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

To find \( T_{10} \) of each arithmetic sequence, we will follow the formulas and calculations relevant to the properties of arithmetic sequences. For an arithmetic sequence, the \( n \)-th term is given by: \[ T_n = a + (n-1)d \] where \( a \) is the first term and \( d \) is the common difference. ### a) Given \( a = 4 \) and \( d = \frac{1}{2} \): Calculating \( T_{10} \): \[ T_{10} = 4 + (10 - 1) \cdot \frac{1}{2} = 4 + 9 \cdot \frac{1}{2} = 4 + 4.5 = 8.5 \] ### b) Given \( T_2 = -10 \) and \( T_6 = -160 \): The common difference \( d \) can be calculated as follows: \[ T_2 = a + (2-1)d = a + d = -10 \] \[ T_6 = a + (6-1)d = a + 5d = -160 \] We now have a system of equations: 1. \( a + d = -10 \) 2. \( a + 5d = -160 \) Subtract the first from the second: \[ (a + 5d) - (a + d) = -160 + 10 \] \[ 4d = -150 \Rightarrow d = -37.5 \] Plugging \( d \) back into the first equation: \[ a - 37.5 = -10 \Rightarrow a = 27.5 \] Now, calculating \( T_{10} \): \[ T_{10} = 27.5 + (10-1)(-37.5) = 27.5 - 337.5 = -310 \] ### c) Given \( T_5 = 8 \) and \( d = 3 \): Calculating \( T_1 \): \[ T_5 = a + (5-1)d = a + 4d = 8 \] \[ a + 4 \cdot 3 = 8 \Rightarrow a + 12 = 8 \Rightarrow a = -4 \] Now, calculating \( T_{10} \): \[ T_{10} = -4 + (10-1) \cdot 3 = -4 + 27 = 23 \] ### d) Given \( T_2 = 4x \) and \( T_5 = 10x \): From the first term: \[ T_2 = a + d = 4x \] \[ T_5 = a + 4d = 10x \] The equations become: 1. \( a + d = 4x \) 2. \( a + 4d = 10x \) Subtracting: \[ (a + 4d) - (a + d) = 10x - 4x \] \[ 3d = 6x \Rightarrow d = 2x \] Substituting \( d \) into the first equation: \[ a + 2x = 4x \Rightarrow a = 2x \] Now calculating \( T_{10} \): \[ T_{10} = a + (10-1)d = 2x + 9(2x) = 2x + 18x = 20x \] In summary: - \( T_{10} \text{ for } a) = 8.5 \) - \( T_{10} \text{ for } b) = -310 \) - \( T_{10} \text{ for } c) = 23 \) - \( T_{10} \text{ for } d) = 20x \)