A series of books was published at 10-year intervals. When the 10th book was…

For n terms in an arithmetic progression (AP) with first term a and common difference d, the sum of the terms is S = n/2 x [2a + (n-1)d]. Equivalently, the sum equals n times the average of the first and last terms.

1. Let x be the year the first book was published. Since 10 books are published at 10-year intervals, their publication years form an AP: x, x+10, x+20, ..., x+90.

2. The sum of these 10 terms is 10x + (0+10+20+...+90) = 10x + 450.

3. This sum equals the given total, so 10x + 450 = 19560.

4. Solving: 10x = 19560 - 450 = 19110, so x = 1911.

As an independent check, the average of the 10 terms must equal 19560/10 = 1956. In an AP, the average of all terms equals the average of the first and last term, i.e. (x + (x+90))/2 = x+45. Setting x+45 = 1956 gives x = 1911, confirming the result independently.

The first book was published in 1911.