We are given the expression \[ \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \] and it is stated that this expression is the geometric mean between a and b.
The geometric mean (G.M.) between a and b is \[ \sqrt{ab}. \]
Therefore, we have \[ \frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \sqrt{ab}. \]
Squaring both sides, we get \[ \left( \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \right)^2 = ab. \]
\[ \frac{(a^{n+1} + b^{n+1})^2}{(a^n + b^n)^2} = ab. \]
Cross-multiplying, \[ (a^{n+1} + b^{n+1})^2 = ab (a^n + b^n)^2. \]
Expanding both sides:
\[ a^{2n+2} + 2a^{n+1}b^{n+1} + b^{2n+2} \]
\[ = ab(a^{2n} + 2a^n b^n + b^{2n}) \]
\[ = a^{2n+1}b + 2a^{n+1}b^{n+1} + ab^{2n+1}. \]
Cancelling the common term \( 2a^{n+1}b^{n+1} \) from both sides, we get
\[ a^{2n+2} + b^{2n+2} = a^{2n+1}b + ab^{2n+1}. \]
Factorising both sides:
\[ a^{2n+1}(a - b) = b^{2n+1}(a - b). \]
Since \( a \neq b \), dividing both sides by \( (a - b) \), we obtain
\[ a^{2n+1} = b^{2n+1}. \]
This is possible only if \[ 2n + 1 = 0. \]
Solving, \[ n = -\frac{1}{2}. \]
Thus, the required value of n is \[ \boxed{-\frac{1}{2}}. \]
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab) n .