28. $\sum_{r=1}^{10} D_r$ is
Step-by-Step Solution
Key Concept: The determinant D_r is a linear function of r. The sum of a linear function over an arithmetic progression is given by the sum of the first and last terms multiplied by the number of terms divided by 2.
The determinant is $D_r = \begin{vmatrix} 1 & 10 & 2r \\ 70 & 17 & 3r+1 \\ 1 & 1 & 1 \end{vmatrix}$. Expanding along the third row: $D_r = 1(10(3r+1) - 17(2r)) - 1(1(3r+1) - 70(2r)) + 1(17 - 700) = (30r + 10 - 34r) - (3r + 1 - 140r) - 683 = -4r + 10 + 137r - 1 - 683 = 133r - 674$. Then $\sum_{r=1}^{10} D_r = 133 \sum_{r=1}^{10} r - \sum_{r=1}^{10} 674 = 133 \times \frac{10 \times 11}{2} - 6740 = 133 \times 55 - 6740 = 7315 - 6740 = 575$.
Correct Answer: 575.00