{1}{a}}{3} = \frac{a - b}{3 ab}$ .
$\frac{1}{H _{1}} = \frac{1}{a} + d^{'} = \frac{1}{a} + \frac{a - b}{3 ab} = \frac{3 b + a - b}{3 ab} = \frac{2 b + a}{3 ab}$ .
So, $H_{1} = \frac{3 ab}{2 b + a}$ .

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H_{2} Explicitly

$\frac{1}{H _{2}} = \frac{1}{a} + 2 d^{'} = \frac{1}{a} + \frac{2 a - 2 b}{3 ab} = \frac{3 b + 2 a - 2 b}{3 ab} = \frac{b + 2 a}{3 ab}$ .
So, $H_{2} = \frac{3 ab}{b + 2 a}$ .

The Final Calculation

Product $H_{1} H_{2} = \frac{3 ab}{2 b + a} \cdot \frac{3 ab}{b + 2 a} = \frac{9 a ^{2} b ^{2}}{( 2 b + a ) ( b + 2 a )}$ .
Calculate $\frac{ab}{H _{1} H _{2}} = ab \cdot \frac{( 2 b + a ) ( b + 2 a )}{9 a ^{2} b ^{2}} = \frac{( 2 b + a ) ( b + 2 a )}{9 ab}$ .
Final Result: $\frac{G _{1} G _{2}}{H _{1} H _{2}} = \frac{A _{1} + A _{2}}{H _{1} + H _{2}} = \frac{( 2 a + b ) ( a + 2 b )}{9 ab}$ .

Key Takeaways

Key Takeaway: Sum of $n$ AMs is $n \cdot \frac{a + b}{2}$ .
Key Takeaway: Product of $n$ GMs is $(ab)^{n /2}$ .
Challenge: Try proving a similar relation for $n$ means.

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Visualized Solution (English)

Arithmetic Means $A_{1}, A_{2}$

Geometric Means $G_{1}, G_{2}$

Harmonic Means $H_{1}, H_{2}$

Comparing the Ratios

Finding $H_{1}$ Explicitly

Finding $H_{2}$ Explicitly

The Final Calculation

Key Takeaways

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Visualized Solution (English)

Arithmetic Means .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​,A2​

Geometric Means .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G1​,G2​

Harmonic Means .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H1​,H2​

Comparing the Ratios

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H1​ Explicitly

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H2​ Explicitly

The Final Calculation

Key Takeaways

Conceptually Similar Problems

.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.

The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a>0,b>0 and c>0, prove that (a+b+c)(a1​+b1​+c1​)≥9.

If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:

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Arithmetic Means $A_{1}, A_{2}$

Geometric Means $G_{1}, G_{2}$

Harmonic Means $H_{1}, H_{2}$

Finding $H_{1}$ Explicitly

Finding $H_{2}$ Explicitly

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