10. (2004 National High School Competition Fujian Province Preliminary Question) 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., $\frac{n}{\sum_{k=1}^{n} \frac{1}{x_{2 k}}}$.
10. (1) Calculate the first few terms: $x_{1}=a, x_{2}=b, x_{3}=\frac{b^{2}}{a}, x_{4}=\frac{b(2 b-a)}{a}, x_{5}=\frac{(2 b-a)^{2}}{a}, x_{6}=$ $\frac{(2 b-a)(3 b-2 a)}{a}, x_{7}=\frac{(3 b-2 a)^{2}}{a}, \cdots$, we conjecture that $x_{2 k-1}=\frac{[(k-1) b-(k-2) a]^{2}}{a}, x_{2 k}=$ $\frac{[(k-1) b-(k-2) a][k b-(k-1) a]}{a}\left(k \in \mathbf{N}_{+}\right)$. In fact, when $k=1$, it is obviously true. Assuming $x_{2 k-1}, x_{2 k}$ as above, then $x_{2 k+1}=\frac{\left(x^{2 k}\right)^{2}}{x_{2 k-1}}=\frac{[k b-(k-1) a]^{2}}{a}, x_{2 k+2}=2 x_{2 k+1}-x_{2 k}=\frac{[k b-(k-1) a][(k+1) b-k a)]}{a}$, thus the general term formula holds, $\left\{x_{n}\right\}$ exists if and only if $x_{n} \neq 0 \Leftrightarrow \frac{b}{a} \neq \frac{n-1}{n}\left(n \in \mathbf{N}_{+}\right)$.
( 2 ) When $b \neq a$, $\frac{1}{x_{2 k}}=\frac{a}{b-a}\left[\frac{1}{(k-1) b-(k-2) a}-\frac{1}{k b-(k-1) a}\right]$, hence $\frac{n}{\sum_{k=1}^{n} \frac{1}{x_{2 k}}}=$ $\frac{n}{\frac{n}{b-a}\left[\frac{1}{a}-\frac{1}{n b-(n-1) a}\right]}=n b-(n-1) a$, when $b=a$, all $x_{n}=a$, the result is also correct.