To find the sum of the given series, we can first notice that the numerator of each fraction is increasing by 5 each time.

So, we can rewrite the series as:

A = (3/200) + (8/200) + (13/200) + (18/200) + ... + (193/200) + (198/200)

= (3 + 8 + 13 + 18 + ... + 193 + 198) / 200

Now, we can find the sum of the numerator:

3 + 8 + 13 + 18 + ... + 193 + 198

This is an arithmetic series with a common difference of 5 and a first term of 3. We can use the formula for the sum of an arithmetic series to find the sum:

Sn = (n/2)(2a + (n-1)d)

where Sn is the sum of the series, n is the number of terms, a is the first term, and d is the common difference.

In this case, n = (198 - 3)/5 + 1 = 40, a = 3, and d = 5.

So, the sum of the numerator is:

Sn = (40/2)(2(3) + (40-1)(5))

= 20(6 + 39(5))

= 20(6 + 195)

= 20(201)

= 4020

Now, we can substitute this value back into the original equation:

A = 4020/200

= 20.1

Therefore, the sum of the series A is 20.1.