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 \( \frac{12285}{4} \) ?

To work with the geometric sequence \( \frac{3}{4} ; \frac{3}{2} ; 3 ; \ldots \), we first identify the first term and the common ratio. 1. The first term \( a = \frac{3}{4} \). To find the common ratio \( r \): \[ r = \frac{\text{second term}}{\text{first term}} = \frac{3/2}{3/4} = \frac{3}{2} \cdot \frac{4}{3} = 2. \] Thus, \( r = 2 \). The formula for the sum of the first \( n \) terms of a geometric sequence is given by: \[ S_n = a \frac{1 - r^n}{1 - r} \text{ for } r \neq 1. \] For the first 10 terms: \[ S_{10} = \frac{3}{4} \frac{1 - 2^{10}}{1 - 2} = \frac{3}{4} \frac{1 - 1024}{-1} = \frac{3}{4} \cdot 1023 = \frac{3069}{4}. \] 2. To find out how many terms will add up to \( \frac{12285}{4} \), we set: \[ S_n = a \frac{1 - r^n}{1 - r}. \] Substituting the values: \[ \frac{12285}{4} = \frac{3}{4} \frac{1 - 2^n}{1 - 2}. \] This simplifies to: \[ \frac{12285}{4} = \frac{3}{4} \cdot (1 - 2^n) \cdot (-1). \] So we get: \[ 12285 = 3(1 - 2^n) \] \[ \Rightarrow \frac{12285}{3} = 1 - 2^n \] \[ \Rightarrow 4095 = 1 - 2^n \] \[ \Rightarrow 2^n = 1 - 4095 = -4094 \] This equation leads to a logical inconsistency since \( 2^n \) can never be negative. This implies there are no terms in the sequence that can add up to \( \frac{12285}{4} \). Thus: (1) The sum of the first 10 terms is \( \frac{3069}{4} \). (2) There are no terms in this sequence that sum to \( \frac{12285}{4} \).