The sequence given is:
2, 5, 11, 20, …
The general term for the n-th row of this arithmetic progression can be expressed as:
\( T_n = \frac{3n^2 - 3n + 4}{2} \)
For the 10th row, we substitute \( n = 10 \):
\( T_{10} = \frac{3(100) - 3(10) + 4}{2} = \frac{300 - 30 + 4}{2} = \frac{274}{2} = 137 \)
Since there are 10 terms in the 10th row, with a common difference \( c.d. = 3 \), the sum of the terms of the 10th row is given by:
\( \text{Sum} = \frac{10}{2} (2 \times 137 + 9 \times 3) \)
Calculating:
\( \text{Sum} = 5 (274 + 27) = 5 \times 301 = 1505 \)
Answer: 1505
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: