If `(p)/(q)=(r)/(s)=(t)/(u)` then `(p^(2)+r^(2)+t^(2))/(q^(2)+s^(2)+u^(2))=`

Options

A. `(prt)/(qsu)`
B. `1`
C. `(p^(3)+r^(3)+t^(3))/(q^(3)+s^(3)+u^(3))`
D. `((p+r+t)/(q+s+u))^(2)`

Answer

To solve the problem, we start with the given condition:

\[ \frac{p}{q} = \frac{r}{s} = \frac{t}{u} \]

Let's denote this common ratio as \(\lambda\). Thus, we can express \(p\), \(r\), and \(t\) in terms of \(q\), \(s\), and \(u\):

1. **Express \(p\), \(r\), and \(t\) in terms of \(\lambda\)**:
\[ p = \lambda q \]
\[ r = \lambda s \]
\[ t = \lambda u \]

2. **Substitute \(p\), \(r\), and \(t\) into the expression**:
We need to find the value of:
\[ \frac{p^2 + r^2 + t^2}{q^2 + s^2 + u^2} \]
Substituting the values of \(p\), \(r\), and \(t\):
\[ = \frac{(\lambda q)^2 + (\lambda s)^2 + (\lambda u)^2}{q^2 + s^2 + u^2} \]

3. **Simplify the numerator**:
The numerator becomes:
\[ = \frac{\lambda^2 q^2 + \lambda^2 s^2 + \lambda^2 u^2}{q^2 + s^2 + u^2} \]

4. **Factor out \(\lambda^2\)**:
We can factor out \(\lambda^2\) from the numerator:
\[ = \frac{\lambda^2 (q^2 + s^2 + u^2)}{q^2 + s^2 + u^2} \]

5. **Cancel out the common terms**:
Since \(q^2 + s^2 + u^2\) appears in both the numerator and the denominator, we can cancel them out:
\[ = \lambda^2 \]

Thus, the final result is:
\[ \frac{p^2 + r^2 + t^2}{q^2 + s^2 + u^2} = \lambda^2 \]