The sum of the first two terms of an arithmetic series is 13 and the sum of the four terms is 46. Determine the first six terms of the series and the sum of the first six terms.

To solve this problem, we need to find the common difference of the arithmetic series first.

Let's assume that the first term of the series is "a" and the common difference is "d".

Given that the sum of the first two terms is 13, we can write the equation as:
a + (a + d) = 13

And the sum of the first four terms is 46, so we have:
a + (a + d) + (a + 2d) + (a + 3d) = 46

Now we can solve these two equations to find the values of "a" and "d".

From the first equation, we have:
2a + d = 13 (equation 1)

From the second equation, we have:
4a + 6d = 46 (equation 2)

To solve these equations, we can multiply equation 1 by 2 and subtract it from equation 2, which gives us:
4a + 6d - (4a + 2d) = 46 - 26

Simplifying, we get:
4d = 20
d = 5

Now, substituting the value of "d" into equation 1, we can solve for "a":
2a + 5 = 13
2a = 8
a = 4

So, the first term of the series, "a", is 4 and the common difference, "d", is 5.

The first six terms of the series would be:
4, 9, 14, 19, 24, 29

To find the sum of the first six terms, we can use the formula for the sum of an arithmetic series:

Sum(n) = (n/2) * (2a + (n-1)d)

In this case, n is 6 (the number of terms), a is 4 (the first term), and d is 5 (the common difference).

Using the formula, we can calculate the sum:
Sum(6) = (6/2) * (2(4) + (6-1)(5))
= 3 * (8 + 5(5))
= 3 * (8 + 25)
= 3 * 33
= 99

Therefore, the sum of the first six terms of the series is 99.