Source: 1976 Euclid Part A Problem 2
-----

The sum of the series $2+5+8+11+14+...+50$ equals

$\textbf{(A) } 90 \qquad \textbf{(B) } 425 \qquad \textbf{(C) } 416 \qquad \textbf{(D) } 442 \qquad \textbf{(E) } 495$
1. **Identify the series and its properties:**
   The given series is \(2, 5, 8, 11, 14, \ldots, 50\). This is an arithmetic series where the first term \(a = 2\) and the common difference \(d = 3\).

2. **Find the number of terms in the series:**
   To find the number of terms \(n\), we use the formula for the \(n\)-th term of an arithmetic series:
   \[
   a_n = a + (n-1)d
   \]
   Given \(a_n = 50\), we solve for \(n\):
   \[
   50 = 2 + (n-1) \cdot 3
   \]
   \[
   50 = 2 + 3n - 3
   \]
   \[
   50 = 3n - 1
   \]
   \[
   51 = 3n
   \]
   \[
   n = \frac{51}{3} = 17
   \]
   So, there are 17 terms in the series.

3. **Calculate the sum of the series:**
   The sum \(S_n\) of the first \(n\) terms of an arithmetic series is given by:
   \[
   S_n = \frac{n}{2} \left(2a + (n-1)d\right)
   \]
   Substituting \(n = 17\), \(a = 2\), and \(d = 3\):
   \[
   S_{17} = \frac{17}{2} \left(2 \cdot 2 + (17-1) \cdot 3\right)
   \]
   \[
   S_{17} = \frac{17}{2} \left(4 + 16 \cdot 3\right)
   \]
   \[
   S_{17} = \frac{17}{2} \left(4 + 48\right)
   \]
   \[
   S_{17} = \frac{17}{2} \cdot 52
   \]
   \[
   S_{17} = 17 \cdot 26
   \]
   \[
   S_{17} = 442
   \]

4. **Verify the result:**
   The sum of the series is indeed 442.

\(\boxed{442}\)