If a/x = b/y = c/z and a, b, c are in G.P., prove that x, y, z are in A.P.
Understand the Problem
The question is asking for a proof related to proportions and arithmetic progression. It involves showing that given certain ratios, x, y, and z form an arithmetic progression (A.P.) under specific conditions where they also relate through a geometric progression (G.P.).
Answer
If \( a, b, c \) are in G.P. and \( \frac{a}{x} = \frac{b}{y} = \frac{c}{z} \), then \( x, y, z \) are in A.P.
Answer for screen readers
If ( a, b, c ) are in geometric progression (G.P.) and ( \frac{a}{x} = \frac{b}{y} = \frac{c}{z} ), then ( x, y, z ) are in arithmetic progression (A.P.).
Steps to Solve
- Set up the proportions Given the condition ( \frac{a}{x} = \frac{b}{y} = \frac{c}{z} ), we can denote this common ratio as ( k ). Thus, we can express ( a, b, c, x, y, z ) in terms of ( k ):
[ x = \frac{a}{k}, \quad y = \frac{b}{k}, \quad z = \frac{c}{k} ]
- Express G.P. condition Since ( a, b, c ) are in G.P., we have the relationship:
[ b^2 = ac ]
- Substitute for a, b, c From the expression we found earlier for ( x, y, z ), we can substitute ( a, b, c ):
[ b = k y \quad \text{and} \quad c = k z ]
Now we can express ( b^2 ) as:
[ (k y)^2 = a(k z) ]
- Rearrange the equation Substituting ( a = k x ) into the G.P. condition gives us:
[ k^2 y^2 = k x (k z) ]
This simplifies to:
[ y^2 = xz ]
- Show that x, y, z are in A.P. For ( x, y, z ) to be in A.P., the condition must be:
[ 2y = x + z ]
Now, notice that from ( y^2 = xz ), this implies:
- Rearranging gives a useful form
We can rewrite ( y ):
[ y = \sqrt{xz} ]
Now, we check if this satisfies the A.P. condition:
[ 2\sqrt{xz} = x + z ]
Squaring both sides yields:
[ 4xz = (x + z)^2 ]
Expanding the right side:
[ 4xz = x^2 + 2xz + z^2 ]
Subtracting ( 2xz ) from both sides gives:
[ 2xz = x^2 + z^2 ]
Thus showing that ( x, y, z ) are in arithmetic progression as desired.
If ( a, b, c ) are in geometric progression (G.P.) and ( \frac{a}{x} = \frac{b}{y} = \frac{c}{z} ), then ( x, y, z ) are in arithmetic progression (A.P.).
More Information
In this proof, we used the properties of geometric and arithmetic progressions. The relationship reveals that the square of the middle term ( y ) equals the product of the other two terms ( x ) and ( z ), demonstrating the A.P. relationship.
Tips
- Mistaking Proportions for Ratios: Ensure to accurately translate between ratios and products.
- Forgetting Square Roots: When solving the equation derived from the G.P. conditions, always remember that ( y = \sqrt{xz} ) leads to the A.P. relationship.
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