Determine \( T_{10} \) of the following arithmetic sequences: \( \begin{array}{ll}\text { a) } a=4 \text { and } d=\frac{1}{2} & \text { b) } T_{2}=-10 \text { and } T_{6}=-160 \\ \text { d) } T_{5}=8 \text { and } d=3 & \text { d) } T_{2}=4 x \text { and } T_{5}=10 x\end{array} \)

Let's calculate \( T_{10} \) for each of the given arithmetic sequences. **a)** For the first sequence, where \( a = 4 \) and \( d = \frac{1}{2} \): The formula for the n-th term of an arithmetic sequence is given by: \[ T_n = a + (n-1)d \] Plugging in the values for \( n = 10 \): \[ T_{10} = 4 + (10-1) \times \frac{1}{2} = 4 + 9 \times \frac{1}{2} = 4 + \frac{9}{2} = 4 + 4.5 = 8.5 \] Thus, \( T_{10} = 8.5 \). **b)** For the second sequence, where \( T_2 = -10 \) and \( T_6 = -160 \): We can find the common difference \( d \) using the formula: \[ T_n = T_1 + (n-1)d \] We can express \( T_2 \) and \( T_6 \) in terms of \( T_1 \): \[ T_2 = T_1 + d = -10 \quad \text{(1)} \] \[ T_6 = T_1 + 5d = -160 \quad \text{(2)} \] Subtract equation (1) from equation (2): \[ (T_1 + 5d) - (T_1 + d) = -160 + 10 \] \[ 4d = -150 \Rightarrow d = -\frac{150}{4} = -37.5 \] Substituting \( d \) back into equation (1): \[ T_1 - 37.5 = -10 \Rightarrow T_1 = -10 + 37.5 = 27.5 \] Now, we can find \( T_{10} \): \[ T_{10} = T_1 + 9d = 27.5 + 9(-37.5) = 27.5 - 337.5 = -310 \] Thus, \( T_{10} = -310 \). **c)** For the third sequence where \( T_5 = 8 \) and \( d = 3 \): Using the general term formula: \[ T_5 = T_1 + 4d = 8 \quad \Rightarrow T_1 + 4 \times 3 = 8 \quad \Rightarrow T_1 + 12 = 8 \Rightarrow T_1 = 8 - 12 = -4 \] Now we can calculate \( T_{10} \): \[ T_{10} = T_1 + 9d = -4 + 9 \times 3 = -4 + 27 = 23 \] So, \( T_{10} = 23 \). **d)** For the last sequence where \( T_2 = 4x \) and \( T_5 = 10x \): Using the formulas: \[ T_2 = T_1 + d = 4x \quad \text{(1)} \] \[ T_5 = T_1 + 4d = 10x \quad \text{(2)} \] Subtracting equation (1) from equation (2): \[ (T_1 + 4d) - (T_1 + d) = 10x - 4x \] \[ 3d = 6x \Rightarrow d = 2x \] Substituting \( d \) into equation (1): \[ T_1 + 2x = 4x \Rightarrow T_1 = 4x - 2x = 2x \] Now, we can find \( T_{10} \): \[ T_{10} = T_1 + 9d = 2x + 9(2x) = 2x + 18x = 20x \] Thus, \( T_{10} = 20x \). In summary: - \( T_{10} \) for sequence (a) is \( 8.5 \) - \( T_{10} \) for sequence (b) is \( -310 \) - \( T_{10} \) for sequence (c) is \( 23 \) - \( T_{10} \) for sequence (d) is \( 20x \)