Step 1: Recall the formula for the sum of the first \( n \) terms of an arithmetic progression (A.P.): \[ S_n = \frac{n}{2} [2a + (n - 1)d], \] where \( n \) represents the number of terms, \( a \) is the first term, and \( d \) is the common difference.
Step 2: Substitute the provided values:
- \( n = 10 \),
- \( a = 3 \),
- \( d = 2 \).
This yields: \[ S_{10} = \frac{10}{2} [2 \cdot 3 + (10 - 1) \cdot 2]. \]
Step 3: Execute the calculation systematically:
- Simplify the term multiplier: \[ S_{10} = 5 [6 + 9 \cdot 2]. \] - Calculate the expression within the brackets: \[ 9 \cdot 2 = 18, \quad 6 + 18 = 24. \] - Perform the final multiplication: \[ S_{10} = 5 \cdot 24 = 120. \]
Step 4: Confirm the result.
The sequence comprises 3, 5, 7, up to the 10th term.
The 10th term is calculated as \( a + (n - 1)d = 3 + 9 \cdot 2 = 21 \). An alternative formula for the sum of an A.P. is \( S_n = \frac{n}{2} (a + l) \), where \( l \) denotes the last term. Applying this:
\[ S_{10} = \frac{10}{2} (3 + 21) = 5 \cdot 24 = 120. \]