An arithmetic and a geometric sequence are combined to form the pattern, which is given by: \( P_{n}=x ; \frac{1}{3} ; 2 x ; \frac{1}{9} ; 3 x ; \frac{1}{27} ; \ldots \) \( \begin{array}{ll}2.2 .1 & \text { Write down the next TWO terms of the pattern. } \\ 2.2 .2 \quad & \begin{array}{l}\text { Determine the general term }\left(T_{n}\right) \text { for the odd terms of this pattern. Write } \\ \text { down your answer in terms of } x .\end{array} \\ 2.2 .3 & \text { Calculate the value of } P_{26} \text {. } \\ 2.2 .4 & \text { If } \sum_{n=1}^{21} P_{n}=33,5, \text { determine the value of } x .\end{array} \) View key concept ©

**2.2.1** The pattern is formed by alternating terms from an arithmetic sequence and a geometric sequence. The given terms are \[ P_1 = x,\quad P_2 = \frac{1}{3},\quad P_3 = 2x,\quad P_4 = \frac{1}{9},\quad P_5 = 3x,\quad P_6 = \frac{1}{27},\dots \] Notice that the odd‐numbered terms are \[ x,\;2x,\;3x,\dots \] and the even–numbered terms are \[ \frac{1}{3},\;\frac{1}{9},\;\frac{1}{27},\dots \] So the next two terms, which are the 7th term (an odd term) and the 8th term (an even term), are \[ P_7 = 4x,\quad P_8 = \frac{1}{81}\quad \text{(since } \frac{1}{81}=\frac{1}{3^4}\text{).} \] --- **2.2.2** The odd–numbered terms form an arithmetic sequence: \[ P_1 = x,\quad P_3 = 2x,\quad P_5 = 3x,\dots \] If we let \( k \) count the odd–numbered terms (so that \( k=1 \) corresponds to \( P_1 \), \( k=2 \) to \( P_3 \), etc.), then the \( k\)th odd term is given by \[ P_{2k-1} = kx,\quad \text{for } k=1,2,3,\dots \] --- **2.2.3** We wish to find \( P_{26} \). Since the pattern alternates, even–numbered terms come from the geometric sequence. The even terms are \[ P_2 = \frac{1}{3},\quad P_4 = \frac{1}{9},\quad P_6 = \frac{1}{27},\dots \] The first even term (when \( k=1 \)) is \( \frac{1}{3} \) and the common ratio is \[ r = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{3}. \] Thus, the general formula for an even–numbered term (when written as \( P_{2k} \)) is \[ P_{2k} = \frac{1}{3^k}. \] Since \( 26 \) is even, we write \( 26 = 2k \) so that \( k = 13 \). Then \[ P_{26} = \frac{1}{3^{13}}. \] --- **2.2.4** We are given that \[ \sum_{n=1}^{21} P_n = 33.5. \] There are 21 terms in total. Notice that the odd–numbered terms (positions 1, 3, …, 21) and even–numbered terms (positions 2, 4, …, 20) occur as follows: - Number of odd–numbered terms: \(\frac{21+1}{2} = 11\). - Number of even–numbered terms: \(\frac{21-1}{2} = 10\). **Sum of the odd–numbered terms:** They are \[ P_1=x,\quad P_3=2x,\quad P_5=3x,\dots,\quad P_{21}=11x. \] Thus, their sum is \[ S_{\text{odd}} = x + 2x + 3x + \cdots + 11x = x(1+2+\cdots+11). \] The sum of the first 11 positive integers is \[ 1+2+\cdots+11 = \frac{11\cdot12}{2}=66. \] Hence, \[ S_{\text{odd}} = 66x. \] **Sum of the even–numbered terms:** They form a geometric sequence: \[ P_2 = \frac{1}{3},\; P_4 = \frac{1}{9},\; \dots,\; P_{20} = \frac{1}{3^{10}}. \] The sum of the first 10 terms of a geometric sequence with first term \( a = \frac{1}{3} \) and common ratio \( r = \frac{1}{3} \) is given by \[ S_{\text{even}} = \frac{a\left(1-r^{10}\right)}{1-r} = \frac{\frac{1}{3}\left(1-\frac{1}{3^{10}}\right)}{1-\frac{1}{3}}. \] Since \( 1-\frac{1}{3}=\frac{2}{3} \), we have \[ S_{\text{even}} = \frac{\frac{1}{3}\left(1-\frac{1}{3^{10}}\right)}{\frac{2}{3}} = \frac{1-\frac{1}{3^{10}}}{2}. \] Now, the total sum of the first 21 terms is \[ 66x + \frac{1-\frac{1}{3^{10}}}{2} = 33.5. \] To solve for \( x \), multiply both sides by 2 to eliminate the fraction: \[ 132x + \left(1-\frac{1}{3^{10}}\right)=67. \] Then, isolate \( x \): \[ 132x = 67 - 1 + \frac{1}{3^{10}} = 66 + \frac{1}{3^{10}}, \] \[ x = \frac{66 + \frac{1}{3^{10}}}{132} = \frac{66\cdot 3^{10}+1}{132\cdot 3^{10}}. \] Thus, the value of \( x \) is \[ x = \frac{66\cdot 3^{10}+1}{132\cdot 3^{10}}. \]