QUESTION 3 A geometric series has a constant ratio of \( \frac{1}{2} \) and a sum to infinity of 6 . \( 3.1 \quad \) Calculate the first term of the series. \( 3.2 \quad \) Calculate the \( 8^{\text {th }} \) term of the series. \( \begin{array}{ll}3.3 & \text { Given: } \sum_{k=1}^{n} 3(2)^{1-k}=5,8125 \quad \text { Calculate the value of } n . \\ 3.4 & \text { If } \sum_{k=1}^{20} 3(2)^{1-k}=p \text {, write down } \sum_{k=1}^{20} 24(2)^{-k} \text { in terms of } p .\end{array} \)

To calculate the first term of the series, we can use the formula for the sum of an infinite geometric series, \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. Given that \( S = 6 \) and \( r = \frac{1}{2} \), we can set up the equation: \[ 6 = \frac{a}{1 - \frac{1}{2}} \] This simplifies to: \[ 6 = \frac{a}{\frac{1}{2}} \implies a = 6 \times \frac{1}{2} = 3. \] For the \( 8^{\text{th}} \) term, we use the formula for the \( n^{\text{th}} \) term of a geometric series, \( T_n = a \cdot r^{n-1} \). With \( a = 3 \) and \( r = \frac{1}{2} \): \[ T_8 = 3 \cdot \left(\frac{1}{2}\right)^{8-1} = 3 \cdot \left(\frac{1}{2}\right)^7 = 3 \cdot \frac{1}{128} = \frac{3}{128}. \] Now, let’s dive into some history of geometric series which dates back to ancient civilizations, where they were used in contexts like modeling population growth and understanding how resources deplete over time. Famous mathematicians like Archimedes explored these concepts, leading to wide-ranging implications in calculus and real-world applications today, from finance to physics. Speaking of applications, one real-world use of geometric series is in calculating the present value of an annuity, which helps financial planners determine how much a series of future payments is worth today. This is vital for making informed decisions about investments and savings, illustrating just how powerful these mathematical concepts are in managing our finances effectively!