50. 令 $a_{1}=r \cos \theta, a_{10}=r \sin \theta, 0 \leqslant r \leqslant \sqrt{10}$ ，则 $$ 9 d=r(\sin \theta-\cos \theta) \Rightarrow d=\frac{1}{9} r(\sin \theta-\cos \theta) $$ 故 $$ \begin{aligned} S & =\left(a_{1}+9 d\right)+\left(a_{1}+10 d\right)+\cdots+\left(a_{1}+18 d\right)=10 a_{1}+135 d \\ & =10 r \cos \theta+15 r(\sin \theta-\cos \theta)=5 r(3 \sin \theta-\cos \theta) \\ & =5 \sqrt{10} r \sin (\theta-\varphi) \leqslant 50 \end{aligned} $$ 即 $S$ 的最大值是 50 ．