If a, b, c are in A.P., a², b², c² are...

MathematicsRelation Between A.M., G.M., and H.M.JEE Advanced 2003Moderate

If $a, b, c$ are in A.P., $a^{2}, b^{2}, c^{2}$ are in H.P., then prove that either $a = b = c$ or $a, b, - c /2$ form a G.P.

Visualized Solution (English)

Defining the A.P. Condition

Given that $a, b, c$ are in A.P.
The common difference is constant: $b - a = c - b$
Rearranging gives the standard A.P. property: $2 b = a + c$

Defining the H.P. Condition

Given that $a^{2}, b^{2}, c^{2}$ are in H.P.
The middle term is the harmonic mean: $b^{2} = \frac{2 a ^{2} c ^{2}}{a ^{2} + c ^{2}}$

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b into the H.P. Equation

From A.P.: $b = \frac{a + c}{2}$
Substitute $b$ into the H.P. equation:
$(\frac{a + c}{2})^{2} = \frac{2 a ^{2} c ^{2}}{a ^{2} + c ^{2}}$

Cross-Multiplication and Simplification

Expand the square: $\frac{( a + c ) ^{2}}{4} = \frac{2 a ^{2} c ^{2}}{a ^{2} + c ^{2}}$
Cross-multiply: $(a + c)^{2} (a^{2} + c^{2}) = 8 a^{2} c^{2}$

Expanding the Binomial

Expand $(a + c)^{2}$ :
$(a^{2} + c^{2} + 2 a c) (a^{2} + c^{2}) = 8 a^{2} c^{2}$

Quadratic Substitution

Let $X = a^{2} + c^{2}$
Substitute $X$ into the equation: $(X + 2 a c) X = 8 a^{2} c^{2}$
Rearrange to quadratic form: $X^{2} + 2 a c X - 8 a^{2} c^{2} = 0$

Factoring the Quadratic

Factor the quadratic: $(X + 4 a c) (X - 2 a c) = 0$
This implies either $X - 2 a c = 0$ or $X + 4 a c = 0$

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a = b = c

Case 1: $X - 2 a c = 0$
Substitute $X$ : $a^{2} + c^{2} - 2 a c = 0 \Rightarrow (a - c)^{2} = 0$
This gives $a = c$
Since $2 b = a + c$ , then $2 b = 2 a \Rightarrow a = b = c$

Case 2: Setting up the G.P.

Case 2: $X + 4 a c = 0$
Substitute $X$ : $a^{2} + c^{2} = - 4 a c$
Add $2 a c$ to both sides: $a^{2} + c^{2} + 2 a c = - 4 a c + 2 a c$
Simplify: $(a + c)^{2} = - 2 a c$

Proving the G.P. Condition

Substitute $a + c = 2 b$ into the equation:
$(2 b)^{2} = - 2 a c \Rightarrow 4 b^{2} = - 2 a c$
Divide by $4$ : $b^{2} = a (- \frac{c}{2})$
This is the condition for $a, b, - \frac{c}{2}$ to be in G.P.

Final Conclusion

Conclusion:
Either $a = b = c$
OR $a, b, - \frac{c}{2}$ form a G.P.
The proof is complete.

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If $a, b, c$ are in A.P., $a^{2}, b^{2}, c^{2}$ are in H.P., then prove that either $a = b = c$ or $a, b, - c /2$ form a G.P.

Visualized Solution (English)

Defining the A.P. Condition

Defining the H.P. Condition

Substituting $b$ into the H.P. Equation

Cross-Multiplication and Simplification

Expanding the Binomial

Quadratic Substitution

Factoring the Quadratic

Case 1: $a = b = c$

Case 2: Setting up the G.P.

Proving the G.P. Condition

Final Conclusion

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.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.

Visualized Solution (English)

Defining the A.P. Condition

Defining the H.P. Condition

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b into the H.P. Equation

Cross-Multiplication and Simplification

Expanding the Binomial

Quadratic Substitution

Factoring the Quadratic

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=b=c

Case 2: Setting up the G.P.

Proving the G.P. Condition

Final Conclusion

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