Given that 7, 11, 15, 19, ... is an arithmetic progression, find the (a) tenth term, (b) sum of the first twelve terms.

Given the arithmetic progression $7, 11, 15, 19, \dots$. The first term is $a = 7$. The common difference $d$ is the difference between consecutive terms: $d = 11 - 7 = 4$.

a) Find the tenth term.

Step 1: Use the formula for the $n$-th term of an arithmetic progression, $T_n = a + (n-1)d$. For the tenth term, $n=10$. $T_{10} = a + (10-1)d$

Step 2: Substitute the values of $a$, $n$, and $d$ into the formula. $T_{10} = 7 + (9) \times 4$

Step 3: Calculate the value. $T_{10} = 7 + 36$ $T_{10} = 43$ The tenth term is $\boxed{43}$.

b) Find the sum of the first twelve terms.

Step 1: Use the formula for the sum of the first $n$ terms of an arithmetic progression, $S_n = \frac{n}{2}[2a + (n-1)d]$. For the sum of the first twelve terms, $n=12$. $S_{12} = \frac{12}{2}[2a + (12-1)d]$

Step 2: Substitute the values of $a$, $n$, and $d$ into the formula. $S_{12} = \frac{12}{2}[2(7) + (11) \times 4]$

Step 3: Simplify the expression inside the brackets. $S_{12} = 6[14 + 44]$ $S_{12} = 6[58]$

Step 4: Calculate the sum. $S_{12} = 348$ The sum of the first twelve terms is $\boxed{348}$.