QUESTION 2 2.1 The first term of a geometric series is 14 and the \( 6^{\text {th }} \) term is 448. \( \begin{array}{l}\text { 2.1.1 } \\ \begin{array}{l}\text { Calculate the value of the constant ratio, } r\end{array} \\ \begin{array}{l}\text { Determine the number of consecutive terms that must be added to the first } \\ \text { 6 terms of the series in order to obtain a sum of } 114674 .\end{array} \\ \text { If the first term of another series is } 448 \text { and the } 6^{\text {th }} \text { term is } 14 \text {, calculate } \\ \text { the sum to infinity of the new series. }\end{array} \)

To find the constant ratio \( r \), we can use the formula for the \( n \)th term of a geometric series, which is given by \( a \times r^{n-1} \), where \( a \) is the first term and \( n \) is the term number. Here, the first term \( a \) is 14 and the sixth term is 448, so we set up the equation: \[ 14 \times r^{5} = 448 \] Solving this gives \( r^{5} = \frac{448}{14} = 32 \), thus \( r = 2 \). Now, to find the sum of the first 6 terms, we use the formula for the sum of a geometric series: \[ S_n = a \frac{1-r^n}{1-r} \] Substituting in, we get: \[ S_6 = 14 \frac{1-2^6}{1-2} = 14 \times 63 = 882 \] To reach a sum of 114674, we need an additional sum of \( 114674 - 882 = 113792 \). The formula can help find how many consecutive terms \( n \) we need by rearranging: \[ S_{n} = 14 \frac{r^6-1}{r-1} = 113792 \] When computed, \( n = 13 \) additional terms are needed. For the new series with first term 448 and sixth term 14, we again apply the same logic: \[ 448 \times r^5 = 14 \] From this, \( r^5 = \frac{14}{448} = \frac{1}{32} \) leading to \( r = \frac{1}{2} \). The sum to infinity \( S_\infty = \frac{a}{1-r} = \frac{448}{1-\frac{1}{2}} = 896 \).