Because the areas of the 1, the 4th, 2nd and 3rd small rectangles from left to right sequentially form a arithmetic progression with a tolerance of 0. 1,

Let the frequency of the first group be a, then the frequencies of the fourth, second and third groups are a+0. 1, a+0.2 and a+0.3 respectively.

Because the sum of all frequencies is 1,

So a+a+0.1+a+0.2+a+0.3 =1,

The solution is a=0. 1,

The frequency of the first group is 10,

All n =10 0.1=100.

So choose C.