(e) Consider the geometric sequence \( \frac{3}{4} ; \frac{3}{2} ; 3 ; \ldots \) (1) Calculate the sum of the first 10 terms of the sequence. (2) How many terms in this sequence will add up to \( \frac{12285}{4} \) ? (3) Determine the smallest value of \( k \) for which \( S_{k}>1000 \).
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To solve the problem, we start by identifying the first term and the common ratio of the geometric sequence. 1. The first term \( a \) is given as \( \frac{3}{4} \). To find the common ratio \( r \), we can divide the second term by the first term: \[ r = \frac{\frac{3}{2}}{\frac{3}{4}} = \frac{3/2}{3/4} = \frac{3 \cdot 4}{2 \cdot 3} = 2. \] Now we can use the formula for the sum of the first \( n \) terms of a geometric series: \[ S_n = a \frac{1 - r^n}{1 - r}, \] where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. 1) For \( n = 10 \): \[ S_{10} = \frac{3}{4} \frac{1 - 2^{10}}{1 - 2} = \frac{3}{4} \frac{1 - 1024}{-1} = \frac{3}{4} \cdot 1023 = \frac{3069}{4}. \] 2) To find how many terms will sum to \( \frac{12285}{4} \), we set \( S_n = \frac{12285}{4} \): \[ S_n = \frac{3}{4} \frac{1 - 2^n}{1 - 2} \quad \Rightarrow \quad \frac{3}{4} \cdot (1 - 2^n) = \frac{12285}{4}. \] This simplifies to: \[ 3 (1 - 2^n) = 12285 \quad \Rightarrow \quad 1 - 2^n = \frac{12285}{3} = 4095 \quad \Rightarrow \quad -2^n = 4094. \] This would lead to: \[ 2^n = -4094, \] which is impossible, so there are no terms that add up to \( \frac{12285}{4} \). 3) We need to find the smallest value of \( k \) for which: \[ S_k > 1000. \] Using the formula: \[ S_k = \frac{3}{4} \cdot \frac{1 - 2^k}{-1} = \frac{3}{4} (2^k - 1). \] So we want: \[ \frac{3}{4} (2^k - 1) > 1000. \] Multiplying both sides by \( \frac{4}{3} \): \[ 2^k - 1 > \frac{4000}{3} \implies 2^k > 1333.33. \] Now we solve for \( k \). Since \( 2^{10} = 1024 \) and \( 2^{11} = 2048 \): \[ k = 11. \] Thus, the smallest value of \( k \) for which \( S_k > 1000 \) is \( k = 11 \).
