14. Let $a, b$ be positive integers, two lines $l_{1}: y=$ $-\frac{b}{2 a} x+b$ and $l_{2}: y=\frac{b}{2 a} x$ intersect at $\left(x_{1}, y_{1}\right)$. For natural number $n(n \geqslant 2)$, the line passing through the points $(0, b)$ and $\left(x_{n-1}, 0\right)$ intersects the line $l_{2}$ at $\left(x_{n}, y_{n}\right)$. Find the general terms of the sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$.
$[2,3]$ is monotonically decreasing, find the range of values for $k$.

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For the first part of the problem, we need to find the intersection point of the lines $l_1$ and $l_2$.

1. **Finding the intersection point $\left(x_1, y_1\right)$:**

   The equations of the lines are:
   \[
   l_1: y = -\frac{b}{2a} x + b
   \]
   \[
   l_2: y = \frac{b}{2a} x
   \]

   Setting the equations equal to each other to find the intersection:
   \[
   -\frac{b}{2a} x + b = \frac{b}{2a} x
   \]
   \[
   b = \frac{b}{2a} x + \frac{b}{2a} x
   \]
   \[
   b = \frac{b}{a} x
   \]
   \[
   x = a
   \]

   Substituting $x = a$ back into the equation of $l_2$:
   \[
   y = \frac{b}{2a} a = \frac{b}{2}
   \]

   Therefore, the intersection point is:
   \[
   \left(x_1, y_1\right) = \left(a, \frac{b}{2}\right)
   \]

2. **Finding the general terms $\left\{x_n\right\}$ and $\left\{y_n\right\}$:**

   For $n \geq 2$, the line passing through $(0, b)$ and $\left(x_{n-1}, 0\right)$ has the equation:
   \[
   y = -\frac{b}{x_{n-1}} x + b
   \]

   This line intersects $l_2$ at $\left(x_n, y_n\right)$. Setting the equations equal:
   \[
   -\frac{b}{x_{n-1}} x + b = \frac{b}{2a} x
   \]
   \[
   b = \frac{b}{2a} x + \frac{b}{x_{n-1}} x
   \]
   \[
   b = b \left( \frac{1}{2a} + \frac{1}{x_{n-1}} \right) x
   \]
   \[
   1 = \left( \frac{1}{2a} + \frac{1}{x_{n-1}} \right) x
   \]
   \[
   x = \frac{1}{\frac{1}{2a} + \frac{1}{x_{n-1}}}
   \]
   \[
   x = \frac{2a x_{n-1}}{2a + x_{n-1}}
   \]

   Therefore, the general term for $x_n$ is:
   \[
   x_n = \frac{2a x_{n-1}}{2a + x_{n-1}}
   \]

   To find $y_n$, substitute $x_n$ into the equation of $l_2$:
   \[
   y_n = \frac{b}{2a} x_n = \frac{b}{2a} \cdot \frac{2a x_{n-1}}{2a + x_{n-1}} = \frac{b x_{n-1}}{2a + x_{n-1}}
   \]

   Therefore, the general term for $y_n$ is:
   \[
   y_n = \frac{b x_{n-1}}{2a + x_{n-1}}
   \]

3. **Finding the range of $k$ for the function to be monotonically decreasing on $[2,3]$:**

   The function is given as:
   \[
   f(x) = kx + \frac{1}{x}
   \]

   To determine if the function is monotonically decreasing, we need to check the derivative:
   \[
   f'(x) = k - \frac{1}{x^2}
   \]

   For the function to be monotonically decreasing on $[2,3]$, the derivative must be less than or equal to zero for all $x$ in this interval:
   \[
   k - \frac{1}{x^2} \leq 0
   \]
   \[
   k \leq \frac{1}{x^2}
   \]

   The maximum value of $\frac{1}{x^2}$ on $[2,3]$ is at $x = 2$:
   \[
   \frac{1}{2^2} = \frac{1}{4}
   \]

   Therefore, the range of $k$ is:
   \[
   k \leq \frac{1}{4}
   \]

   \[
   \boxed{k \leq \frac{1}{4}}
   \]
14. The line $l_{1}$ passes through the points $(2 a, 0)$ and $(0, b)$. It is easy to see that the intersection point of $l_{1}$ and $l_{2}$ is
$$
\left(x_{1}, y_{1}\right)=\left(a, \frac{b}{2}\right).
$$

The line passing through the points $(0, b)$ and $\left(x_{n-1}, 0\right)$ has the equation
$$
\frac{x}{x_{n-1}}+\frac{y}{b}=1,
$$

and its intersection with $l_{2}$ is $\left(x_{n}, y_{n}\right)$. Therefore,
$$
\begin{array}{l}
\frac{x_{n}}{x_{n-1}}+\frac{y_{n}}{b}=1, \text { and } y_{n}=\frac{b}{2 a} x_{n}. \\
\text { Hence } \frac{1}{x_{n}}-\frac{1}{x_{n-1}}=\frac{1}{2 a}(n \geqslant 2).
\end{array}
$$

This shows that the sequence $\left\{\frac{1}{x_{n}}\right\}$ is an arithmetic sequence with the first term $\frac{1}{a}$ and common difference $\frac{1}{2 a}$. Thus, $\frac{1}{x_{n}}=\frac{1}{a}+\frac{n-1}{2 a}=\frac{n+1}{2 a}$, which means
$$
x_{n}=\frac{2 a}{n+1}.
$$

From this, we can obtain $y_{n}=\frac{b}{2 a} x_{n}=\frac{b}{2 a} \cdot \frac{2 a}{n+1}=\frac{b}{n+1}$.