11. 已知成等差数列的四个数之和为 26 , 第二个数与第三个数之积为 40 , 求这四个数.
设这四个数为 $a-3 d, a-d, a+d, a+3 d$, 则由题设得

$\left\{\begin{array}{l}(a-3 d)+(a-d)+(a+d)+(a+3 d)=26, \\ (a-d)(a+d)=40,\end{array}\right.$

$\therefore\left\{\begin{array}{l}4 a=26, \\ a^{2}-d^{2}=40 .\end{array}\right.$ 解得 $\left\{\begin{array}{l}a=\frac{13}{2}, \\ d=\frac{3}{2}\end{array}\right.$ 或 $\left\{\begin{array}{l}a=\frac{13}{2}, \\ d=-\frac{3}{2} .\end{array}\right.$ 所以这四个数为 $2,5,8,11$ 或 $11,8,5,2$.