5.1 \( 2 x-1 ; 4 x-5 \) and \( 3 x+3 \) are the first three terms of an arithmetic sequence. 5.1.1 Show that \( x=4 \) and write down the first two terms of the sequence. 5.1.2 Determine the \( \mathrm{n}^{\mathrm{m}} \) term, \( \mathrm{T}_{\mathrm{a}} \), of the sequence. 5.1.3 If the \( \mathrm{n}^{\mathrm{n}} \) term, \( \mathrm{T}_{\mathrm{a}} \), of the sequence is 43 , calculate \( n \). 5.2 Calculate; \( \sum_{k=1}^{10}(2 k+4) \)

To show that \( x = 4 \), we need to find the common difference of the arithmetic sequence. The first term is \( 2x - 1 \), the second term is \( 4x - 5 \), and the third term is \( 3x + 3 \). Setting the difference between the first and second term equal to the difference between the second and third term gives us the equation: \( (4x - 5) - (2x - 1) = (3x + 3) - (4x - 5) \). Simplifying this leads us to find \( x = 4 \). Plugging \( x \) back, the first two terms are \( 7 \) and \( 11 \). For the \( n^{th} \) term \( T_a \) of an arithmetic sequence, we can use the formula \( T_n = a + (n - 1)d \), where \( a \) is the first term and \( d \) is the common difference. The first term is \( 7 \) and the common difference \( d \) between the terms is \( 4 \) (calculated as \( 11 - 7 \)). Therefore, the formula becomes \( T_n = 7 + (n - 1)4 = 4n + 3 \). Now, to find \( n \) when \( T_n = 43 \), we set up the equation \( 4n + 3 = 43 \). Solving for \( n \) gives us \( 4n = 40 \), thus \( n = 10 \). Now onto the summation! To calculate \( \sum_{k=1}^{10}(2k+4) \), we can simplify the expression first. This sum transforms into \( \sum_{k=1}^{10}(2k) + \sum_{k=1}^{10}(4) \). The first part becomes \( 2 \sum_{k=1}^{10} k = 2 \cdot \frac{10(11)}{2} = 110 \). The second part \( \sum_{k=1}^{10}(4) = 4 \cdot 10 = 40 \). Combining these gives us \( 110 + 40 = 150 \). So, the final answer for the summation is \( 150 \).