2.1 Given the sequence: \( 4: x: 32 \) Determine the value(s) of \( x \) if the sequence is: \( \begin{array}{l}2.1 .1 \quad \text { Arithmetic } \\ 2.1 .2 \quad \text { Geometric } \\ 2.2 \quad \text { Determine the value of } \mathrm{P} \text { if } \mathrm{P}=\sum_{\mathrm{k}=1}^{13} 3^{\mathrm{k}-5}\end{array} \).

To solve the sequences and the sum, let's break it down step by step. ### For the Arithmetic Sequence \( 4: x: 32 \) In an arithmetic sequence, the difference between consecutive terms is constant. Let \( d \) be the common difference. We can express \( x \) in terms of \( d \): 1. From \( 4 \) to \( x \): \( x - 4 = d \) 2. From \( x \) to \( 32 \): \( 32 - x = d \) Setting the two equations for \( d \) equal gives: \[ x - 4 = 32 - x \] \[ 2x = 36 \quad \Rightarrow \quad x = 18 \] **Value of \( x \) in the arithmetic case is \( 18 \).** ### For the Geometric Sequence \( 4: x: 32 \) In a geometric sequence, the ratio between consecutive terms is constant. Let \( r \) be the common ratio. We have: 1. From \( 4 \) to \( x \): \( \frac{x}{4} = r \) 2. From \( x \) to \( 32 \): \( \frac{32}{x} = r \) Setting the two equations for \( r \) equal gives: \[ \frac{x}{4} = \frac{32}{x} \] Cross-multiplying results in: \[ x^2 = 128 \quad \Rightarrow \quad x = \sqrt{128} = 8\sqrt{2} \] Thus, the two values of \( x \) in the geometric case are \( 8\sqrt{2} \) or approximately \( 11.31 \). **Values of \( x \) in the geometric case are \( 8\sqrt{2} \) or approximately \( 11.31 \).** ### For the Sum \( P = \sum_{k=1}^{13} 3^{k-5} \) This can be rewritten as: \[ P = 3^{-5} \sum_{k=1}^{13} 3^{k} = 3^{-5} (3^1 + 3^2 + ... + 3^{13}) \] This is a geometric series where the first term \( a = 3 \) and the common ratio \( r = 3 \). The sum of the first \( n \) terms of a geometric series is given by: \[ S_n = a \frac{r^n - 1}{r - 1} \] In this case: \[ S_{13} = 3 \frac{3^{13} - 1}{3 - 1} = \frac{3(3^{13} - 1)}{2} = \frac{3^{14} - 3}{2} \] Then plugging back into \( P \): \[ P = 3^{-5} \cdot \frac{3^{14} - 3}{2} = \frac{3^{14-5} - 3^{-5}}{2} = \frac{3^{9} - \frac{1}{243}}{2} \] Calculating gives: \[ = \frac{19683 - \frac{1}{243}}{2} \] Calculating \( 19683 \) (which is \( 3^9 \)) and subtracting gives a large \( P \) value approximately \( 9840.5 \) after simplification. Hence, the series sum provides \( P = 9840.5 \).