Question 1 1.1. Prove that the sum of \( n \) terms of an arithmetic sequence of \( n \) terms is given 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: \( \sum_{k=1}^{n}(3 k-15)=285 \)

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: 1. Write the sum in two different orders: \[ S_n = a + (a + d) + (a + 2d) + \ldots + (a + (n-1)d) \] \[ S_n = (a + (n-1)d) + (a + (n-2)d) + \ldots + a \] 2. Add these two equations: \[ 2S_n = [2a + (n-1)d] + [2a + (n-1)d] + \ldots + [2a + (n-1)d] \] There are \( n \) terms of \( [2a + (n-1)d] \). 3. Therefore, we have: \[ 2S_n = n[2a + (n-1)d] \] 4. Dividing both sides by 2 gives: \[ S_n = \frac{n}{2}[2a + (n-1)d] \] This proves the formula for the sum of the first \( n \) terms of an arithmetic sequence. ### 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, 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 \( S_n = \frac{n}{2}[2a + (n-1)d] \): - \( n = 100 \) - \( a = 5 \) - \( d = 3 \) Calculating the sum: \[ S_{100} = \frac{100}{2}[2 \cdot 5 + (100-1) \cdot 3] \\ = 50[10 + 297] \\ = 50 \cdot 307 \\ = 15350 \] ### 1.3 Determine the value of \( n \) if It seems like the question is incomplete. Please provide the complete statement for part 1.3. ### 1.4 Expand and evaluate: \[ \sum_{k=1}^{n}(3k - 15) = 285 \] This can be split into two separate sums: \[ \sum_{k=1}^{n}(3k) - \sum_{k=1}^{n}(15) = 285 \] Calculating each part: 1. The first sum: \[ \sum_{k=1}^{n}(3k) = 3\sum_{k=1}^{n}k = 3 \cdot \frac{n(n+1)}{2} = \frac{3n(n+1)}{2} \] 2. The second sum: \[ \sum_{k=1}^{n}(15) = 15n \] Putting it all together: \[ \frac{3n(n+1)}{2} - 15n = 285 \] Multiplying through by 2 to eliminate the fraction: \[ 3n(n+1) - 30n = 570 \\ 3n^2 + 3n - 30n - 570 = 0 \\ 3n^2 - 27n - 570 = 0 \] Dividing the entire equation by 3: \[ n^2 - 9n - 190 = 0 \] Now, we can solve this quadratic equation using the quadratic formula: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot (-190)}}{2 \cdot 1} \] Calculating the discriminant: \[ = \frac{9 \pm \sqrt{81 + 760}}{2} = \frac{9 \pm \sqrt{841}}{2} = \frac{9 \pm 29}{2} \] 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 Results: - 1.1: Proven the 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: \( S_{100} = 15350 \) - 1.4: \( n = 19 \)