QUESTION 2 2.1. Given the arithmetic sequence \( 1 ;-1 ;-3 ; \ldots \) Determine: 2.1.1. Sum to \( n \) terms. 2.1.2. The maximum number of terms we need in order to have a sum larger than -99 . 2.1.3. If this linear sequence forms the first differences of a quadratic sequence with first term 10 , calculate the \( 75^{\text {th }} \) term of the quadratic sequence. If \( a+1 ; a-1 ; 2 a-5 \) are the first three terms of a convergent geometric series, calculate: 2.2.1 The value of \( a \). 2.2.2 The sum to infinity of the series.

Let's solve the problem step by step. ### 2.1 Arithmetic Sequence The given arithmetic sequence is \( 1, -1, -3, \ldots \). #### 2.1.1 Sum to \( n \) terms The first term \( a = 1 \) and the common difference \( d = -2 \) (since \(-1 - 1 = -2\)). The formula for the sum \( S_n \) of the first \( n \) terms of an arithmetic sequence is given by: \[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \] Substituting the values of \( a \) and \( d \): \[ S_n = \frac{n}{2} \times (2 \cdot 1 + (n - 1)(-2)) \] Now, let's simplify this expression: \[ S_n = \frac{n}{2} \times (2 - 2(n - 1)) = \frac{n}{2} \times (2 - 2n + 2) = \frac{n}{2} \times (4 - 2n) \] \[ S_n = n(2 - n) \] #### 2.1.2 Maximum number of terms for sum larger than -99 We need to find \( n \) such that: \[ S_n > -99 \] Substituting the expression for \( S_n \): \[ n(2 - n) > -99 \] Rearranging gives: \[ n^2 - 2n - 99 < 0 \] Now, we will solve the quadratic equation \( n^2 - 2n - 99 = 0 \) using the quadratic formula: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -2, c = -99 \): \[ n = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-99)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 396}}{2} = \frac{2 \pm \sqrt{400}}{2} = \frac{2 \pm 20}{2} \] Calculating the roots: \[ n_1 = \frac{22}{2} = 11, \quad n_2 = \frac{-18}{2} = -9 \] Since \( n \) must be positive, we take \( n = 11 \). The inequality \( n(n - 2) < 99 \) holds for \( n < 11 \) and \( n > -9 \). Thus, the maximum number of terms needed is \( n = 11 \). #### 2.1.3 \( 75^{\text{th}} \) term of the quadratic sequence The first differences of the quadratic sequence are given by the arithmetic sequence \( 1, -1, -3, \ldots \). The first term of the quadratic sequence is \( 10 \). The \( n^{\text{th}} \) term of the quadratic sequence can be expressed as: \[ Q_n = Q_1 + (n-1)D_n \] where \( D_n \) is the sum of the first \( n-1 \) terms of the first differences. The sum of the first \( n-1 \) terms of the arithmetic sequence is: \[ D_n = S_{n-1} = (n-1)(2 - (n-1)) = (n-1)(3 - n) \] Now, substituting \( n = 75 \): \[ D_{75} = (75 - 1)(3 - 75) = 74 \cdot (-72) = -5328 \] Thus, the \( 75^{\text{th}} \) term of the quadratic sequence is: \[ Q_{75} = 10 + D_{75} = 10 - 5328 = -5318 \] ### 2.2 Geometric Series The first three terms of the geometric series are \( a + 1, a - 1, 2a - 5 \). #### 2.2.1 Value of \( a \) For a geometric series, the ratio between consecutive terms must be constant: \[ \frac{a - 1}{a + 1} = \frac{2a - 5}{a - 1} \] Cross-multiplying gives: \[ (a - 1)^2 = (a + 1)(2a - 5) \] Expanding both sides: \[ a^2 - 2a + 1 = 2a^2 - 5a + 2a - 5 \] Simplifying: \[ a^2 - 2a + 1 = 2a^2 - 3a - 5 \] Rearranging gives: \[ 0 = a^2 - a - 6 \] Factoring: \[ 0 = (a - 3)(a + 2) \] Thus, \( a = 3 \) or \( a = -2 \). #### 2.2.2 Sum to infinity of the series The sum to infinity \( S \) of a geometric series is given by: \[ S = \frac{a_1}{1 - r} \] where \( a_1 \) is the first term and \( r \) is the common ratio. Using \( a = 3 \): The first term is \( 3 + 1 = 4 \) and the second term is \( 3 - 1 = 2 \), so the common ratio \( r \) is: \[ r = \frac{2}{4} = \frac{1}{2} \] Now, substituting into the sum formula: \[ S = \frac{4}{1 - \frac{1}{2}} = \frac{4}{\frac{1}{2}} = 8 \] ### Final Answers - **2.1.1**: \( S_n = n(2 - n) \) - **2.1.2**: Maximum number of terms \( n = 11 \) - **2.1.3**: \( 75^{\text{th}} \) term of the quadratic sequence \( Q_{75} = -5318 \) - **2.2.1**: \( a = 3 \) or \( a = -2 \) - **2.2.2**: Sum to infinity \( S = 8 \) (using \( a = 3 \))