QUESTION 2 2.1 The arithmetic series \( 3+5+7+\ldots+103 \) is given. 2.1.1 Determine the general term \( 2.1 .2 \quad \) Calculate the number of terms in the given series \( 2.1 .3 \quad \) Hence, write the series in sigma notation 2.2 Calculate \( \boldsymbol{n} \) if : 2.3 \( \sum_{n=2}^{n}(2 \boldsymbol{n}+\mathbf{1})>\mathbf{1 9 2} \) The sum of the \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) terms of a geometric series is 30 , and the sum to infinity is 54 . Determine the \( 1^{\text {st }} \) term \( \boldsymbol{a} \), and the common ratio, \( \boldsymbol{r} \), with \( \boldsymbol{r}>\mathbf{0} \).

Let's solve the problem step by step. ### 2.1 Arithmetic Series \( 3 + 5 + 7 + \ldots + 103 \) #### 2.1.1 Determine the general term The series is an arithmetic series where the first term \( a = 3 \) and the common difference \( d = 5 - 3 = 2 \). The general term \( T_n \) of an arithmetic series can be expressed as: \[ T_n = a + (n - 1)d \] Substituting the values of \( a \) and \( d \): \[ T_n = 3 + (n - 1) \cdot 2 = 3 + 2n - 2 = 2n + 1 \] #### 2.1.2 Calculate the number of terms in the given series To find the number of terms \( n \), we set the last term equal to 103: \[ T_n = 103 \] Substituting the expression for \( T_n \): \[ 2n + 1 = 103 \] Now, solve for \( n \): \[ 2n = 103 - 1 = 102 \\ n = \frac{102}{2} = 51 \] #### 2.1.3 Write the series in sigma notation The series can be expressed in sigma notation as: \[ \sum_{n=1}^{51} (2n + 1) \] ### 2.2 Calculate \( n \) if \( \sum_{n=2}^{n}(2n + 1) > 192 \) First, we need to calculate the sum \( \sum_{n=2}^{k} (2n + 1) \) for some \( k \). The sum can be calculated as follows: \[ \sum_{n=2}^{k} (2n + 1) = \sum_{n=2}^{k} 2n + \sum_{n=2}^{k} 1 \] The first part is: \[ \sum_{n=2}^{k} 2n = 2 \sum_{n=2}^{k} n = 2 \left( \frac{k(k + 1)}{2} - 1 \right) = k(k + 1) - 2 \] The second part is: \[ \sum_{n=2}^{k} 1 = k - 1 \] Combining these: \[ \sum_{n=2}^{k} (2n + 1) = k(k + 1) - 2 + (k - 1) = k^2 + k - 3 \] Now, we set up the inequality: \[ k^2 + k - 3 > 192 \] This simplifies to: \[ k^2 + k - 195 > 0 \] Now, we will solve the quadratic equation \( k^2 + k - 195 = 0 \) using the quadratic formula: \[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 780}}{2} = \frac{-1 \pm \sqrt{781}}{2} \] Calculating \( \sqrt{781} \): \[ \sqrt{781} \approx 27.93 \] Thus: \[ k \approx \frac{-1 + 27.93}{2} \approx 13.465 \] Since \( k \) must be an integer, we check \( k = 14 \) and \( k = 13 \). Calculating for \( k = 14 \): \[ 14^2 + 14 - 195 = 196 + 14 - 195 = 15 > 0 \] Calculating for \( k = 13 \): \[ 13^2 + 13 - 195 = 169 + 13 - 195 = -13 < 0 \] Thus, \( k = 14 \) is the smallest integer satisfying the inequality. ### 2.3 Geometric Series The first term \( a \) and the common ratio \( r \) satisfy the following conditions: 1. \( a + ar = 30 \) 2. \( \frac{a}{1 - r} = 54 \) From the first equation: \[ a(1 + r) = 30 \quad \text{(1)} \] From the second equation: \[ a = 54(1 - r) \quad \text{(2)} \] Substituting (2) into (1): \[ 54(1 - r)(1 + r) = 30 \] Expanding: \[ 54(1 - r^2) = 30 \] This simplifies to: \[ 54 - 54r^2 = 30 \\ 54r^2 = 54 - 30 = 24 \\ r^2 = \frac{24}{54} = \frac{4}{9} \\ r = \frac{2}{3} \quad \text{(since \( r > 0 \))} \] Now substituting \( r \) back into (2) to find \( a \): \[ a = 54(1 - \frac{2}{3}) = 54 \cdot \frac{1}{3} = 18 \] ### Final Answers - General term: \( 2n + 1 \) - Number of terms: \( 51 \) - Sigma notation: \( \sum_{n=1}^{51} (2n + 1) \) - \( n \): \( 14 \) - First term \( a \): \( 18 \) - Common ratio \( r \): \( \frac{2}{3} \)