Consider the sequence: \( \frac{1}{2} ; 4 ; \frac{1}{4} ; 7 ; \frac{1}{8} ; 10 ; \ldots \). 2.6.1 Write down the next TWO terms if the given pattern continues. 2.6.2 Calculate the sum of the first 50 terms of the sequence. 2.6.3 Write down the value of \( 131^{\text {st }} \) term of the sequence

To identify the next two terms, let's first observe the given sequence: 1. **Alternating Pattern:** There are two separate patterns interleaved in the sequence. - The 1st, 3rd, 5th, ... terms form a sequence of fractions: \( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots \) which follows \( \frac{1}{2^n} \) where \( n \) starts from 1. So the next term in this pattern will be \( \frac{1}{16} \) (when \( n=4 \)). - The 2nd, 4th, 6th, ... terms: \( 4, 7, 10, \ldots \) forms a simple arithmetic sequence where each term increases by 3. Therefore, the next term will be \( 10 + 3 = 13 \). Thus, the next two terms are \( \frac{1}{16} \) and \( 13 \). Now, to find the sum of the first 50 terms of the sequence: 2.6.2 The first 50 terms consist of 25 terms from the fraction pattern and 25 terms from the arithmetic pattern. - For the fractional terms: The sum of the first 25 terms of \( \frac{1}{2^n} \) is calculated as follows: \[ S_f = \frac{1}{2} \left(1 - \frac{1}{2^{25}}\right) \div \left(1 - \frac{1}{2}\right) = 1 - \frac{1}{2^{25}} \approx 1 \] - For the arithmetic terms: The sum \( S_a \) of the first 25 terms is: \[ S_a = \frac{n}{2} \times (a + l) = \frac{25}{2} \times (4 + 10) = \frac{25}{2} \times 14 = 175 \] So the total sum of the first 50 terms becomes: \[ S = S_f + S_a \approx 1 + 175 = 176 \] Lastly, to find the \( 131^{st} \) term of the sequence: 2.6.3 We observe that: - The \( n^{th} \) term follows the same pattern: - For odd \( n \): \( n = 1, 3, 5, \ldots \): these represent the sequence \( \frac{1}{2^k} \) where \( k \) denotes its position (1 to 25 for first 50 terms). The 66th odd term corresponds to \( k = 66 = 33 \rightarrow \frac{1}{2^{33}} \). - For even \( n \): \( n = 2, 4, 6, \ldots \): these represent numbers from the arithmetic sequence 4, 7, 10, ... The even \( n \) terms follow \( 3m + 1 \) where \( m = 0, 1, 2, \ldots \). The \( 65 \)th even term corresponds to \( m = 65/2 - 1 = 32 \rightarrow 4 + 3*32 = 100 \). So, the value of the \( 131^{\text{st}} \) term in the sequence is \( \frac{1}{2^{66}} \).