6. For $b>a>0$, take points $A_{k}\left(x_{k}\right.$, $\left.y_{k}\right)(k=1,2, \cdots, n)$ in the first quadrant, such that $a, x_{1}, x_{2}, \cdots, x_{n}, b$ form an arithmetic sequence, and $a, y_{1}, y_{2}, \cdots, y_{n}, b$ form a geometric sequence. Then the relationship between the points $A_{1}, A_{2}, \cdots, A_{n}$ and the ray $l: y=x(x>0)$ is $(\quad)$.
(A) All points are above the ray $l$
(B) All points are on the ray $l$
(C) All points are below the ray $l$
(D) Cannot be determined
6. (C).

From the general terms of arithmetic and geometric sequences, we have
$$
x_{k}=a+\frac{b-a}{n+1} \cdot k, \quad y_{k}=a\left(\frac{b}{a}\right)^{\frac{k}{n+1}}
$$

Thus,
$$
\begin{aligned}
x_{k} & =\frac{k b+(n+1-k) a}{n+1} \\
& =a \frac{k\left(\frac{b}{a}\right)+(n+1-k)}{n+1} \\
& >a \sqrt[n+1]{\left(\frac{b}{a}\right)^{k} \cdot 1^{n+1-k}} \text{ (inequality is strict) } \\
& =a\left(\frac{b}{a}\right)^{\frac{k}{n+1}}=y_{k} . \quad (k=1,2, \cdots, n)
\end{aligned}
$$

This shows that $A_{k}$ is below the ray $y=x(x>0)$. Note: Special values of $a$ and $b$ can help with understanding, such as:
$$
\begin{array}{l}
a=1, x_{1}=3 \frac{1}{3}, x_{2}=5 \frac{2}{3}, b=8, \\
a=1, y_{1}=2, y_{2}=4, b=8 .
\end{array}
$$