1,4,7,10,13,16

To get from the first term 1 to the 2nd term 4 we add 3.

To get from the 2nd term 4 to the 3rd term 7 we add 3.  Therefore
To get from the 1st term 1 to the 3rd term 7 we must add 3 2 times.

To get from the 3rd term 7 to the 4th term 10 we add 3.  Therefore
To get from the 1st term 1 to the 4th term 10 we must add 3 3 times.

To get from the 4th term 10 to the 5th term 13 we add 3.  Therefore
To get from the 1st term 1 to the 5th term 13 we must add 3 4 times.

To get from the 5th term 13 to the 6th term 16 we add 3.  Therefore
To get from the 1st term 1 to the 6th term 16 we must add 3 5 times.


We assume that this pattern continues forever:

To get from the (n-1)th term an-1 to the nth term an we add 3.  Therefore
To get from the 1st term 1 to the nth term an we must add 3 n-1 times.

and we end up with 

To get from the 199th term a199 to the 200th term a200 we add 3.  Therefore
To get from the 1st term 1 to the 200th term a200 we must add 3 199 times.

Therefore 

a200 = 1 + 199×3 = 1 + 597 = 598

Actually the formula for the nth term when there is a common
difference d, such as the 3 that is added each time, is

an = a1 + (n-1)d 

where a1 is the first term and d is the common difference
and n is the number of terms.

This is called an "arithmetic sequence".

So we could have just substituted a1 = 1, d = 3 and n = 200
into

an = a1 + (n-1)d

a200 = 1 + (200-1)3

a200 = 1 + (199)3

a200 = 1 + 597

a200 = 598

Edwin