For what value of `n , (a ^(n +3) + b ^(n +3))/(a ^(n +2) + b ^(n +2)), a ne b` in the A.M. of a and b.

Answer

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

\[ \frac{a^{(n+3)} + b^{(n+3)}}{a^{(n+2)} + b^{(n+2)}} \]

is the arithmetic mean (A.M.) of \( a \) and \( b \), which is given by

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

### Step-by-step Solution:

1. **Set up the equation**: We want to equate the given expression to the arithmetic mean. Thus, we have:

\[ \frac{a^{(n+3)} + b^{(n+3)}}{a^{(n+2)} + b^{(n+2)}} = \frac{a + b}{2}. \]

2. **Cross-multiply**: To eliminate the fraction, we cross-multiply:

\[ 2(a^{(n+3)} + b^{(n+3)}) = (a + b)(a^{(n+2)} + b^{(n+2)}). \]

3. **Expand the right-hand side**: We expand the right-hand side:

\[ 2(a^{(n+3)} + b^{(n+3)}) = a \cdot a^{(n+2)} + a \cdot b^{(n+2)} + b \cdot a^{(n+2)} + b \cdot b^{(n+2)}. \]

This simplifies to:

\[ 2(a^{(n+3)} + b^{(n+3)}) = a^{(n+3)} + ab^{(n+2)} + ba^{(n+2)} + b^{(n+3)}. \]

4. **Rearranging the equation**: We can rearrange the equation to group similar terms:

\[ 2a^{(n+3)} + 2b^{(n+3)} - a^{(n+3)} - b^{(n+3)} - ab^{(n+2)} - ba^{(n+2)} = 0. \]

This simplifies to:

\[ a^{(n+3)} + b^{(n+3)} - (ab^{(n+2)} + ba^{(n+2)}) = 0. \]

5. **Factoring out common terms**: We can factor out \( a^{(n+2)} \) and \( b^{(n+2)} \):

\[ a^{(n+2)}(a - b) + b^{(n+2)}(b - a) = 0. \]

Since \( a \neq b \), we can divide by \( a - b \):

\[ b^{(n+2)} - a^{(n+2)} = 0. \]

6. **Setting the exponents equal**: This leads us to:

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

Since \( a \neq b \), the only way this can hold true is if \( n + 2 = 0 \).

7. **Solve for \( n \)**: Thus, we find:

\[ n + 2 = 0 \implies n = -2. \]

### Final Answer:
The value of \( n \) is \( -2 \).