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 \( \sum_{k=1}^{p} \frac{1}{16}(2)^{k-2}>31 \) ?

Let's solve each of the problems step by step. --- ## **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**. ---