MathematicsRelation Between A.M., G.M., and H.M.JEE Advanced 2003Moderate
Visualized Solution (Hindi)
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: 2b=a+c
Defining the H.P. Condition
- Given that a2,b2,c2 are in H.P.
- The middle term is the harmonic mean: b2=a2+c22a2c2
Substituting b into the H.P. Equation
- From A.P.: b=2a+c
- Substitute b into the H.P. equation:
- (2a+c)2=a2+c22a2c2
Cross-Multiplication and Simplification
- Expand the square: 4(a+c)2=a2+c22a2c2
- Cross-multiply: (a+c)2(a2+c2)=8a2c2
Expanding the Binomial
- Expand (a+c)2:
- (a2+c2+2ac)(a2+c2)=8a2c2
Quadratic Substitution
- Let X=a2+c2
- Substitute X into the equation: (X+2ac)X=8a2c2
- Rearrange to quadratic form: X2+2acX−8a2c2=0
Factoring the Quadratic
- Factor the quadratic: (X+4ac)(X−2ac)=0
- This implies either X−2ac=0 or X+4ac=0
Case 1: a=b=c
- Case 1: X−2ac=0
- Substitute X: a2+c2−2ac=0⇒(a−c)2=0
- This gives a=c
- Since 2b=a+c, then 2b=2a⇒a=b=c
Case 2: Setting up the G.P.
- Case 2: X+4ac=0
- Substitute X: a2+c2=−4ac
- Add 2ac to both sides: a2+c2+2ac=−4ac+2ac
- Simplify: (a+c)2=−2ac
Proving the G.P. Condition
- Substitute a+c=2b into the equation:
- (2b)2=−2ac⇒4b2=−2ac
- Divide by 4: b2=a(−2c)
- This is the condition for a,b,−2c to be in G.P.
Final Conclusion
- Conclusion:
- Either a=b=c
- OR a,b,−2c form a G.P.
- The proof is complete.
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