【Analysis and Solution】
Arithmetic sequence.
(Method One)
The subsequent box contains $x$ more candies than the previous one.
Box 1 contains $(11-x)$ candies, Box 2 contains 11 candies, Box 3 contains $(11+x)$ candies, Box 4 contains $(11+2 x)$ candies, …, Box 10 contains $(11+8 x)$ candies;

According to the problem, we have $(11-x)+11+(11+x)+(11+2 x)+\cdots+(11+8 x)=320$;
Solving this, we get $x=6$;
The subsequent box contains 6 more candies than the previous one.
(Method Two)
Since Box 2 contains 11 candies;
Therefore, Boxes 1 and 3 together contain $11 \times 3=33$ candies;
Therefore, Boxes 4 to 10 together contain $320-33=287$ candies;
Therefore, Box 7 contains $287 \div 7=41$ candies;
The subsequent box contains $(41-11) \div(7-2)=6$ more candies than the previous one.
【Question 26】
There are 320 candies in 10 boxes numbered 1-10. It is known that each box contains the same additional number of candies compared to the previous one. If the 2nd box contains 11 candies, then the number of additional candies in each subsequent box compared to the previous one is $\qquad$. 

保留源文本的换行和格式，直接输出翻译结果。