Let a and b be two positive unequal numbers , then `(a^(n+1)+b^(n+1))/(a^(n)+b^(n))` represents

Options

A. their A.M if n = 0
B. their G.M if `n = 1/2 `
C. their H.M if `n=-1`
D. none of these

Answer

To solve the problem, we need to analyze the expression given:

\[ \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \]

where \( a \) and \( b \) are two positive unequal numbers.

### Step 1: Substitute values for \( n \)

Let's evaluate the expression for specific values of \( n \).

1. **For \( n = 0 \)**:
\[ \frac{a^{0+1} + b^{0+1}}{a^0 + b^0} = \frac{a^1 + b^1}{a^0 + b^0} = \frac{a + b}{1 + 1} = \frac{a + b}{2} \]
This represents the **Arithmetic Mean (AM)** of \( a \) and \( b \).

### Step 2: Check for \( n = 1 \)

2. **For \( n = 1 \)**:
\[ \frac{a^{1+1} + b^{1+1}}{a^1 + b^1} = \frac{a^2 + b^2}{a + b} \]
This does not represent a standard mean directly but can be simplified further.

### Step 3: Check for \( n = -1 \)

3. **For \( n = -1 \)**:
\[ \frac{a^{-1+1} + b^{-1+1}}{a^{-1} + b^{-1}} = \frac{a^0 + b^0}{\frac{1}{a} + \frac{1}{b}} = \frac{1 + 1}{\frac{b + a}{ab}} = \frac{2}{\frac{a + b}{ab}} = \frac{2ab}{a + b} \]
This represents the **Harmonic Mean (HM)** of \( a \) and \( b \).

### Conclusion

From the evaluations, we find:

- For \( n = 0 \), the expression represents the Arithmetic Mean of \( a \) and \( b \).
- For \( n = -1 \), the expression represents the Harmonic Mean of \( a \) and \( b \).

Thus, the expression \( \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \) can represent both the Arithmetic Mean and the Harmonic Mean depending on the value of \( n \).

### Final Answer

The expression represents:
- Arithmetic Mean when \( n = 0 \)
- Harmonic Mean when \( n = -1 \)