2) 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 \( S_{10}=310 \). 3) Using your knowledge of series: a) Prove that the sum of the first \( n \), odd natural numbers is \( n^{2} \). b) Prove that the sum of the first \( n \), even natural numbers is \( n^{2}+n \). 4) The sum of the first 5 terms of an arithmetic sequence is 0 and the fiffth term is 8 . Determine \( \mathbf{t} \) common difference and the first term. 5) In a geometric series the first term is 36 , the last term is 16 and the sum of the series is 76 . Determine the common ratio and the number of terms in the sequence.

Let's solve each part of the problem step by step. ### Problem 2 Given the arithmetic sequence where \( T_{1}=(1-x)^{2} \) and \( T_{2}=1+x^{2} \): #### a) Write down an expression for the common difference in terms of \( x \). The common difference \( d \) in an arithmetic sequence is given by: \[ d = T_{2} - T_{1} \] Substituting the expressions for \( T_{1} \) and \( T_{2} \): \[ d = (1 + x^{2}) - (1 - x)^{2} \] Now, let's simplify \( (1 - x)^{2} \): \[ (1 - x)^{2} = 1 - 2x + x^{2} \] Thus, we have: \[ d = (1 + x^{2}) - (1 - 2x + x^{2}) = 1 + x^{2} - 1 + 2x - x^{2} = 2x \] So, the expression for the common difference is: \[ d = 2x \] #### b) Determine the value(s) of \( x \) if \( S_{10}=310 \). The sum of the first \( n \) terms of an arithmetic sequence is given by: \[ S_{n} = \frac{n}{2} (2a + (n-1)d) \] For \( n = 10 \) and \( S_{10} = 310 \): \[ 310 = \frac{10}{2} (2T_{1} + 9d) \] This simplifies to: \[ 310 = 5(2T_{1} + 9d) \] Dividing both sides by 5: \[ 62 = 2T_{1} + 9d \] Substituting \( T_{1} = (1-x)^{2} \) and \( d = 2x \): \[ 62 = 2(1-x)^{2} + 9(2x) \] Expanding this: \[ 62 = 2(1 - 2x + x^{2}) + 18x \] \[ 62 = 2 - 4x + 2x^{2} + 18x \] \[ 62 = 2x^{2} + 14x + 2 \] Rearranging gives: \[ 2x^{2} + 14x + 2 - 62 = 0 \] \[ 2x^{2} + 14x - 60 = 0 \] Dividing the entire equation by 2: \[ x^{2} + 7x - 30 = 0 \] Now, we can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] Here, \( a = 1, b = 7, c = -30 \): \[ x = \frac{-7 \pm \sqrt{7^{2} - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1} \] \[ x = \frac{-7 \pm \sqrt{49 + 120}}{2} \] \[ x = \frac{-7 \pm \sqrt{169}}{2} \] \[ x = \frac{-7 \pm 13}{2} \] Calculating the two possible values for \( x \): 1. \( x = \frac{6}{2} = 3 \) 2. \( x = \frac{-20}{2} = -10 \) Thus, the values of \( x \) are \( 3 \) and \( -10 \). ### Problem 3 #### a) Prove that the sum of the first \( n \), odd natural numbers is \( n^{2} \). The first \( n \) odd natural numbers are \( 1, 3, 5, \ldots, (2n-1) \). The sum can be expressed as: \[ S = 1 + 3 + 5 + \ldots + (2n-1) \] This can be rewritten as: \[ S = \sum_{k=1}^{n} (2k - 1) \] Using the formula for the sum of an arithmetic series: \[ S = n \cdot \frac{(1 + (2n - 1))}{2} = n \cdot \frac{(2n)}{2} = n^2 \] Thus, the sum of the first \( n \) odd natural numbers is \( n^{2} \). #### b) Prove that the sum of the first \( n \), even natural numbers is \( n^{2}+n \). The first \( n \) even natural numbers are \( 2, 4, 6, \ldots, 2n \). The sum can be expressed as: \[ S = 2 + 4 + 6 + \ldots + 2n \] Factoring out 2 gives: \[ S = 2(1 + 2 + 3 + \ldots + n) = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) \] Thus, the sum of the first \( n \) even natural numbers is \( n(n + 1) = n^{2} + n \). ### Problem 4 The sum of the first 5 terms of an arithmetic sequence is 0, and the fifth term is 8. Let the first term be \( a \) and the common difference be \( d \). The sum of the first 5 terms is: \[ S_5 = \frac{5}{2} (2a + 4d) = 0 \] This simplifies to: \[ 2a + 4d = 0 \quad \text{(1)} \] The fifth term is given by: \[ T_5 = a + 4d = 8 \quad \text{(2)} \] Now we can solve these two equations simultaneously. From equation (1): \[ 2a = -4d \implies a = -2d \] Substituting \( a = -2d \) into equation (2): \[ -2d + 4d = 8 \] \[ 2d = 8 \implies d = 4 \] Now substituting \( d = 4 \) back into \( a = -2d \): \[ a = -2(4) = -8 \] Thus, the first term \( a \) is \( -8 \) and the common difference \( d \) is \( 4 \). ### Problem 5