`"If " (a^(n)+b^(n))/(a^(n-1)+b^(n-1))"` is the A.M. between" a and b, then find the value of n.

Explanation

To solve the problem, we need to determine the value of \( n \) such that the expression

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

is the arithmetic mean (A.M.) of \( a \) and \( b \).

### Step 1: Set up the equation
The arithmetic mean of \( a \) and \( b \) is given by:

\[ \text{A.M.} = \frac{a + b}{2} \]

We set the given expression equal to the arithmetic mean:

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

### Step 2: Cross-multiply
Cross-multiplying gives us:

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

### Step 3: Expand the right-hand side
Now, we expand the right-hand side:

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

This simplifies to:

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

### Step 4: Set up the equation
Now we can set our equation:

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

### Step 5: 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 6: Factor the equation
We can factor this equation:

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

This can be rewritten as:

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

### Step 7: Analyze the factors
From the equation, we have:

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

This gives us two cases:

1. \( a - b = 0 \) (which we discard since \( a \) and \( b \) are distinct)
2. \( a^{n-1} - b^{n-1} = 0 \)

### Step 8: Solve for \( n \)
The second case implies:

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

Taking the logarithm of both sides (assuming \( a \neq b \)), we can conclude that:

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

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

\[ \boxed{1} \]