SOLUTION: Let a, b, c, p, q, r be positive real numbers such that a, b, c are in geometric sequence and {{{ a^p = b^q = c^r }}} then which one of the following condition holds: A- p, q, r a

Algebra ->  Sequences-and-series -> SOLUTION: Let a, b, c, p, q, r be positive real numbers such that a, b, c are in geometric sequence and {{{ a^p = b^q = c^r }}} then which one of the following condition holds: A- p, q, r a      Log On


   



Question 1184306: Let a, b, c, p, q, r be positive real numbers such that a, b, c are in geometric sequence and +a%5Ep+=+b%5Eq+=+c%5Er+ then which one of the following condition holds:
A- p, q, r are in geometric sequence
B- p, q, r are in arithmetic sequence
C- p, q, r are in harmonic sequence
D- p^2, q^2, r^2 are in arithmetic sequence
E- p^2, q^2, r^2 are geometric sequence
..
[Note: ^2 means power 2]

Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Here's how to determine the correct relationship between p, q, and r:
1. **Express the geometric sequence:**
Since a, b, and c are in a geometric sequence, we can write $b^2 = ac$.
2. **Express the given relationship:**
We are given that $ap = bq = cr$. Let's call this common value k. Then:
* $ap = k \Rightarrow a = \frac{k}{p}$
* $bq = k \Rightarrow b = \frac{k}{q}$
* $cr = k \Rightarrow c = \frac{k}{r}$
3. **Substitute into the geometric sequence equation:**
Substitute the expressions for a, b, and c into the equation $b^2 = ac$:
$(\frac{k}{q})^2 = (\frac{k}{p})(\frac{k}{r})$
$\frac{k^2}{q^2} = \frac{k^2}{pr}$
4. **Simplify:**
Since k is a positive real number, we can divide both sides by $k^2$:
$\frac{1}{q^2} = \frac{1}{pr}$
$q^2 = pr$
5. **Interpret the result:**
The equation $q^2 = pr$ means that p, q, and r are in a geometric sequence.
Therefore, the correct condition is that p, q, and r are in geometric sequence.
Final Answer: The final answer is $\boxed{A}$