a) A = 2 + 5 + 8 + 11 + ... + 120

This is an arithmetic sequence with a common difference of 3. To find the sum of the sequence, we can use the formula for the sum of an arithmetic series:

Sn = n/2 * (a1 + an)

where:
Sn = sum of the first n terms
n = number of terms
a1 = first term
an = nth term

In this case, the first term (a1) is 2, the last term (an) is 120, and the common difference is 3. We need to find the number of terms in the sequence.

an = a1 + (n-1)d
120 = 2 + (n-1)3
120 = 2 + 3n - 3
120 = 3n - 1
121 = 3n
n = 40

Now we can plug in the values into the formula:

A = 40/2 * (2 + 120)
A = 20 * 122
A = 2440

Therefore, the sum of the sequence is 2440.

b) B = (1 - 1/2) * (1 - 1/3) * (1 - 1/4) * ... * (1 - 1/2011) * (1 - 1/2012)

This is a product of terms in the form (1 - 1/n). We can simplify this expression by expanding each term and then multiplying them together.

B = (1 - 1/2) * (1 - 1/3) * (1 - 1/4) * ... * (1 - 1/2011) * (1 - 1/2012)
B = (1/2) * (2/3) * (3/4) * ... * (2011/2012) * (2012/2013)
B = 1/2013

Therefore, the value of B is 1/2013.