Let the first term be \( a \) and the common difference be \( d \). The 5th term is given by: \[ a + 4d = 24 \quad \text{(i)} \] The 12th term is given by: \[ a + 11d = 94 \quad \text{(ii)} \] Subtract (i) from (ii): \[ (a + 11d) - (a + 4d) = 94 - 24 \] \[ 7d = 70 \Rightarrow d = 10 \] Substitute \( d = 10 \) in equation (i): \[ a + 4(10) = 24 \Rightarrow a = 24 - 40 = -16 \] Now, the sum of first 20 terms: \[ S_{20} = \frac{20}{2} \left[2a + (20 - 1)d\right] \] \[ = 10 [2(-16) + 19(10)] = 10 [-32 + 190] = 10 \times 158 = 1580 \]
The correct option is (D): \(1580\)