awk sum every 4th number – field

So my input file is:

1;a;b;2;c;d;3;e;f;4;g;h;5 
1;a;b;2;c;d;9;e;f;101;g;h;9 
3;a;b;1;c;d;3;e;f;10;g;h;5

I want to sum the numbers then write it to a file (so i need every 4th field).

I tried many sum examples on the net but i didnt found answer for my problem.

My ouput file should looks:

159

Thanks!

Update:

a;b;**2**;c;d;g
3;e;**3**;s;g;k
h;5;**2**;d;d;l

The problem is the same.

I want to sum the 3th numbers (But in the line it is 3th).

So 2+3+2.

Output: 7

Answer

Apparently you want to print every 3rd field, not every 4th. The following code loops through all fields, suming each one in a 3k+1 position.

$ awk -F";" '{for (i=1; i<=NF; i+=3) sum+=$i} END{print sum}' file
159

The value is printed after processing the whole file, in the END {} block.