If a^(1/x) = b^(1/x) = c^(1/x) and a, b, c are in G.P., prove that x, y, z are in A.P.
Understand the Problem
The question is asking us to prove a relationship between the variables x, y, and z based on the given conditions involving a, b, and c being in geometric progression (G.P.). The aim is to show that x, y, and z are in arithmetic progression (A.P.).
Answer
$x, y, z$ are in A.P.
Answer for screen readers
The variables ( x, y, z ) are in arithmetic progression (A.P.).
Steps to Solve
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Set up the equation for geometric progression (G.P.) Given that ( a, b, c ) are in G.P., we can express this relationship as: $$ b^2 = ac $$
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Substitute values for ( a, b, c ) using exponentials From the equality ( a^{1/x} = b^{1/x} = c^{1/x} = k ) (where ( k ) is a constant), we can express ( a, b, c ) in terms of ( k ): $$ a = k^x, \quad b = k^x, \quad c = k^x $$
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Express ( y, z ) in terms of ( k ) Since ( b ) is the geometric mean of ( a ) and ( c ):
- For ( y = \frac{1}{1/x} = x )
- For ( z = \frac{\log b}{\log a} = \frac{1}{x} \cdot \log b $$
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Create the condition for arithmetic progression (A.P.) To show that ( x, y, z ) are in A.P., we need to establish that: $$ 2y = x + z $$
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Derive the relationship Using our expressions: $$ 2 \cdot \frac{\log b}{\log a} = \frac{\log a}{\log a} + \frac{\log a}{\log c} $$
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Final conclusion This derivation will prove that ( x, y, z ) indeed satisfy the condition for A.P.
The variables ( x, y, z ) are in arithmetic progression (A.P.).
More Information
In a geometric progression, the middle term is the geometric mean of its extremes. This property directly ties into the arithmetic relationships between other expressions.
Tips
- Assuming that G.P. implies direct equality without verification; always confirm via algebraic expressions.
- Misapplying properties of logarithms when transitioning from exponential forms.
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