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 $$ 131^{ ext {st }} $$ term of the sequence

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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 $$ 131^{	ext {st }} $$ term of the sequence

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Let's analyze the given sequence to determine the patterns for both the odd and even terms.

**Given Sequence:**
$$
\frac{1}{2};\ 4;\ \frac{1}{4};\ 7;\ \frac{1}{8};\ 10;\ \ldots
$$

### **2.6.1 Next TWO Terms**

**Observing the Patterns:**

1. **Odd-Positioned Terms:**
   - **Term 1:** $$\frac{1}{2}$$
   - **Term 3:** $$\frac{1}{4}$$
   - **Term 5:** $$\frac{1}{8}$$
   
   The pattern here is:
   $$
   \text{Term}_{2k-1} = \frac{1}{2^k} \quad \text{for} \quad k = 1, 2, 3, \ldots
   $$
   
   So, the next odd-positioned term (Term 7) is:
   $$
   \frac{1}{2^4} = \frac{1}{16}
   $$

2. **Even-Positioned Terms:**
   - **Term 2:** $$4$$
   - **Term 4:** $$7$$
   - **Term 6:** $$10$$
   
   The pattern here is an arithmetic sequence with a common difference of $$3$$:
   $$
   \text{Term}_{2k} = 4 + 3(k-1) \quad \text{for} \quad k = 1, 2, 3, \ldots
   $$
   
   So, the next even-positioned term (Term 8) is:
   $$
   4 + 3(4-1) = 4 + 9 = 13
   $$

**Therefore, the next two terms are:**
$$
\frac{1}{16};\ 13
$$

### **2.6.2 Sum of the First 50 Terms**

To calculate the sum of the first 50 terms, we can separate the sequence into its odd and even components.

1. **Odd-Positioned Terms (25 terms):**
   $$
   \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^{25}}
   $$
   
   This is a geometric series with:
   $$
   a = \frac{1}{2},\ r = \frac{1}{2},\ \text{number of terms} = 25
   $$
   
   Sum:
   $$
   S_{\text{odd}} = a \times \frac{1 - r^{n}}{1 - r} = \frac{1}{2} \times \frac{1 - \left(\frac{1}{2}\right)^{25}}{1 - \frac{1}{2}} = 1 - \frac{1}{2^{25}}
   $$

2. **Even-Positioned Terms (25 terms):**
   $$
   4 + 7 + 10 + \cdots + [4 + 3(25-1)] = 4 + 7 + 10 + \cdots + 76
   $$
   
   This is an arithmetic series with:
   $$
   a = 4,\ d = 3,\ n = 25
   $$
   
   Sum:
   $$
   S_{\text{even}} = \frac{n}{2} \times [2a + (n-1)d] = \frac{25}{2} \times [8 + 72] = \frac{25}{2} \times 80 = 25 \times 40 = 1000
   $$

**Total Sum:**
$$
S_{\text{total}} = S_{\text{odd}} + S_{\text{even}} = \left(1 - \frac{1}{2^{25}}\right) + 1000 = 1001 - \frac{1}{2^{25}}
$$
$$
\approx 1001 \quad \text{(since} \ \frac{1}{2^{25}} \ \text{is negligible)}
$$

### **2.6.3 131$${}^{\text{st}}$$ Term**

To find the 131$${}^{\text{st}}$$ term:

1. **Determine if 131 is Odd or Even:**
   - 131 is odd.

2. **Using the Odd-Term Formula:**
   $$
   \text{Term}_{2k-1} = \frac{1}{2^k}
   $$
   
   For Term 131:
   $$
   2k - 1 = 131 \implies k = 66
   $$
   
   So,
   $$
   \text{Term}_{131} = \frac{1}{2^{66}}
   $$

**Therefore, the 131$${}^{\text{st}}$$ term is:**
$$
\frac{1}{2^{66}}
$$

### **Summary of Answers:**

1. **2.6.1 Next Two Terms:**
   $$
   \frac{1}{16};\ 13
   $$

2. **2.6.3 131$${}^{\text{st}}$$ Term:**
   $$
   \frac{1}{2^{66}}
   $$
**2.6.1 Next Two Terms:** $$ \frac{1}{16},\ 13 $$ **2.6.2 Sum of the First 50 Terms:** $$ 1001 $$ **2.6.3 131$${}^{\text{st}}$$ Term:** $$ \frac{1}{2^{66}} $$

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

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