Write the first six terms of the arithmetic sequence with the given properties. a = 3, d = 2

a. 3,8,13,18,23,28
b. 3,5,7,9,11,13
c. 3,6,8,10,12,14
d. 2,7,12,17,22,27
e. 2,4,6,8,10,12

d and e are obviously out, since they do not start with 3.

Of the others, which has terms that increase by 2?

Well then, is it letter A?

To find the first six terms of an arithmetic sequence, you need to use the formula for the nth term of an arithmetic sequence given by:

a_n = a + (n-1)d

where:
a_n is the nth term of the sequence
a is the first term of the sequence
d is the common difference between terms
n is the position of the term in the sequence

In this case, the first term (a) is 3, and the common difference (d) is 2. Let's calculate the values for n = 1, 2, 3, 4, 5, and 6.

For n = 1:
a_1 = 3 + (1-1)2 = 3

For n = 2:
a_2 = 3 + (2-1)2 = 3 + 2 = 5

For n = 3:
a_3 = 3 + (3-1)2 = 3 + 4 = 7

For n = 4:
a_4 = 3 + (4-1)2 = 3 + 6 = 9

For n = 5:
a_5 = 3 + (5-1)2 = 3 + 8 = 11

For n = 6:
a_6 = 3 + (6-1)2 = 3 + 10 = 13

So, the first six terms of the arithmetic sequence with a(1)=3 and d=2 are: 3, 5, 7, 9, 11, 13.

Therefore, the correct answer is b. 3, 5, 7, 9, 11, 13.