4 The sixth term of an arithmetic series is 32 and the sum of the first nine terms is 234. Determine the first three terms and the \( n^{\text {th }} \) term.

To tackle this problem, we can start by using the formula for the \( n^{th} \) term of an arithmetic sequence, which is given by: \[ a_n = a_1 + (n - 1)d \] Here, \( a_n \) is the \( n^{th} \) term, \( a_1 \) is the first term, and \( d \) is the common difference. According to the information provided: 1. For the sixth term (\( n = 6 \)), we have: \[ a_6 = a_1 + 5d = 32 \quad \text{(1)} \] 2. The sum of the first \( n \) terms of an arithmetic series can be calculated by the formula: \[ S_n = \frac{n}{2} \times (2a_1 + (n - 1)d) \] For the sum of the first nine terms (\( n = 9 \)), we have: \[ S_9 = \frac{9}{2} \times (2a_1 + 8d) = 234 \quad \text{(2)} \] Now let's simplify equation (2): \[ \frac{9}{2} \times (2a_1 + 8d) = 234 \] Multiplying both sides by \( 2 \): \[ 9 \times (2a_1 + 8d) = 468 \] Dividing both sides by \( 9 \): \[ 2a_1 + 8d = 52 \quad \text{(3)} \] Now we have two equations: (1) and (3). We can solve these equations simultaneously. From equation (1): \[ a_1 + 5d = 32 \quad \Rightarrow \quad a_1 = 32 - 5d \quad \text{(4)} \] Substituting equation (4) into equation (3): \[ 2(32 - 5d) + 8d = 52 \] Expanding this gives: \[ 64 - 10d + 8d = 52 \] Simplifying: \[ 64 - 2d = 52 \quad \Rightarrow \quad -2d = 52 - 64 \quad \Rightarrow \quad -2d = -12 \quad \Rightarrow \quad d = 6 \] Now substituting \( d = 6 \) back into equation (4) to find \( a_1 \): \[ a_1 = 32 - 5(6) = 32 - 30 = 2 \] Now we have \( a_1 = 2 \) and \( d = 6 \). The first three terms of the arithmetic series are: - First term \( a_1 = 2 \) - Second term \( a_2 = a_1 + d = 2 + 6 = 8 \) - Third term \( a_3 = a_1 + 2d = 2 + 2(6) = 2 + 12 = 14 \) So the first three terms are \( 2, 8, 14 \). To find the \( n^{th} \) term, use the formula \( a_n = a_1 + (n - 1)d \): \[ a_n = 2 + (n - 1) \cdot 6 = 2 + 6n - 6 = 6n - 4 \] Thus, the \( n^{th} \) term of the arithmetic series is \( 6n - 4 \). **Final Answers:** First three terms: \( 2, 8, 14 \) \( n^{th} \) term: \( 6n - 4 \)