In the arithmetic mean of a and b is `(a^(n)+b^(n))/(a^(n-1)+b^(n-1))` then n= ?

Options

A. 0
B. 1
C. -1
D. None of these

Explanation

To solve the problem, we need to find the value of \( n \) such that the arithmetic mean of \( a \) and \( b \) is given by the expression:

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

The arithmetic mean of \( a \) and \( b \) is defined as:

\[ \frac{a + b}{2} \]

We will set these two expressions equal to each other:

\[ \frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \frac{a + b}{2} \]

### Step 1: Cross-multiply the equation

Cross-multiplying gives us:

\[ 2(a^n + b^n) = (a + b)(a^{n-1} + b^{n-1}) \]

### Step 2: Expand the right-hand side

Expanding the right-hand side:

\[ 2(a^n + b^n) = a \cdot a^{n-1} + a \cdot b^{n-1} + b \cdot a^{n-1} + b \cdot b^{n-1} \]

This simplifies to:

\[ 2(a^n + b^n) = a^n + a b^{n-1} + b a^{n-1} + b^n \]

### Step 3: Rearrange the equation

Rearranging gives us:

\[ 2a^n + 2b^n - a^n - b^n - ab^{n-1} - ba^{n-1} = 0 \]

This simplifies to:

\[ a^n + b^n - ab^{n-1} - ba^{n-1} = 0 \]

### Step 4: Factor the equation

We can factor this equation as follows:

\[ a^n - ab^{n-1} + b^n - ba^{n-1} = 0 \]

### Step 5: Set up for further simplification

Rearranging gives us:

\[ a^n - b^n = ab^{n-1} - ba^{n-1} \]

### Step 6: Factor out common terms

Factoring out common terms gives us:

\[ a^n - b^n = (a - b)(b^{n-1} + ab^{n-2} + a^2b^{n-3} + \ldots + a^{n-1}) \]

### Step 7: Analyze the equation

For the equality to hold, we need \( n = 1 \) or \( a = b \). If \( a \neq b \), we can conclude that:

\[ a^{n-1} + b^{n-1} = 0 \]

### Step 8: Conclusion

Thus, the value of \( n \) that satisfies the original equation is:

\[ \boxed{1} \]