59 The arithmetic sequence $\left\{a_{n}\right\}$ has a positive common difference, $n>1$, then we have
A. $a_{1} a_{n+1}=a_{2} a_{n}$
B. $a_{1} a_{n+1} \geqslant a_{2} a_{n}$
C. $a_{1} a_{n+1}>a_{2} a_{n}$
D. $a_{1} a_{n+1}<a_{2} a_{n}$
59 D. Let the common difference of the sequence $\left\{a_{n}\right\}$ be $d$. Since $d>0, n>1$, we have
$$
a_{n}=a_{1}+(n-1) d>a_{1} .
$$

Therefore,
$$
\begin{aligned}
a_{1} a_{n+1}-a_{2} a_{n} & =a_{1}\left(a_{n}+d\right)-\left(a_{1}+d\right) a_{n} \\
& =d\left(a_{1}-a_{n}\right)<0,
\end{aligned}
$$

which means
$$
a_{1} a_{n+1}<a_{2} a_{n} .
$$