The sum of the first 4 termis of an arithmetic sequence is 24 . The \( 7^{\text {th }} \) term exceeds the \( 3^{\text {rd }} \) term by 24. Determine the sum of the first 12 terms. opyright Kevin Smith / Berlut Books CC

To find the sum of the first 12 terms of the arithmetic sequence, we start with the information given. Let the first term be \( a \) and the common difference be \( d \). The sum of the first four terms can be expressed as: \[ S_4 = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d = 24 \] From this, we simplify to: \[ 2a + 3d = 12 \quad (1) \] Next, we know the 7th term \( (a + 6d) \) exceeds the 3rd term \( (a + 2d) \) by 24: \[ a + 6d = (a + 2d) + 24 \] This simplifies to: \[ 4d = 24 \Rightarrow d = 6 \quad (2) \] Substituting \( d = 6 \) into equation (1): \[ 2a + 3(6) = 12 \Rightarrow 2a + 18 = 12 \Rightarrow 2a = -6 \Rightarrow a = -3 \] Now we have \( a = -3 \) and \( d = 6 \). To find the sum of the first 12 terms \( S_{12} \), we use the sum formula for an arithmetic sequence: \[ S_n = \frac{n}{2} \times (2a + (n-1)d) \] Substituting \( n = 12 \): \[ S_{12} = \frac{12}{2} \times (2(-3) + (12-1)(6)) \] \[ = 6 \times (-6 + 11 \times 6) = 6 \times (-6 + 66) = 6 \times 60 = 360 \] Thus, the sum of the first 12 terms is \( 360 \).