\( \begin{array}{ll}8.2 & \text { Determine the value of } \mathrm{P} \text { if } \quad \mathrm{P}=\sum_{k=1}^{13} 3^{k-5} \\ 8.3 \quad \begin{array}{l}\text { Prove that for any arithmetic sequence of which the first term } a \text { and the constant } \\ \text { Difference is } d \text {, the sum to } n \text { terms can be expressed as }\end{array} \\ S_{n}=\frac{n}{2}(2 a+(n-1) d)\end{array} \)

To find the value of \( P \), we need to evaluate the series \( P = \sum_{k=1}^{13} 3^{k-5} \). This series can be simplified by recognizing it as a geometric series. The first term when \( k = 1 \) is \( 3^{-4} \) (since \( 3^{1-5} = 3^{-4} \)), and the last term when \( k = 13 \) is \( 3^{8} \). The total number of terms is 13, and the common ratio \( r \) is 3. The sum of a geometric series can be found with the formula \( S_n = a \frac{(r^n - 1)}{r - 1} \), which gives us \( 3^{-4} \frac{(3^{13} - 1)}{2} \). For the sum of an arithmetic sequence, the formula \( S_n = \frac{n}{2}(2a + (n-1)d) \) is derived from the fact that the sum can be viewed as adding the first and last terms, then the second and second-to-last terms, and so forth. Each pair sums to the same total, with the number of pairs being \( n/2 \). The average of the first and last terms is \( \frac{2a + (n-1)d}{2} \), multiplied by the number of terms \( n \), leading to the modern cleaned-up version we commonly use today. It combines both positions seamlessly into a tidy package—perfect for any arithmetic scenario!