15. (20 points) In the infinite sequence $\left\{x_{n}\right\}(n \geqslant 1)$, for each odd $n$, $x_{n}, x_{n+1}, x_{n+2}$ form a geometric sequence, and for each even $n$, $x_{n}, x_{n+1}, x_{n+2}$ form an arithmetic sequence. Given $x_{1}=a, x_{2}=b$.
(1) Find the general term formula of the sequence, and what necessary and sufficient conditions must the real numbers $a, b$ satisfy for such an infinite sequence to exist?
(2) Find the harmonic mean of $x_{2}, x_{4}, \cdots, x_{2 n}$, i.e., the value of $\frac{n}{n} 1$.
15. (1) Observe the first few terms: $a, b, \frac{b^{2}}{a}, \frac{b(2 b-a)}{a}$, $\frac{(2 b-a)^{2}}{a}, \frac{(2 b-a)(3 b-2 a)}{a}, \frac{(3 b-2 a)^{2}}{a}, \cdots$, and guess
$$
\begin{array}{l}
x_{2 k-1}=\frac{[(k-1) b-(k-2) a]^{2}}{a}, \\
x_{2 k}=\frac{[(k-1) b-(k-2) a][k b-(k-1) a]}{a}(k \geqslant 1) .
\end{array}
$$

Prove the general term formula by mathematical induction on $k$.
When $k=1$, it is obviously true.
Assume $x_{2 k-1}$ and $x_{2 k}$ are as above, then
$$
\begin{array}{c}
x_{2 k+1}=\frac{\left(x_{2 k}\right)^{2}}{x_{2 k-1}}=\frac{[k b-(k-1) a]^{2}}{a}, \\
x_{2 k+2}=2 x_{2 k+1}-x_{2 k} \\
=\frac{[k b-(k-1) a][(k+1) b-k a]}{a} .
\end{array}
$$

Therefore, the formula holds.
The necessary and sufficient condition for the existence of such an infinite sequence is:
All $x_{n} \neq 0 \Leftrightarrow \frac{b}{a} \notin\left\{\left.\frac{n}{n+1} \right\rvert\, n \in \mathbf{N}\right\}$.
(2) When $b \neq a$,
$$
\frac{1}{x_{2 k}}=\frac{a}{b-a}\left[\frac{1}{(k-1) b-(k-2) a}-\frac{1}{k b-(k-1) a}\right] \text {. }
$$

Thus, $\frac{n}{\sum_{k=1}^{n} \frac{1}{x_{2 k}}}=\frac{n}{\frac{a}{b-a}\left[\frac{1}{a}-\frac{1}{n b-(n-1) a}\right]}$
$$
=n b-(n-1) a \text {. }
$$

When $b=a$, all $x_{n}=a$, and the result is also correct.