\( a_{1}=2 ; d=8 \) da: \( \sum_{k=10}^{k=25} a_{k}=? \)

To solve the arithmetic series given \( a_{1}=2 \) and \( d=8 \), we can find the general term \( a_k \) by using the formula \( a_k = a_1 + (k-1)d \). This gives us: \[ a_k = 2 + (k-1) \cdot 8 = 2 + 8k - 8 = 8k - 6 \] Now we plug in the limits from \( k=10 \) to \( k=25 \) into our formula: \[ \sum_{k=10}^{25} a_k = \sum_{k=10}^{25} (8k - 6) \] Splitting the summation gives: \[ \sum_{k=10}^{25} (8k) - \sum_{k=10}^{25} 6 \] Calculating each part: 1. **For** \( \sum_{k=10}^{25} 8k \): This can be factored out as \( 8 \sum_{k=10}^{25} k \). The sum \( \sum_{k=10}^{25} k \) can be calculated using the formula for the sum of an arithmetic series: \[ \text{Number of terms} = 25 - 10 + 1 = 16 \] \[ \text{Sum} = \frac{n}{2} (a + l) = \frac{16}{2} \cdot (10 + 25) = 8 \cdot 35 = 280 \] Thus, \[ \sum_{k=10}^{25} 8k = 8 \cdot 280 = 2240 \] 2. **For** \( \sum_{k=10}^{25} 6 \): This is simply \( 6 \cdot \text{Number of terms} = 6 \cdot 16 = 96 \). Now we combine both parts: \[ \sum_{k=10}^{25} a_k = 2240 - 96 = 2144 \] Therefore, the final answer is: \[ \sum_{k=10}^{25} a_k = 2144 \]