Find n so that:
(i) `(a^(n+1)+b^(n+1))/(a^(n)+b^(n))` (ii) `(a^(n)+b^(n))/(a^(n-1)+b^(n-10)` may be A.M. between a and b.

Answer

To solve the given problem, we need to find the value of \( n \) such that the two expressions are the arithmetic mean (A.M.) between \( a \) and \( b \).

### Part (i)

We start with the expression:

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

**Step 1:** Cross-multiply to eliminate the fraction:

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

**Step 2:** Expand both sides:

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

**Step 3:** Rearrange the equation to bring all terms to one side:

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

**Step 4:** Combine like terms:

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

**Step 5:** Factor the equation:

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

This can be factored as:

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

**Step 6:** Since \( a \) and \( b \) are not equal, we can set:

\[ a^n = b^n \]

This implies:

\[ \left(\frac{a}{b}\right)^n = 1 \]

**Step 7:** The only solution is:

\[ n = 0 \]

### Part (ii)

Now, we consider the second expression:

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

**Step 1:** Cross-multiply:

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

**Step 2:** Expand both sides:

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

**Step 3:** Rearrange the equation:

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

**Step 4:** Combine like terms:

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

**Step 5:** Factor the equation:

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

This can be factored as:

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

**Step 6:** Since \( a \) and \( b \) are not equal, we can set:

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

This implies:

\[ \left(\frac{a}{b}\right)^{n-1} = 1 \]

**Step 7:** The only solution is:

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

### Final Answers

- For part (i), \( n = 0 \)
- For part (ii), \( n = 1 \)