If a/b=c/d=e/f, then the value of `((a^np+c^nq+e^nr)/(b^np+d^nq+f^nr))^(1/n)`

Options

A. `(ad)/(bc)`
B. `(af)/(be)`
C. `(ck)/(dk)`
D. none of these

Answer

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

If \( \frac{a}{b} = \frac{c}{d} = \frac{e}{f} \), let's denote this common ratio as \( k \). Therefore, we can express \( a, c, e \) in terms of \( b, d, f \) respectively:

\[ a = bk, \quad c = dk, \quad e = fk \]

Now, we need to find the value of:

\[ \left( \frac{a^{np} + c^{nq} + e^{nr}}{b^{np} + d^{nq} + f^{nr}} \right)^{\frac{1}{n}} \]

### Step 1: Substitute \( a, c, e \)

Substituting the expressions for \( a, c, e \):

\[ = \left( \frac{(bk)^{np} + (dk)^{nq} + (fk)^{nr}}{b^{np} + d^{nq} + f^{nr}} \right)^{\frac{1}{n}} \]

### Step 2: Simplify the numerator

The numerator can be simplified as follows:

\[ = \left( \frac{b^{np} k^{np} + d^{nq} k^{nq} + f^{nr} k^{nr}}{b^{np} + d^{nq} + f^{nr}} \right)^{\frac{1}{n}} \]

### Step 3: Factor out \( k^{np} \)

Now, factor out \( k^{np} \) from the numerator:

\[ = \left( \frac{k^{np} \left( b^{np} + d^{nq} \frac{k^{nq}}{k^{np}} + f^{nr} \frac{k^{nr}}{k^{np}} \right)}{b^{np} + d^{nq} + f^{nr}} \right)^{\frac{1}{n}} \]

### Step 4: Simplify the expression

Notice that \( \frac{k^{nq}}{k^{np}} = k^{nq - np} \) and \( \frac{k^{nr}}{k^{np}} = k^{nr - np} \). Thus, we can rewrite the expression:

\[ = \left( k^{np} \cdot \frac{b^{np} + d^{nq} k^{nq - np} + f^{nr} k^{nr - np}}{b^{np} + d^{nq} + f^{nr}} \right)^{\frac{1}{n}} \]

### Step 5: Apply the exponent

Now, we apply the exponent \( \frac{1}{n} \):

\[ = k^{\frac{np}{n}} \cdot \left( \frac{b^{np} + d^{nq} k^{nq - np} + f^{nr} k^{nr - np}}{b^{np} + d^{nq} + f^{nr}} \right)^{\frac{1}{n}} \]

### Step 6: Final simplification

Since \( k^{\frac{np}{n}} = k^{p} \), we can express the final answer as:

\[ = k^{p} \cdot \left( \frac{b^{np} + d^{nq} k^{nq - np} + f^{nr} k^{nr - np}}{b^{np} + d^{nq} + f^{nr}} \right)^{\frac{1}{n}} \]

### Conclusion

Since all terms in the numerator and denominator are proportional to \( k \), we can conclude that the final answer is simply:

\[ = k \]