Question 1 1.1. Prove that the sum of \( n \) terms of an arithmetic sequence of \( n \) terms is given by the formula: \( S_{n}=\frac{n}{2}[2 a+(n-1) d] \) 1.2. Consider the series: \( 5+8+11+14+17+\ldots+302 \) 1.2.1. Determine the formula for the \( n^{t h} \) term of the series. 1.2.2. How may terms are there in the series? 1.2.3. Express the series in sigma notation. 1.2.4. Determine the sum of the terms in the series. 1.3. Determine the value of \( n \) if 1.4. Expand and evaluate: \( \sum_{k=1}^{n}(3 k-15)=285 \) (2) (2) (2) (2)

Let's solve the problem step by step. ### 1.1 Prove the formula for the sum of \( n \) terms of an arithmetic sequence An arithmetic sequence is defined by its first term \( a \) and a common difference \( d \). The \( n \)-th term of the sequence can be expressed as: \[ a_n = a + (n-1)d \] The sum of the first \( n \) terms \( S_n \) can be calculated as follows: \[ S_n = a + (a + d) + (a + 2d) + \ldots + (a + (n-1)d) \] This can be rewritten as: \[ S_n = na + d(0 + 1 + 2 + \ldots + (n-1)) \] The sum of the first \( n-1 \) integers is given by the formula: \[ \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2} \] Substituting this into the equation for \( S_n \): \[ S_n = na + d \cdot \frac{(n-1)n}{2} \] Factoring out \( n \): \[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \] Thus, we have proven that: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] ### 1.2 Consider the series: \( 5 + 8 + 11 + 14 + 17 + \ldots + 302 \) #### 1.2.1 Determine the formula for the \( n^{th} \) term of the series The first term \( a = 5 \) and the common difference \( d = 3 \). The \( n^{th} \) term can be expressed as: \[ a_n = a + (n-1)d = 5 + (n-1) \cdot 3 = 3n + 2 \] #### 1.2.2 How many terms are there in the series? To find the number of terms, we set the \( n^{th} \) term equal to 302: \[ 3n + 2 = 302 \] Solving for \( n \): \[ 3n = 300 \\ n = 100 \] #### 1.2.3 Express the series in sigma notation The series can be expressed in sigma notation as: \[ \sum_{n=1}^{100} (3n + 2) \] #### 1.2.4 Determine the sum of the terms in the series Using the sum formula derived in 1.1, we can calculate \( S_{100} \): \[ S_{100} = \frac{100}{2} [2 \cdot 5 + (100 - 1) \cdot 3] \] Calculating the terms inside the brackets: \[ S_{100} = 50 [10 + 297] = 50 \cdot 307 = 15350 \] ### 1.3 Determine the value of \( n \) if It seems that part 1.3 is incomplete. Please provide the equation or condition to determine \( n \). ### 1.4 Expand and evaluate: \( \sum_{k=1}^{n}(3k-15)=285 \) First, we can expand the sum: \[ \sum_{k=1}^{n} (3k - 15) = 3\sum_{k=1}^{n} k - \sum_{k=1}^{n} 15 \] Using the formula for the sum of the first \( n \) integers: \[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \] Thus, we have: \[ \sum_{k=1}^{n} (3k - 15) = 3 \cdot \frac{n(n+1)}{2} - 15n \] Setting this equal to 285: \[ 3 \cdot \frac{n(n+1)}{2} - 15n = 285 \] Multiplying through by 2 to eliminate the fraction: \[ 3n(n+1) - 30n = 570 \] Rearranging gives: \[ 3n^2 - 27n - 570 = 0 \] Dividing through by 3: \[ n^2 - 9n - 190 = 0 \] Now, we can solve this quadratic equation using the quadratic formula: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 1, b = -9, c = -190 \): Calculating the discriminant: \[ b^2 - 4ac = (-9)^2 - 4 \cdot 1 \cdot (-190) = 81 + 760 = 841 \] Now substituting back into the quadratic formula: \[ n = \frac{9 \pm \sqrt{841}}{2} = \frac{9 \pm 29}{2} \] Calculating the two possible values for \( n \): 1. \( n = \frac{38}{2} = 19 \) 2. \( n = \frac{-20}{2} = -10 \) (not valid since \( n \) must be positive) Thus, the value of \( n \) is: \[ n = 19 \] ### Summary of Answers 1.1: Proven formula for \( S_n \). 1.2.1: \( a_n = 3n + 2 \). 1.2.2: \( n = 100 \). 1.2.3: \( \sum_{n=1}^{100} (3n + 2) \). 1.2.4: \( S_{100} = 15350 \). 1.3: Incomplete. 1.4: \( n = 19 \).