Given:
\[ d \rightarrow \text{common difference.} \]
The general term:
\[ A_k = kd \left[ 2a + (2k - 1)d \right] \]
Given:
\[ A_3 = -153 \] \[ \Rightarrow 153 = 13d \left[ 2a + 5d \right] \]
Simplifying:
\[ 51 = d \left[ 2a + 5d \right] \quad \dots (1) \]
Also, given:
\[ A_5 = -435 \] \[ 435 = 5d \left[ 2a + 9d \right] \]
Simplifying:
\[ 87 = d \left[ 2a + 9d \right] \quad \dots (2) \]
Subtracting equation (1) from equation (2):
\[ 36 = 4d^2 \] \[ d = 3, \quad a = 1 \]
Finally:
\[ a_{17} - A_7 = 49 - \left[ -7.3 \left[ 2 + 39 \right] \right] = 910 \]
Let $ a_1, a_2, a_3, \ldots $ be in an A.P. such that $$ \sum_{k=1}^{12} 2a_{2k - 1} = \frac{72}{5}, \quad \text{and} \quad \sum_{k=1}^{n} a_k = 0, $$ then $ n $ is:
The sum $ 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + ... $ upto $ \infty $ terms, is equal to
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: