In an arithmetic sequence \( T_{1}=(1-x)^{2} \) and \( T_{2}=1+x^{2}: \) a) Write down an expression for the common difference in terms of \( x \). b) Determine the value(s) of \( x \) if \( S_{10}=310 \).

Let's solve the problem step by step. ### Part (a): Common Difference In an arithmetic sequence, the common difference \( d \) is the difference between consecutive terms. Given: - \( T_{1} = (1 - x)^2 \) - \( T_{2} = 1 + x^2 \) The common difference \( d \) is: \[ d = T_{2} - T_{1} = (1 + x^2) - (1 - x)^2 \] Expanding \( (1 - x)^2 \): \[ (1 - x)^2 = 1 - 2x + x^2 \] Substituting back: \[ d = 1 + x^2 - (1 - 2x + x^2) = 1 + x^2 - 1 + 2x - x^2 = 2x \] **Answer to Part (a):** \[ d = 2x \] ### Part (b): Determining \( x \) When \( S_{10} = 310 \) The sum of the first \( n \) terms of an arithmetic sequence is given by: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] where \( a = T_{1} = (1 - x)^2 \) and \( d = 2x \). For \( n = 10 \) and \( S_{10} = 310 \): \[ 310 = \frac{10}{2} [2(1 - x)^2 + 9(2x)] \] \[ 310 = 5 [2(1 - 2x + x^2) + 18x] \] \[ 310 = 5 [2 + 14x + 2x^2] \] \[ 310 = 10x^2 + 70x + 10 \] \[ 10x^2 + 70x + 10 - 310 = 0 \] \[ 10x^2 + 70x - 300 = 0 \] Divide the entire equation by 10: \[ x^2 + 7x - 30 = 0 \] Solving the quadratic equation: \[ x = \frac{-7 \pm \sqrt{49 + 120}}{2} = \frac{-7 \pm \sqrt{169}}{2} = \frac{-7 \pm 13}{2} \] Thus, the solutions are: \[ x = \frac{6}{2} = 3 \quad \text{and} \quad x = \frac{-20}{2} = -10 \] **Answer to Part (b):** \[ x = 3 \quad \text{and} \quad x = -10 \] ### Final Answer **a)** The common difference is \( d = 2x \). **b)** The possible values of \( x \) are \( x = 3 \) and \( x = -10 \).