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:

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

### Step 1: Write the formula for the arithmetic mean
The arithmetic mean (AM) of two numbers \( a \) and \( b \) is given by:

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

### Step 2: Set the two expressions equal to each other
According to the problem, we can set the arithmetic mean equal to the given expression:

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

### Step 3: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:

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

### Step 4: Expand the right-hand side
Now, we expand 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:

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

### Step 5: Rearrange the equation
Now, we can rearrange the equation:

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

This simplifies to:

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

### Step 6: Factor the equation
We can factor the left-hand side:

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

This can be rewritten as:

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

### Step 7: Set each factor to zero
Since \( a \) and \( b \) are different, we can set the terms equal to zero:

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

### Step 8: Solve for \( n \)
The above equation implies that:

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

This equality holds true if \( n - 1 = 0 \) (since \( a \) and \( b \) are different). Therefore:

\[ n - 1 = 0 \implies n = 1 \]

### Final Answer
Thus, the value of \( n \) is:

\[ \boxed{1} \]