LCTIVITIESIASSESSMENTS L. 2 Calculate the sum of the first 12 terms of the geometric sequence $$ \frac{1}{4} ;-\frac{1}{2} ; 1 $$; L 3 Calculate $$ \sum_{m=3}^{11} 8\left(\frac{1}{2}\right)^{m-4} $$ L. 4 Determine $$ m $$ if: $$ \sum_{p=1}^{m}(-8)\left(\frac{1}{2}\right)^{p-1}=-15 \frac{3}{4} $$ L. 5 What is the least value of $$ p $$ for which the series $$ \sum_{k=1}^{p} \frac{1}{16}(2)^{k-2}31 $$ ?

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LCTIVITIESIASSESSMENTS L. 2 Calculate the sum of the first 12 terms of the geometric sequence $$ \frac{1}{4} ;-\frac{1}{2} ; 1 $$; L 3 Calculate $$ \sum_{m=3}^{11} 8\left(\frac{1}{2}\right)^{m-4} $$ L. 4 Determine $$ m $$ if: $$ \sum_{p=1}^{m}(-8)\left(\frac{1}{2}\right)^{p-1}=-15 \frac{3}{4} $$ L. 5 What is the least value of $$ p $$ for which the series $$ \sum_{k=1}^{p} \frac{1}{16}(2)^{k-2}>31 $$ ?

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## **L.2 Calculate the sum of the first 12 terms of the geometric sequence $$ \frac{1}{4} ;-\frac{1}{2} ; 1 $$**

### **Solution:**

1. **Identify the first term ($$ a_1 $$) and the common ratio ($$ r $$):**
   - $$ a_1 = \frac{1}{4} $$
   - To find $$ r $$, divide the second term by the first term:
     $$
     r = \frac{-\frac{1}{2}}{\frac{1}{4}} = -2
     $$

2. **Use the formula for the sum of the first $$ n $$ terms of a geometric series:**
   $$
   S_n = a_1 \times \frac{1 - r^n}{1 - r}
   $$
   
3. **Plug in the values:**
   $$
   S_{12} = \frac{1}{4} \times \frac{1 - (-2)^{12}}{1 - (-2)} = \frac{1}{4} \times \frac{1 - 4096}{3} = \frac{1}{4} \times \frac{-4095}{3} = \frac{-4095}{12} = -341.25
   $$

### **Answer:**
The sum of the first 12 terms is **-341.25**.

---

## **L.3 Calculate $$ \sum_{m=3}^{11} 8\left(\frac{1}{2}\right)^{m-4} $$**

### **Solution:**

1. **Simplify the exponent:**
   $$
   8\left(\frac{1}{2}\right)^{m-4} = 8 \times 2^{-(m-4)} = 8 \times 2^{4-m} = 128 \times \left(\frac{1}{2}\right)^m
   $$
   
2. **Compute each term from $$ m = 3 $$ to $$ m = 11 $$:**
   
   | $$ m $$ | $$ 8\left(\frac{1}{2}\right)^{m-4} $$ |
   |---------|---------------------------------------|
   | 3       | $$ 8 \times 2 = 16 $$                 |
   | 4       | $$ 8 \times 1 = 8 $$                  |
   | 5       | $$ 8 \times \frac{1}{2} = 4 $$        |
   | 6       | $$ 8 \times \frac{1}{4} = 2 $$        |
   | 7       | $$ 8 \times \frac{1}{8} = 1 $$        |
   | 8       | $$ 8 \times \frac{1}{16} = 0.5 $$     |
   | 9       | $$ 8 \times \frac{1}{32} = 0.25 $$    |
   | 10      | $$ 8 \times \frac{1}{64} = 0.125 $$   |
   | 11      | $$ 8 \times \frac{1}{128} = 0.0625 $$ |

3. **Sum all the terms:**
   $$
   16 + 8 + 4 + 2 + 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 31.9375
   $$
   
   Alternatively, expressed as a fraction:
   $$
   31 \frac{15}{16}
   $$

### **Answer:**
The sum is **31.9375** or $$ 31 \frac{15}{16} $$.

---

## **L.4 Determine $$ m $$ if: $$ \sum_{p=1}^{m}(-8)\left(\frac{1}{2}\right)^{p-1} = -15 \frac{3}{4} $$**

### **Solution:**

1. **Identify the first term ($$ a_1 $$) and the common ratio ($$ r $$):**
   - $$ a_1 = -8 \times \left(\frac{1}{2}\right)^{0} = -8 $$
   - $$ r = \frac{1}{2} $$

2. **Use the sum formula for a geometric series:**
   $$
   S_m = a_1 \times \frac{1 - r^m}{1 - r} = -8 \times \frac{1 - \left(\frac{1}{2}\right)^m}{1 - \frac{1}{2}} = -16 \times \left(1 - \left(\frac{1}{2}\right)^m\right)
   $$
   
3. **Set the sum equal to $$-15.75$$ and solve for $$ m $$:**
   $$
   -16 \times \left(1 - \left(\frac{1}{2}\right)^m\right) = -15.75 \
   \Rightarrow 1 - \left(\frac{1}{2}\right)^m = \frac{15.75}{16} = 0.984375 \
   \Rightarrow \left(\frac{1}{2}\right)^m = 1 - 0.984375 = 0.015625 \
   $$
   
4. **Solve for $$ m $$:**
   $$
   \left(\frac{1}{2}\right)^m = \frac{1}{64} \
   \Rightarrow 2^m = 64 \
   \Rightarrow m = 6
   $$

### **Answer:**
The value of $$ m $$ is **6**.

---

## **L.5 What is the least value of $$ p $$ for which the series $$ \sum_{k=1}^{p} \frac{1}{16}(2)^{k-2} >31 $$ ?**

### **Solution:**

1. **Rewrite the general term:**
   $$
   \frac{1}{16} \times 2^{k-2} = \frac{2^{k-2}}{16} = \frac{2^{k}}{64} = \frac{1}{64} \times 2^k
   $$
   
   However, it's more straightforward to treat this as a geometric series:
   
   - First term ($$ a_1 $$): For $$ k = 1 $$, $$ \frac{1}{16} \times 2^{-1} = \frac{1}{32} $$
   - Common ratio ($$ r $$): $$ 2 $$

2. **Use the sum formula for a geometric series:**
   $$
   S_p = a_1 \times \frac{r^p - 1}{r - 1} = \frac{1}{32} \times \frac{2^p - 1}{1} = \frac{2^p - 1}{32}
   $$
   
3. **Set the sum greater than 31 and solve for $$ p $$:**
   $$
   \frac{2^p - 1}{32} > 31 \
   \Rightarrow 2^p - 1 > 992 \
   \Rightarrow 2^p > 993
   $$
   
4. **Find the smallest integer $$ p $$ such that $$ 2^p > 993 $$:**
   $$
   2^{9} = 512 < 993 \
   2^{10} = 1024 > 993 \
   $$
   
   Thus, $$ p = 10 $$

### **Answer:**
The least value of $$ p $$ is **10**.

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**Solutions:** 1. **L.2:** The sum of the first 12 terms is **-341.25**. 2. **L.3:** The sum is **31.9375** or $$ 31 \frac{15}{16} $$. 3. **L.4:** The value of $$ m $$ is **6**. 4. **L.5:** The least value of $$ p $$ is **10**.

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