If $(x + 1) + (x + 2) + ... + (x + 20) = 174 + 176 + 178 + ... + 192$, then what is the value of $x$?

$\mathrm{(A) \ } 80 \qquad \mathrm{(B) \ } 81 \qquad \mathrm {(C) \ } 82 \qquad \mathrm{(D) \ } 83 \qquad \mathrm{(E) \ } 84$

1. First, we need to express the left-hand side of the equation, which is the sum of the arithmetic series $(x + 1) + (x + 2) + \ldots + (x + 20)$. The sum of an arithmetic series can be calculated using the formula:
   \[
   S = \frac{n}{2} \left( a + l \right)
   \]
   where \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term. Here, \( n = 20 \), \( a = x + 1 \), and \( l = x + 20 \).

2. Applying the formula:
   \[
   S = \frac{20}{2} \left( (x + 1) + (x + 20) \right) = 10 \left( 2x + 21 \right) = 20x + 210
   \]

3. Next, we need to express the right-hand side of the equation, which is the sum of the arithmetic series \( 174 + 176 + 178 + \ldots + 192 \). The first term \( a \) is 174, the common difference \( d \) is 2, and the last term \( l \) is 192. We need to find the number of terms \( n \).

4. The number of terms \( n \) in an arithmetic series can be found using the formula:
   \[
   l = a + (n-1)d
   \]
   Solving for \( n \):
   \[
   192 = 174 + (n-1) \cdot 2 \implies 192 = 174 + 2n - 2 \implies 192 = 172 + 2n \implies 20 = 2n \implies n = 10
   \]

5. Now, we calculate the sum of the series:
   \[
   S = \frac{n}{2} \left( a + l \right) = \frac{10}{2} \left( 174 + 192 \right) = 5 \left( 366 \right) = 1830
   \]

6. Equating the two sums:
   \[
   20x + 210 = 1830
   \]

7. Solving for \( x \):
   \[
   20x = 1830 - 210 \implies 20x = 1620 \implies x = \frac{1620}{20} \implies x = 81
   \]

The final answer is \( \boxed{81} \).