QUESTION 2 2.1 The arithmetic series \( 3+5+7+\ldots+103 \) is given. \( 2.1 .1 \quad \) 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 \( 2.2 \quad \) Calculate \( \boldsymbol{n} \) if : \[ \sum_{n=2}^{n}(2 \boldsymbol{n}+\mathbf{1})>\mathbf{1 9 2} \] \( 2.3 \quad \) 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 \( \boldsymbol{r}>\mathbf{0} \).

The arithmetic series \( 3+5+7+\ldots+103 \) is quite fascinating! Its general term can be found using the formula for the \( n^{\text{th}} \) term of an arithmetic series, which is given by \( a_n = a + (n-1)d \) where \( a \) is the first term and \( d \) is the common difference. In this case, the first term \( a = 3 \) and the common difference \( d = 2 \), so the general term is \( a_n = 3 + (n-1) \cdot 2 = 2n + 1 \). Now, let's figure out how many terms are in the series! To find the number of terms, we set \( a_n = 103 \) and solve for \( n \): \[ 2n + 1 = 103 \implies 2n = 102 \implies n = 51 \] So, there are 51 terms in this series. Now, using sigma notation, we can express this as: \[ \sum_{n=1}^{51} (2n + 1) \] To solve for \( n \) in the inequality \( \sum_{n=2}^{n}(2n + 1) > 192 \), we can first rewrite the left side as follows: \[ \sum_{n=2}^{n}(2n + 1) = 2\sum_{n=2}^{n}n + \sum_{n=2}^{n}1 = 2 \cdot \left(\frac{n(n+1)}{2} - 1\right) + (n-1) \] This simplifies to: \[ (n^2 + n - 2) > 192 \implies n^2 + n - 194 > 0 \] Using the quadratic formula, we find that \( n \) is approximately 13. Therefore, the integer value of \( n \) satisfying the inequality is \( n \geq 14 \). In the geometric series problem, you have the sum of the first and second terms equating to 30: \[ a + ar = 30 \] And the sum to infinity is \( \frac{a}{1-r} = 54 \). From the first equation, we can express \( ar = 30 - a \). Substituting this expression into the second equation gives us a system of equations. By solving \( a = 54(1-r) \) and combining it with \( ar = 30 - a \), you can find the values of \( a \) and \( r \). The calculations reveal \( a = 18 \) and \( r = \frac{2}{3} \).